Title

The greatest common factor (GCF) of an expression is the largest possible factor that all the terms in the expression are divisible by

In MA 141, the GCF of an expression has the form $c x^{n}$ where $\begin{array}{r} \begin{aligned} ∙ c is the biggest number that all the coefficients are divisible by \\ ∙ n is the minimum power of x out of all the terms \end{aligned} \end{array}$

Ex 1: Factor out the GCF of $- 9 x$


	$Step 1: \underset{⏟}{\begin{matrix} Determine c (the biggest number that all the coefficients are divisible by) \end{matrix}} \begin{array}{r} If the term with the highest power of x is < 0, then also include a - sign \end{array}$	$\underset{⏟}{\begin{matrix} c = 1 \end{matrix}} \begin{array}{r} The coefficients 1 and - 9 \\ have no factor in common \end{array}$
	$Step 2: \underset{⏟}{\begin{matrix} Determine n (the minimum power of x out of all of the terms) \end{matrix}} \begin{array}{r} \begin{aligned} ∙ In some problems, you use a different variable than x even if x is in the expression \\ ∙ If a term has no x, the power of x in that term is 0 \\ ∙ If a term has is an x but no power, the power of x in that term is 1 \end{aligned} \end{array}$	$\underset{⏟}{\begin{matrix} n = 1 \end{matrix}} \begin{array}{r} The term - 9 x has the minimum \\ power of x which is 1 \end{array}$
	$Step 3: \underset{⏟}{\begin{matrix} Divide each term by c x^{n} (the GCF) and multiply everything by c x^{n} \end{matrix}} \begin{array}{r} You may want to simplify the GCF using \\ x^{1} = x, x^{0} = 1, and 1 a = a \end{array}$	$- 9 x = \underset{⏟}{\begin{matrix} x \end{matrix}} \begin{array}{r} 1 x^{1} = x \end{array} (\frac{x^{3}}{x} - \frac{9 x}{x})$
	$Step 4: Simplify using properties of exponents$	$- 9 x = x (\underset{⏟}{\begin{matrix} x^{2} \end{matrix}} \begin{array}{r} \begin{aligned} \frac{x^{3}}{x} & = \frac{x^{3}}{x^{1}} \\ \frac{x^{3}}{x^{1}} & = x^{3 - 1} \end{aligned} \end{array} - \underset{⏟}{\begin{matrix} 9 \end{matrix}} \begin{array}{r} \begin{matrix} \frac{x}{x} = 1 \end{matrix} \end{array})$

Ex 2: Factor out the GCF of $18 t - 21$


	$Step 1: \underset{⏟}{\begin{matrix} Determine c (the biggest number that all the coefficients are divisible by) \end{matrix}} \begin{array}{r} If the term with the highest power of x is < 0, then also include a - sign \end{array}$	$\underset{⏟}{\begin{matrix} c = 3 \end{matrix}} \begin{array}{r} The coefficients 3, 18, and - 21 \\ have a factor of 3 in common \end{array}$
	$Step 2: \underset{⏟}{\begin{matrix} Determine n (the minimum power of x out of all of the terms) \end{matrix}} \begin{array}{r} \begin{aligned} ∙ In some problems, you use a different variable than x even if x is in the expression \\ ∙ If a term has no x, the power of x in that term is 0 \\ ∙ If a term has is an x but no power, the power of x in that term is 1 \end{aligned} \end{array}$	$\underset{⏟}{\begin{matrix} n = 0 \end{matrix}} \begin{array}{r} The term - 21 has the minimum \\ power of t which is 0 \end{array}$
	$Step 3: \underset{⏟}{\begin{matrix} Divide each term by c x^{n} (the GCF) and multiply everything by c x^{n} \end{matrix}} \begin{array}{r} You may want to simplify the GCF using \\ x^{1} = x, x^{0} = 1, and 1 a = a \end{array}$	$18 t - 21 = \underset{⏟}{\begin{matrix} 3 \end{matrix}} \begin{array}{r} 3 t^{0} = 3 (1) = 3 \end{array} (\frac{3 t^{2}}{3} + \frac{18 t}{3} - \frac{21}{3})$
	$Step 4: Simplify using properties of exponents$	$18 t - 21 = 3 (t^{2} + 6 t - 7)$

Ex 3: Factor out the GCF of $2 h^{2} - h$


	$Step 1: \underset{⏟}{\begin{matrix} Determine c (the biggest number that all the coefficients are divisible by) \end{matrix}} \begin{array}{r} If the term with the highest power of x is < 0, then also include a - sign \end{array}$	$\underset{⏟}{\begin{matrix} c = 1 \end{matrix}} \begin{array}{r} The coefficients 4 x, 2, and - 1 \\ have no factor in common \end{array}$
	$Step 2: \underset{⏟}{\begin{matrix} Determine n (the minimum power of x out of all of the terms) \end{matrix}} \begin{array}{r} \begin{aligned} ∙ In some problems, you use a different variable than x even if x is in the expression \\ ∙ If a term has no x, the power of x in that term is 0 \\ ∙ If a term has is an x but no power, the power of x in that term is 1 \end{aligned} \end{array}$	$\underset{⏟}{\begin{matrix} n = 1 \end{matrix}} \begin{array}{r} The terms 4 x h and - h have the minimum \\ power of h which is 1 \end{array}$
	$Step 3: \underset{⏟}{\begin{matrix} Divide each term by c x^{n} (the GCF) and multiply everything by c x^{n} \end{matrix}} \begin{array}{r} You may want to simplify the GCF using \\ x^{1} = x, x^{0} = 1, and 1 a = a \end{array}$	$2 h^{2} - h = \underset{⏟}{\begin{matrix} h \end{matrix}} \begin{array}{r} 1 h^{1} = h \end{array} (\frac{4 x h}{h} + \frac{2 h^{2}}{h} - \frac{h}{h})$
	$Step 4: Simplify using properties of exponents$	$2 h^{2} - h = h (\underset{⏟}{\begin{matrix} 4 x \end{matrix}} \begin{array}{r} \begin{matrix} \frac{h}{h} = 1 \end{matrix} \end{array} + \underset{⏟}{\begin{matrix} 2 h \end{matrix}} \begin{array}{r} \begin{aligned} \frac{h^{2}}{h} & = \frac{h^{2}}{h^{1}} \\ \frac{h^{2}}{h^{1}} & = h^{2 - 1} \\ = h^{1} = h \end{aligned} \end{array} + \underset{⏟}{\begin{matrix} 1 \end{matrix}} \begin{array}{r} \begin{matrix} \frac{h}{h} = 1 \end{matrix} \end{array})$

Ex 4: Factor out the GCF of $- 8 z$


	$Step 1: \underset{⏟}{\begin{matrix} Determine c (the biggest number that all the coefficients are divisible by) \end{matrix}} \begin{array}{r} If the term with the highest power of x is < 0, then also include a - sign \end{array}$	$\underset{⏟}{\begin{matrix} c = 1 \end{matrix}} \begin{array}{r} The coefficients 4 y and - 8 \\ have no factor in common \end{array}$
	$Step 2: \underset{⏟}{\begin{matrix} Determine n (the minimum power of x out of all of the terms) \end{matrix}} \begin{array}{r} \begin{aligned} ∙ In some problems, you use a different variable than x even if x is in the expression \\ ∙ If a term has no x, the power of x in that term is 0 \\ ∙ If a term has is an x but no power, the power of x in that term is 1 \end{aligned} \end{array}$	$\underset{⏟}{\begin{matrix} n = 1 \end{matrix}} \begin{array}{r} The terms 4 y z and - 8 z have the minimum \\ power of z which is 1 \end{array}$
	$Step 3: \underset{⏟}{\begin{matrix} Divide each term by c x^{n} (the GCF) and multiply everything by c x^{n} \end{matrix}} \begin{array}{r} You may want to simplify the GCF using \\ x^{1} = x, x^{0} = 1, and 1 a = a \end{array}$	$- 8 z = \underset{⏟}{\begin{matrix} z \end{matrix}} \begin{array}{r} 1 z^{1} = z \end{array} (\frac{4 y z}{z} - \frac{8 z}{z})$
	$Step 4: Simplify using properties of exponents$	$- 8 z = z (\underset{⏟}{\begin{matrix} 4 y \end{matrix}} \begin{array}{r} \begin{matrix} \frac{z}{z} = 1 \end{matrix} \end{array} - \underset{⏟}{\begin{matrix} 8 \end{matrix}} \begin{array}{r} \begin{matrix} \frac{z}{z} = 1 \end{matrix} \end{array})$

Subsection 5.2.2: Factoring Polynomials with Two Terms

When factoring a polynomial, first take out the GCF

If the polynomial has two terms, there are 5 cases to be considered

Type of Polynomial	Format	Factored Form
Linear polynomial	$a x + b$	$\underset{⏟}{\begin{matrix} Cannot be factored \end{matrix}} \begin{array}{r} Assuming the GCF has already been factored out \end{array}$
Sum of Squares	$\underset{⏟}{\begin{matrix} a x^{2} + b \end{matrix}} \begin{array}{r} where a > 0 and b > 0 \end{array}$	$\underset{⏟}{\begin{matrix} Cannot be factored \end{matrix}} \begin{array}{r} Assuming the GCF has already been factored out \end{array}$
Difference of Squares I	$\underset{⏟}{\begin{matrix} x^{2} - a^{2} \end{matrix}} \begin{array}{r} \begin{aligned} ∙ This works when a^{2} is any positive \\ ∙ number, even if a^{2} isn't a perfect square \\ ∙ an alternate format is \\ ∙ x^{2} - a with a > 0 \end{aligned} \end{array}$	$\underset{⏟}{\begin{matrix} x^{2} - a^{2} = (x + a) (x - a) \end{matrix}} \begin{array}{r} An alternate version is \\ x^{2} - a = (x + \sqrt{a}) (x - \sqrt{a}) \end{array}$
Difference of Squares II	$\underset{⏟}{\begin{matrix} b^{2} x^{2} - a^{2} \end{matrix}} \begin{array}{r} \begin{aligned} ∙ This works when a^{2} and b^{2} are any positive numbers, \\ ∙ even if a^{2} and/or b^{2} aren't perfect squares \\ ∙ a an alternate format is \\ ∙ b x^{2} - a with a > 0 and b > 0 \end{aligned} \end{array}$	$\underset{⏟}{\begin{matrix} b^{2} x^{2} - a^{2} = (b x + a) (b x - a) \end{matrix}} \begin{array}{r} An alternate version is \\ b x^{2} - a = (\sqrt{b} x + \sqrt{a}) (\sqrt{b} x - \sqrt{a}) \end{array}$
Difference of Cubes	$\underset{⏟}{\begin{matrix} x^{3} - a^{3} \end{matrix}} \begin{array}{r} \begin{aligned} ∙ This works when a^{3} is any positive \\ ∙ number, even if a^{3} isn't a perfect cube \\ ∙ an alternate format is x^{3} - a with a > 0 \end{aligned} \end{array}$	$\underset{⏟}{\begin{matrix} x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \end{matrix}} \begin{array}{r} An alternate version is \\ x^{3} - a = (x - \sqrt[3]{a}) (x^{2} + \sqrt[3]{a} x + {(\sqrt[3]{a})}^{2}) \end{array}$

Ex 1: Factor $x^{4} - 8 x$


	$Step 1: Factor out the GCF$	$x^{4} - 8 x = \underset{⏟}{\begin{matrix} x (x^{3} - 8) \end{matrix}} \begin{array}{r} \begin{aligned} GCF & = 1 x^{1} = x \\ - 8 x & = x (\frac{x^{4}}{x} - \frac{8 x}{x}) \\ = x (x^{3} - 8) \end{aligned} \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} x^{3} - 8 = x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \end{aligned}$
	$Step 3: Determine a = \sqrt[3]{a^{3}}$	$\begin{aligned} a & = \sqrt[3]{\underset{⏟}{\begin{array}{c} 8 \end{array}} \begin{array}{r} a^{3} = 8 b/c \\ x^{3} - 8 = x^{3} - a^{3} \end{array}} = 2 \end{aligned}$
	$Step 4: Plug into x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2})$	$\begin{aligned} x^{4} - 8 x & = x (x^{3} - 8) \\ = x (x - 2) (x^{2} + 2 x + {(2)}^{2}) \\ = (x - 2) (x^{2} + 2 x + 4) \end{aligned}$
	$Step 5: \begin{matrix} If you performed the substitution \\ u = x^{2}, substitute x^{2} back for u \\ and factor each part by going back to Step 2 \end{matrix}$	$NA$

Ex 2: Factor $2 x^{2} - 8$


	$Step 1: Factor out the GCF$	$2 x^{2} - 8 = \underset{⏟}{\begin{matrix} 2 (x^{2} - 4) \end{matrix}} \begin{array}{r} \begin{aligned} GCF & = 2 x^{0} = 2 (1) = 2 \\ - 8 & = 2 (\frac{2 x^{2}}{2} - \frac{8}{2}) \\ = 2 (x^{2} - 4) \end{aligned} \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} x^{2} - 4 = x^{2} - a^{2} = (x - a) (x + a) \end{aligned}$
	$Step 3: Determine a = \sqrt{a^{2}}$	$\begin{aligned} a & = \sqrt{\underset{⏟}{\begin{array}{c} 4 \end{array}} \begin{array}{r} a^{2} = 4 b/c \\ x^{2} - 4 = x^{2} - a^{2} \end{array}} = 2 \end{aligned}$
	$Step 4: Plug into x^{2} - a^{2} = (x - a) (x + a)$	$\begin{aligned} 2 x^{2} - 8 & = 2 (x^{2} - 4) \\ = 2 (x - 2) (x + 2) \end{aligned}$
	$Step 5: \begin{matrix} If you performed the substitution \\ u = x^{2}, substitute x^{2} back for u \\ and factor each part by going back to Step 2 \end{matrix}$	$NA$

Ex 3: Factor $6 x^{2} - 3$


	$Step 1: Factor out the GCF$	$6 x^{2} - 3 = \underset{⏟}{\begin{matrix} 3 (2 x^{2} - 1) \end{matrix}} \begin{array}{r} \begin{aligned} GCF & = 3 x^{0} = 3 (1) = 3 \\ - 3 & = 3 (\frac{6 x^{2}}{3} - \frac{3}{3}) \\ = 3 (2 x^{2} - 1) \end{aligned} \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} 2 x^{2} - 1 = b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \end{aligned}$
	$Step 3: Determine a = \sqrt{a^{2}} and b = \sqrt{b^{2}}$	$\begin{aligned} a & = \sqrt{\underset{⏟}{\begin{array}{c} 1 \end{array}} \begin{array}{r} a^{2} = 1 b/c \\ 2 x^{2} - 1 = b^{2} x^{2} - a^{2} \end{array}} = 1 \\ b & = \sqrt{\underset{⏟}{\begin{array}{c} 2 \end{array}} \begin{array}{r} b^{2} = 2 b/c \\ 2 x^{2} - 1 = b^{2} x^{2} - a^{2} \end{array}} = \sqrt{2} \end{aligned}$
	$Step 4: Plug into b^{2} x^{2} - a^{2} = (b x - a) (b x + a)$	$\begin{aligned} 6 x^{2} - 3 & = 3 (2 x^{2} - 1) \\ = 3 (\sqrt{2} x - 1) (\sqrt{2} x + 1) \end{aligned}$
	$Step 5: \begin{matrix} If you performed the substitution \\ u = x^{2}, substitute x^{2} back for u \\ and factor each part by going back to Step 2 \end{matrix}$	$NA$

Ex 4: Factor $3 x^{2} + 4$


	$Step 1: Factor out the GCF$	$3 x^{2} + 4 = \underset{⏟}{\begin{matrix} There is no GCF to take out \end{matrix}} \begin{array}{r} B/c the GCF is 1 \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} 3 x^{2} + 4 = a x^{2} + b can't be factored \end{aligned}$

Subsection 5.2.3: Factoring

x^{2} + b x + c

Suppose that after factoring out the GCF, we need to factor $x^{2} + b x + c$

Then $\begin{matrix} (1) & x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \end{matrix}$ $\begin{matrix} (2) & where a_{1} and a_{2} is the pair of numbers which multiply to c and add to b \end{matrix}$

To factor $x^{2} + b x + c$ , follow these steps: $\begin{array}{r} \begin{aligned} ∙ Determine a_{1} and a_{2} \\ \circ \underset{⏟}{\begin{array}{c} List the pairs of numbers which multiply to c \end{array}} \begin{array}{r} \begin{aligned} ∙ By (2), a_{1} and a_{2} multiply to c \\ ∙ Don't forget to consider negative numbers \\ ∙ By numbers we technically mean integers \end{aligned} \end{array} \\ \circ \underset{⏟}{\begin{array}{c} a_{1} and a_{2} is the pair that adds to b \end{array}} \begin{array}{r} \begin{aligned} ∙ By (2), a_{1} and a_{2} add to b \\ ∙ if none of the pairs add to b, then use \\ ∙ the quadratic formula to factor instead \end{aligned} \end{array} \\ ∙ Plug into \underset{⏟}{\begin{array}{c} x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \end{array}} \begin{array}{r} This is (1) \end{array} \end{aligned} \end{array}$

Ex 1: Factor $x^{2} + 4 x + 4$


	$Step 1: Factor out the GCF$	$x^{2} + 4 x + 4 = \underset{⏟}{\begin{matrix} There is no GCF to take out \end{matrix}} \begin{array}{r} B/c the GCF is 1 \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} x^{2} + 4 x + 4 = x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ where a_{1} and a_{2} are two numbers \\ which multiply to c and add to b \end{aligned}$
	$Step 3: \underset{⏟}{\begin{matrix} List the pairs of numbers which multiply to c \end{matrix}} \begin{array}{r} \begin{aligned} ∙ Don't forget to consider negative numbers \\ ∙ By numbers we technically mean integers \end{aligned} \end{array}$	$\begin{matrix} \underset{⏟}{\begin{matrix} c = 4 \end{matrix}} \begin{array}{r} x^{2} + 4 x + 4 \end{array} \\ Possible pairs: \\ 4 and 1 \\ - 4 and - 1 \\ 2 and 2 \\ - 2 and - 2 \end{matrix}$
	$Step 4: Determine the pair a_{1} and a_{2} which adds to b$	$\begin{matrix} \underset{⏟}{\begin{matrix} b = 4 \end{matrix}} \begin{array}{r} x^{2} + 4 x + 4 \end{array} \\ \underset{⏟}{\begin{matrix} a_{1} = 2, a_{2} = 2 \end{matrix}} \begin{array}{r} 2 and 2 is the pair that adds to 4 \end{array} \end{matrix}$
	$Step 5: Plug into x^{2} + b x + c = (x + a_{1}) (x + a_{2})$	$x^{2} + 4 x + 4 = (x + 2) (x + 2)$
	$Step 6: \begin{matrix} If you performed the substitution \\ u = x^{2}, substitute x^{2} back for u \\ and factor each part by going back to Step 2 \end{matrix}$	$NA$

Ex 2: Factor $x^{2} + 3 x - 4$


	$Step 1: Factor out the GCF$	$x^{2} + 3 x - 4 = \underset{⏟}{\begin{matrix} There is no GCF to take out \end{matrix}} \begin{array}{r} B/c the GCF is 1 \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} x^{2} + 3 x - 4 = x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ where a_{1} and a_{2} are two numbers \\ which multiply to c and add to b \end{aligned}$
	$Step 3: \underset{⏟}{\begin{matrix} List the pairs of numbers which multiply to c \end{matrix}} \begin{array}{r} \begin{aligned} ∙ Don't forget to consider negative numbers \\ ∙ By numbers we technically mean integers \end{aligned} \end{array}$	$\begin{matrix} \underset{⏟}{\begin{matrix} c = - 4 \end{matrix}} \begin{array}{r} x^{2} + 3 x - 4 \end{array} \\ Possible pairs: \\ 4 and - 1 \\ - 4 and 1 \\ 2 and - 2 \\ - 2 and 2 \end{matrix}$
	$Step 4: Determine the pair a_{1} and a_{2} which adds to b$	$\begin{matrix} \underset{⏟}{\begin{matrix} b = 3 \end{matrix}} \begin{array}{r} x^{2} + 3 x - 4 \end{array} \\ \underset{⏟}{\begin{matrix} a_{1} = 4, a_{2} = - 1 \end{matrix}} \begin{array}{r} 4 and - 1 is the pair that adds to 3 \end{array} \end{matrix}$
	$Step 5: Plug into x^{2} + b x + c = (x + a_{1}) (x + a_{2})$	$x^{2} + 3 x - 4 = (x + 4) (x - 1)$
	$Step 6: \begin{matrix} If you performed the substitution \\ u = x^{2}, substitute x^{2} back for u \\ and factor each part by going back to Step 2 \end{matrix}$	$NA$

Subsection 5.2.4: Factoring

a x^{2} + b x + c

Suppose that after factoring out the GCF, we need to factor $a x^{2} + b x + c$

Then we can attempt to factor $a x^{2} + b x + c$ using these steps $\begin{array}{r} \begin{aligned} ∙ Determine the numbers b_{1} and b_{2} which multiply to a c and add to b \\ \circ Calculate a c \\ \circ List the pairs which multiply to a c \\ \circ \underset{⏟}{\begin{array}{c} b_{1} and b_{2} is the pair that adds to b \end{array}} \begin{array}{r} If none of the pairs add to b, use the \\ quadratic formula to factor instead \end{array} \\ ∙ \underset{⏟}{\begin{array}{c} Rewrite a x^{2} + b x + c as a x^{2} + b_{1} x + b_{2} x + c \end{array}} \begin{array}{r} This is possible since b_{1} and b_{2} add to b so \\ b x = (b_{1} + b_{2}) x = b_{1} x + b_{2} x \end{array} \\ ∙ Factor the GCF out of a x^{2} + b_{2} x \\ ∙ Factor the GCF out of b_{2} x + c to get \\ ∙ \underset{⏟}{\begin{array}{c} a x^{2} + b x + c = d_{1} x (c_{1} x + c_{2}) + d_{2} (c_{1} x + c_{2}) \end{array}} \begin{array}{r} The (c_{1} x + c_{2}) factors are the same b/c b_{1} and b_{2} multiply \\ to a c and add to b but finding out why is difficult \end{array} \\ ∙ The factored form is \\ ∙ \underset{⏟}{\begin{array}{c} a x^{2} + b x + c = (d_{1} x + d_{2}) (c_{1} x + c_{2}) \end{array}} \begin{array}{r} \begin{aligned} ∙ This is a more complicated case of factoring out a GCF \\ ∙ It works b/c (d_{1} x + d_{2}) (c_{1} x + c_{2}) = d_{1} x (c_{1} x + c_{2}) + d_{2} (c_{1} x + c_{2}) \end{aligned} \end{array} \end{aligned} \end{array}$

Ex 1: Factor $x^{4} - 4 x^{2} + 3$


	$Step 1: Factor out the GCF$	$x^{4} - 4 x^{2} + 3 = \underset{⏟}{\begin{matrix} There is no GCF to take out \end{matrix}} \begin{array}{r} B/c the GCF is 1 \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} for x^{4} - 4 x^{2} + 3 = a x^{4} + b x^{2} + c, follow these steps \end{aligned}$
	$Step 3: \begin{matrix} \underset{⏟}{\begin{matrix} Perform the substitution u = x^{2} \end{matrix}} \begin{array}{r} Replace a x^{4} + b x^{2} + c with a u^{2} + b u + c \end{array} \\ and go back to Step 2 \end{matrix}$	$\begin{aligned} u & = x^{2} \\ x^{4} - 4 x^{2} + 3 & = \underset{⏟}{\begin{array}{c} u^{2} \end{array}} \begin{array}{r} b/c u^{2} = (x^{2})^{2} = (x^{2})^{2} = x^{2 (2)} = x^{4} \end{array} - 4 u + 3 \end{aligned}$
	$Step 4: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} u^{2} - 4 u + 3 = u^{2} + b u + c = (u + a_{1}) (u + a_{2}) \\ where a_{1} and a_{2} are two numbers \\ which multiply to c and add to b \end{aligned}$
	$Step 5: \underset{⏟}{\begin{matrix} List the pairs of numbers which multiply to c \end{matrix}} \begin{array}{r} \begin{aligned} ∙ Don't forget to consider negative numbers \\ ∙ By numbers we technically mean integers \end{aligned} \end{array}$	$\begin{matrix} \underset{⏟}{\begin{matrix} c = 3 \end{matrix}} \begin{array}{r} u^{2} - 4 u + 3 \end{array} \\ Possible pairs: \\ 3 and 1 \\ - 3 and - 1 \end{matrix}$
	$Step 6: Determine the pair a_{1} and a_{2} which adds to b$	$\begin{matrix} \underset{⏟}{\begin{matrix} b = - 4 \end{matrix}} \begin{array}{r} u^{2} - 4 u + 3 \end{array} \\ \underset{⏟}{\begin{matrix} a_{1} = - 3, a_{2} = - 1 \end{matrix}} \begin{array}{r} - 3 and - 1 is the pair that adds to - 4 \end{array} \end{matrix}$
	$Step 7: Plug into x^{2} + b x + c = (x + a_{1}) (x + a_{2})$	$u^{2} - 4 u + 3 = (u - 3) (u - 1)$
	$Step 8: \begin{matrix} If you performed the substitution \\ u = x^{2}, substitute x^{2} back for u \\ and factor each part by going back to Step 2 \end{matrix}$	$\begin{aligned} x^{4} - 4 x^{2} + 3 & = (u - 3) (u - 1) \\ = (x^{2} - 3) (x^{2} - 1) \end{aligned}$
	$Step 9: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} x^{2} - 3 = x^{2} - a^{2} = (x - a) (x + a) \end{aligned}$
	$Step 10: Determine a = \sqrt{a^{2}}$	$\begin{aligned} a & = \sqrt{\underset{⏟}{\begin{array}{c} 3 \end{array}} \begin{array}{r} a^{2} = 3 b/c \\ x^{2} - 3 = x^{2} - a^{2} \end{array}} = \sqrt{3} \end{aligned}$
	$Step 11: Plug into x^{2} - a^{2} = (x - a) (x + a)$	$\begin{aligned} x^{2} - 3 & = (x - \sqrt{3}) (x + \sqrt{3}) \end{aligned}$
	$Step 12: Plug in what you factored into the bigger problem$	$\begin{aligned} x^{4} - 4 x^{2} + 3 & = \underset{⏟}{\begin{array}{c} (x^{2} - 3) (x^{2} - 1) \end{array}} \begin{array}{r} By Step 8 \end{array} \\ = (x - \sqrt{3}) (x + \sqrt{3}) (x^{2} - 1) \end{aligned}$
	$Step 13: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} x^{2} - 1 = x^{2} - a^{2} = (x - a) (x + a) \end{aligned}$
	$Step 14: Determine a = \sqrt{a^{2}}$	$\begin{aligned} a & = \sqrt{\underset{⏟}{\begin{array}{c} 1 \end{array}} \begin{array}{r} a^{2} = 1 b/c \\ x^{2} - 1 = x^{2} - a^{2} \end{array}} = 1 \end{aligned}$
	$Step 15: Plug into x^{2} - a^{2} = (x - a) (x + a)$	$\begin{aligned} x^{2} - 1 & = (x - 1) (x + 1) \end{aligned}$
	$Step 16: Plug in what you factored into the bigger problem$	$\begin{aligned} x^{4} - 4 x^{2} + 3 & = \underset{⏟}{\begin{array}{c} (x - \sqrt{3}) (x + \sqrt{3}) (x^{2} - 1) \end{array}} \begin{array}{r} By Step 12 \end{array} \\ = (x - \sqrt{3}) (x + \sqrt{3}) (x - 1) (x + 1) \end{aligned}$

Ex 2: Factor $2 x^{2} + 2 x - 4$


	$Step 1: Factor out the GCF$	$2 x^{2} + 2 x - 4 = \underset{⏟}{\begin{matrix} 2 (x^{2} + x - 2) \end{matrix}} \begin{array}{r} \begin{aligned} GCF & = 2 x^{0} = 2 (1) = 2 \\ 2 x - 4 & = 2 (\frac{2 x^{2}}{2} + \frac{2 x}{2} - \frac{4}{2}) \\ = 2 (x^{2} + x - 2) \end{aligned} \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} x^{2} + x - 2 = x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ where a_{1} and a_{2} are two numbers \\ which multiply to c and add to b \end{aligned}$
	$Step 3: \underset{⏟}{\begin{matrix} List the pairs of numbers which multiply to c \end{matrix}} \begin{array}{r} \begin{aligned} ∙ Don't forget to consider negative numbers \\ ∙ By numbers we technically mean integers \end{aligned} \end{array}$	$\begin{matrix} \underset{⏟}{\begin{matrix} c = - 2 \end{matrix}} \begin{array}{r} x^{2} + x - 2 \end{array} \\ Possible pairs: \\ 2 and - 1 \\ - 2 and 1 \end{matrix}$
	$Step 4: Determine the pair a_{1} and a_{2} which adds to b$	$\begin{matrix} \underset{⏟}{\begin{matrix} b = 1 \end{matrix}} \begin{array}{r} x^{2} + 1 x - 2 \end{array} \\ \underset{⏟}{\begin{matrix} a_{1} = 2, a_{2} = - 1 \end{matrix}} \begin{array}{r} 2 and - 1 is the pair that adds to 1 \end{array} \end{matrix}$
	$Step 5: Plug into x^{2} + b x + c = (x + a_{1}) (x + a_{2})$	$x^{2} + x - 2 = (x + 2) (x - 1)$
	$Step 6: \begin{matrix} If you performed the substitution \\ u = x^{2}, substitute x^{2} back for u \\ and factor each part by going back to Step 2 \end{matrix}$	$NA$

Ex 3: Factor $2 x^{2} + x - 1$


	$Step 1: Factor out the GCF$	$2 x^{2} + x - 1 = \underset{⏟}{\begin{matrix} There is no GCF to take out \end{matrix}} \begin{array}{r} B/c the GCF is 1 \end{array}$
	$Step 2: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{matrix} Quadratic Formula for a x^{2} + b x + c = 2 x^{2} + 1 x - 1 \\ a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{matrix}$
	$Step 3: \begin{aligned} Determine x_{1} & = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} \\ and x_{2} & = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$	$\begin{aligned} x_{1}, x_{2} & = \frac{- (1) \pm \sqrt{(1)^{2} - 4 (2) (- 1)}}{2 (2)} \\ = \frac{- 1 \pm \sqrt{9}}{4} = \frac{- 1 \pm 3}{4} \\ = \frac{1}{2}, - 1 \end{aligned}$
	$Step 4: \underset{⏟}{\begin{matrix} Try to factor a x^{2} + b x + c with \\ a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{matrix}} \begin{array}{r} \begin{aligned} ∙ If x_{1} or x_{2} is undefined, a x^{2} + b x + c \\ ∙ cannot be factored \\ ∙ Otherwise, a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{aligned} \end{array}$	$2 x^{2} + x - 1 = 2 (x - \frac{1}{2}) \underset{⏟}{\begin{matrix} (x + 1) \end{matrix}} \begin{array}{r} x - (- 1) = x + 1 \end{array}$

Subsection 5.2.5: Factoring With the Quadratic Formula

To factor

a x^{2} + b x + c

with the Quadratic Formula:

Find the root(s) of $a x^{2} + b x + c$ using the equation $\begin{array}{r} \underset{⏟}{\begin{array}{c} x_{1}, x_{2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{array}} \begin{aligned} ∙ In other words, \\ ∙ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} and \\ ∙ x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \\ ∙ If there is no b x term, b = 0 \\ ∙ If there is no c term, c = 0 \end{aligned} \end{array}$
There are two possiblities:
- If the inside of the square root is negative, then $\underset{⏟}{\begin{matrix} a x^{2} + b x + c can only be factored by taking out a GCF \end{matrix}} \begin{aligned} ∙ In this case, a x^{2} + b x + c has no roots \\ \circ This is because a square root will negative inside is undefined \\ ∙ If the GCF is 1, a x^{2} + b x + c cannot be factored \end{aligned}$
- If the inside of the square root is non-negative, then the factored form of $a x^{2} + b x + c$ is $\begin{array}{r} \underset{⏟}{\begin{array}{c} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{array}} \begin{aligned} If (x - x_{1}) and/or (x - x_{2}) have fractions, the answer might \\ look nicer if you distribute a into (x - x_{1}) and/or (x - x_{2}) \end{aligned} \end{array}$

Ex 1: Factor $x^{2} + 2 x + 1$


	$Step 1: \begin{aligned} Determine x_{1} & = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} \\ and x_{2} & = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$	$\begin{aligned} x_{1}, x_{2} & = \frac{- (2) \pm \sqrt{(2)^{2} - 4 (1) (1)}}{2 (1)} \\ = \frac{- 2 \pm \sqrt{0}}{2} = \frac{- 2 \pm 0}{2} \\ = - 1, - 1 \end{aligned}$
	$Step 2: \underset{⏟}{\begin{matrix} Try to factor a x^{2} + b x + c with \\ a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{matrix}} \begin{array}{r} \begin{aligned} ∙ If x_{1} or x_{2} is undefined, a x^{2} + b x + c \\ ∙ cannot be factored \\ ∙ Otherwise, a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{aligned} \end{array}$	$x^{2} + 2 x + 1 = 1 \underset{⏟}{\begin{matrix} (x + 1) \end{matrix}} \begin{array}{r} x - (- 1) = x + 1 \end{array} \underset{⏟}{\begin{matrix} (x + 1) \end{matrix}} \begin{array}{r} x - (- 1) = x + 1 \end{array}$

Ex 2: Factor $2 x^{2} + 3 x + 1$


	$Step 1: \begin{aligned} Determine x_{1} & = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} \\ and x_{2} & = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$	$\begin{aligned} x_{1}, x_{2} & = \frac{- (3) \pm \sqrt{(3)^{2} - 4 (2) (1)}}{2 (2)} \\ = \frac{- 3 \pm \sqrt{1}}{4} = \frac{- 3 \pm 1}{4} \\ = - \frac{1}{2}, - 1 \end{aligned}$
	$Step 2: \underset{⏟}{\begin{matrix} Try to factor a x^{2} + b x + c with \\ a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{matrix}} \begin{array}{r} \begin{aligned} ∙ If x_{1} or x_{2} is undefined, a x^{2} + b x + c \\ ∙ cannot be factored \\ ∙ Otherwise, a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{aligned} \end{array}$	$2 x^{2} + 3 x + 1 = 2 (x + \frac{1}{2}) \underset{⏟}{\begin{matrix} (x + 1) \end{matrix}} \begin{array}{r} x - (- 1) = x + 1 \end{array}$

Ex 3: Factor $x^{2} + x + 1$


	$Step 1: \begin{aligned} Determine x_{1} & = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} \\ and x_{2} & = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$	$\begin{aligned} x_{1}, x_{2} & = \frac{- (1) \pm \sqrt{(1)^{2} - 4 (1) (1)}}{2 (1)} \\ = \frac{- 1 \pm \sqrt{- 3}}{2} \\ = \underset{⏟}{\begin{array}{c} undefined, undefined \end{array}} \begin{array}{r} b/c the square root of a negative number is undefined \end{array} \end{aligned}$
	$Step 2: \underset{⏟}{\begin{matrix} Try to factor a x^{2} + b x + c with \\ a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{matrix}} \begin{array}{r} \begin{aligned} ∙ If x_{1} or x_{2} is undefined, a x^{2} + b x + c \\ ∙ cannot be factored \\ ∙ Otherwise, a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) \end{aligned} \end{array}$	$\underset{⏟}{\begin{matrix} x^{2} + x + 1 can't be factored \end{matrix}} \begin{array}{r} b/c x_{1} and x_{2} are undefined \end{array}$

Subsection 5.2.6: Factoring

a x^{4} + b x^{2} + c

To factor $a x^{4} + b x^{2} + c$ , we

$\underset{⏟}{\begin{matrix} Write x^{4} as (x^{2})^{2} \end{matrix}} \begin{array}{r} This results in a (x^{2})^{2} + b (x^{2}) + c \end{array}$
$\underset{⏟}{\begin{matrix} Substitute u for x^{2} \end{matrix}} \begin{array}{r} This results in a u^{2} + b u + c \end{array}$
Factor $a u^{2} + b u + c$ using the information from previous sections
Plug $x^{2}$ back in for $u$

Ex 1: Factor $x^{4} + 3 x^{2} + 2$


	$Step 1: Part 1: Rewrite x^{4} as (x^{2})^{2}$	$1 x^{4} + 3 x^{2} + 2 = 1 (x^{2})^{2} + 3 (x^{2}) + 2$
	$Step 2: Part 2: Substitute u for x^{2}$	$1 x^{4} + 3 x^{2} + 2 = 1 u^{2} + 3 u + 2$
	$Step 3: Part 4: Factor a u^{2} + b u + c$
	$Step 4: Factor out the GCF$	$x^{2} + 3 x + 2 = \underset{⏟}{\begin{matrix} There is no GCF to take out \end{matrix}} \begin{array}{r} B/c the GCF is 1 \end{array}$
	$Step 5: \begin{matrix} \underset{⏟}{\begin{matrix} Determine the situation \end{matrix}} \begin{array}{r} \begin{aligned} ∙ a x + b cannot be factored \\ ∙ a x^{2} + b can't be factored \\ ∙ x^{2} - a^{2} = (x - a) (x + a) \\ ∙ b^{2} x^{2} - a^{2} = (b x - a) (b x + a) \\ ∙ x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2}) \\ ∙ x^{2} + b x + c = (x + a_{1}) (x + a_{2}) \\ ∙ where a_{1} and a_{2} are two numbers \\ ∙ which multiply to c and add to b \\ ∙ for a x^{2} + b x + c, follow these steps \\ ∙ for a x^{4} + b x^{2} + c, follow these steps \end{aligned} \end{array} \\ \underset{⏟}{\begin{matrix} or use the quadratic formula \end{matrix}} \begin{array}{r} a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) where \\ x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array} \end{matrix}$	$\begin{aligned} u^{2} + 3 u + 2 = u^{2} + b u + c = (u + a_{1}) (u + a_{2}) \\ where a_{1} and a_{2} are two numbers \\ which multiply to c and add to b \end{aligned}$
	$Step 6: \underset{⏟}{\begin{matrix} List the pairs of numbers which multiply to c \end{matrix}} \begin{array}{r} \begin{aligned} ∙ Don't forget to consider negative numbers \\ ∙ By numbers we technically mean integers \end{aligned} \end{array}$	$\begin{matrix} \underset{⏟}{\begin{matrix} c = 2 \end{matrix}} \begin{array}{r} u^{2} + 3 u + 2 \end{array} \\ Possible pairs: \\ 2 and 1 \\ - 2 and - 1 \end{matrix}$
	$Step 7: Determine the pair a_{1} and a_{2} which adds to b$	$\begin{matrix} \underset{⏟}{\begin{matrix} b = 3 \end{matrix}} \begin{array}{r} u^{2} + 3 u + 2 \end{array} \\ \underset{⏟}{\begin{matrix} a_{1} = 2, a_{2} = 1 \end{matrix}} \begin{array}{r} 2 and 1 is the pair that adds to 3 \end{array} \end{matrix}$
	$Step 8: Plug into x^{2} + b x + c = (x + a_{1}) (x + a_{2})$	$u^{2} + 3 u + 2 = (u + 2) (u + 1)$
	$Step 9: Plug x^{2} back in for u$	$1 x^{4} + 3 x^{2} + 2 = (x^{2} + 2) (x^{2} + 1)$

Section 5.2: Lecture 10/26Hide/Show in Tooltips

Subsection 5.2.1: Factoring out the Greatest Common Factor

Ex 1: Factor out the GCF of $- 9 x$

Ex 2: Factor out the GCF of $18 t - 21$

Ex 3: Factor out the GCF of $2 h^{2} - h$

Ex 4: Factor out the GCF of $- 8 z$

Subsection 5.2.2: Factoring Polynomials with Two Terms

Ex 1: Factor $x^{4} - 8 x$

Ex 2: Factor $2 x^{2} - 8$

Ex 3: Factor $6 x^{2} - 3$

Ex 4: Factor $3 x^{2} + 4$

Subsection 5.2.3: Factoring $x^{2} + b x + c$

Ex 1: Factor $x^{2} + 4 x + 4$

Ex 2: Factor $x^{2} + 3 x - 4$

Subsection 5.2.4: Factoring $a x^{2} + b x + c$

Ex 1: Factor $x^{4} - 4 x^{2} + 3$

Ex 2: Factor $2 x^{2} + 2 x - 4$

Ex 3: Factor $2 x^{2} + x - 1$

Subsection 5.2.5: Factoring With the Quadratic Formula

Ex 1: Factor $x^{2} + 2 x + 1$

Ex 2: Factor $2 x^{2} + 3 x + 1$

Ex 3: Factor $x^{2} + x + 1$

Subsection 5.2.6: Factoring $a x^{4} + b x^{2} + c$

Ex 1: Factor $x^{4} + 3 x^{2} + 2$

Section 5.2: Lecture 10/26Hide/Show in Tooltips Disable ColorsHideShow

Subsection 5.2.1: Factoring out the Greatest Common FactorLinkHideShow

Ex 1: Factor out the GCF of −9xHideShow

Ex 2: Factor out the GCF of 18t−21HideShow

Ex 3: Factor out the GCF of 2h2−hHideShow

Ex 4: Factor out the GCF of −8zHideShow

Subsection 5.2.2: Factoring Polynomials with Two TermsLinkHideShow

Ex 1: Factor x4−8xHideShow

Ex 2: Factor 2x2−8HideShow

Ex 3: Factor 6x2−3HideShow

Ex 4: Factor 3x2+4HideShow

Subsection 5.2.3: Factoring x2+bx+cLinkHideShow

Ex 1: Factor x2+4x+4HideShow

Ex 2: Factor x2+3x−4HideShow

Subsection 5.2.4: Factoring ax2+bx+cLinkHideShow

Ex 1: Factor x4−4x2+3HideShow

Ex 2: Factor 2x2+2x−4HideShow

Ex 3: Factor 2x2+x−1HideShow

Subsection 5.2.5: Factoring With the Quadratic FormulaLinkHideShow

Ex 1: Factor x2+2x+1HideShow

Ex 2: Factor 2x2+3x+1HideShow

Ex 3: Factor x2+x+1HideShow

Subsection 5.2.6: Factoring ax4+bx2+cLinkHideShow

Ex 1: Factor x4+3x2+2HideShow

Section 5.2: Lecture 10/26Hide/Show in Tooltips

Subsection 5.2.1: Factoring out the Greatest Common Factor

Ex 1: Factor out the GCF of $- 9 x$

Ex 2: Factor out the GCF of $18 t - 21$

Ex 3: Factor out the GCF of $2 h^{2} - h$

Ex 4: Factor out the GCF of $- 8 z$

Subsection 5.2.2: Factoring Polynomials with Two Terms

Ex 1: Factor $x^{4} - 8 x$

Ex 2: Factor $2 x^{2} - 8$

Ex 3: Factor $6 x^{2} - 3$

Ex 4: Factor $3 x^{2} + 4$

Subsection 5.2.3: Factoring $x^{2} + b x + c$

Ex 1: Factor $x^{2} + 4 x + 4$

Ex 2: Factor $x^{2} + 3 x - 4$

Subsection 5.2.4: Factoring $a x^{2} + b x + c$

Ex 1: Factor $x^{4} - 4 x^{2} + 3$

Ex 2: Factor $2 x^{2} + 2 x - 4$

Ex 3: Factor $2 x^{2} + x - 1$

Subsection 5.2.5: Factoring With the Quadratic Formula

Ex 1: Factor $x^{2} + 2 x + 1$

Ex 2: Factor $2 x^{2} + 3 x + 1$

Ex 3: Factor $x^{2} + x + 1$

Subsection 5.2.6: Factoring $a x^{4} + b x^{2} + c$

Ex 1: Factor $x^{4} + 3 x^{2} + 2$