Factoring Quadratic Equations

Objectives

This lesson presents the necessary background material to:

Factor expressions of the form $x^{2} + bx + c$
Factor expressions of the form ${ax}^{2} + bx + c$

Introduction

If we observe the quadratic expression $x^{2} + 3 x + 2$ we can readily see that there is no common factor to the three terms. However, in the tutorial on multiplying binomial expressions we learned that

$(x + 1) (x + 2) = x^{2} + 3 x + 2$
Hence there are factors of $x^{2} + 3 x + 2$ , however we will need new techniques for factoring such expressions. This lesson lesson will present the new methods we need to factor expressions of the form

${ax}^{2} + bx + c$

Factoring x² + bx + c

We found in the tutorial on multiplying binomial expressions that

$(x + a) (x + b) = x^{2} + (a + b) x + ab$
Hence if we start with an expression of the form $x^{2} + cx + d$ , our goal is to find $a$ and $b$ such that $a + b = c$ and $ab = d$ . If we can find such $a$ and $b$ then the expression

$x^{2} + cx + d = x^{2} + (a + b) x + ab = (x + a) (x + b)$
and we have succeeded in factoring $x^{2} + cx + d$ . From this we can obtain the following set of steps to factor quadratic expressions with x²coefficient equal to 1.

Example 1:

Factor: $x^{2} + 7 x + 12$

Solution:

Step 1. Find all pairs of (a, b) such that ab = 12:

(12,1), (6,2), (4,3), (-12,-1), (-6,-2), (-4,-3)

As: $(x + a) (x + b) = (x + b) (x + a)$
$a$ and $b$ are interchangeable and we do not repeat pairs such as (12,1) and (1,12).

Step 2. Determine if one of these pairs satisfies a + b = 7:
$12 + 1 \neq 7; 6 + 2 \neq 7; 4 + 3 = 7$
So a= 4, b= 3 satisfies a + b = 7 and ab = 12. Hence:

$x^{2} + 7 x + 12 = (x + 4) (x + 3)$

Step 3. Verify $(x + 4) (x + 3) = x^{2} + 3 x + 4 x + 12$
$(x + 4) (x + 3) = x^{2} + 7 x + 12$

Example 2:

Factor: $x^{2} - 4 x - 21$

Solution:

Note: For convenience of our method we will rewrite differences as sums with negative numbers so the problem becomes factor

$x^{2} + (- 4 x) + (- 21)$

Step 1. Find all pairs of (a,b) such that ab = -21:

(21,-1), (-21,1), (7,-3), (-7,3)

Step 2. Determine if one of these pairs satisfies
              a + b = -4:

$21 + (- 1) \neq - 4; - 21 + 1 \neq - 4;$ $7 + (- 3) \neq - 4; - 7 + 3 = - 4 .$
              So a= -7, b= 3 satisfies a + b = -4 and ab = -21.

              Hence:

$x^{2} - 4 x - 21 = (x + (- 7)) (x + 3)$

Step 3. Verify $(x + (- 7)) (x + 3) = x^{2} + 3 x + (- 7 x) + (- 21)$
$(x - 7) (x + 3) = x^{2} - 4 x - 21$

Practice 1.

Click on the link below to practice factoring quadratic expressions with leading coefficient equal to 1.

Factoring ax² + bx + c

We can multiply the following expressions together to yield the following::

$(ax + b) (cx + d) = {acx}^{2} + (ad + cb) x + bd$
Hence if we start with an expression of the form ${ex}^{2} + fx + g$ , our goal is to find $a$ , $b$ , $c$ and $d$ such that $ac = e$ , $ad + bc = f$ and $bd = g$ . If we can find such $a$ , $b$ , $c$ y $d$ then the expression

$(ax + b) (cx + d) = {acx}^{2} + (ab + cd) x + bd = {ex}^{2} + fx + g$
and we have factored: ${ex}^{2} + fx + g$ . From this we can obtain the following set of steps for factoring quadratic expressions.

Example 1:

Factor: ${2x}^{2} + 11 x + 5$

Solution:

Step 1. Find all pairs of (a,c) such that ac = 2:

(2,1), (1,2), (-2,-1), (-1,-2)

As: $(ax + b) (cx + d) \neq (cx + b) (ax + d)$
$a$ and $c$ nare not interchangeable and we repeat pairs such as (2,1) y (1,2).

Step 2. Find all pairs of (b,d) such that bd = 5:
$(5, 1)$
As: $(ax + b) (cx + d) = (cx + b) (ax + d)$
if we interchange $a$ and $c$ , we do not need to interchange $b$ and $d$ as interchanging both creates an equivalent expression.

Also, as : $(ax + b) (cx + d) = (- ax + (- b)) (- cx + (- d))$
we do not need to include both (x, y) and (-x,-y). One is enough.

Step 3. Substitute all combinations of (a,c) and (b,d) into the expression :
$(ax + b) (cx + d)$
multiply the expressions together and determine if one of these is equal to :

${2x}^{2} + 11 x + 5$

(a,c)	(b,d)	(ax+b)(cx+d)	Product
(2,1)	(5,1)	(2x+5)(1x+1)	2x²+7x+5
(1,2)	(5,1)	(x+5)(2x+1)	2x²+11x+5
(-2,-1)	(5,1)	(-2x+5)(-1x+1)	2x²-7x+5
(-1,-2)	(5,1)	(-x+5)(-2x+1)	2x²-11x+5

Solution: ${2 x}^{2} + 11 x + 5 = (2 x + 1) (x + 5)$

Example 2:

Factor: ${3x}^{2} + 5 x - 12$

Solution:

Step 1. Find all pairs of (a,c) such that ac = 3:

(3,1), (1,3), (-3,-1), (-1,-3)

Step 2. Find all pairs of (b,d) such that bd = -12:

(12,-1), (6,-2), (4,-3)

Paso 3: Substitute all combinations of (a,c) and (b,d) into the expression:
$(ax + b) (cx + d)$
multiply the expressions together and determine if one of these is equal to :

${3x}^{2} + 5 x - 12$

(a,c)	(b,d)	(ax+b)(cx+d)	Product
(3,1)	(12,-1)	(3x+12)(x-1)	3x²+9x -12
(3,1)	(6,-2)	(3x+6)(x-2)	3x²+0x-12
(3,1)	(4,-3)	(3x+4)(1x-3)	3x²-5x -12
(1,3)	(12,-1)	(x+12)(3x-1)	3x²+35x -12
(1,3)	(6,-2)	(x+6)(3x-2)	3x²+16x-12
(1,3)	(4,-3)	(x+4)(3x-3)	3x²+9x-12
(-3,-1)	(12,-1)	(-3x+12)(-1x-1)	3x²-9x-12
(-3,-1)	(6,-2)	(-3x+6)(-x-2)	3x²+0x-12
(-3,-1)	(4,-3)	(-3x+4)(-1x-3)	3x²+5x-12
(-1,-3)	(12,-1)	(-x+12)(-3x-1)	3x²-23x-12
(-1,-3)	(6,-2)	(-x+6)(-3x-2)	3x²-16x-12
(-1,-3)	(4,-3)	(-x+4)(-3x-3)	3x²-9x-12

Solution: ${3 x}^{2} + 5 x - 12 = (3 x - 4) (x + 3)$

Practice 2.

Click on the link below to practice factoring quadratic expressions with leading coefficient not equal to 1.

Summary

Now that you've completed this lesson, you should be able to:

Factor expressions of the form $x^{2} + bx + c$
Factor expressions of the form ${ax}^{2} + bx + c$

Factoring Quadratic Equations

Objectives

Introduction

Factoring x2 + bx + c

Example 1:

Solution:

Example 2:

Solution:

Practice 1.

Factoring ax2 + bx + c

Example 1:

Solution:

Solution: 2⁢x 2+11⁢x+5= 2⁢x+1 x+5

Example 2:

Solution:

Solution: 3⁢x2+5⁢ x-12=3⁢x-4x+3

Practice 2.

Summary

Factoring x² + bx + c

Factoring ax² + bx + c

Solution: ${2 x}^{2} + 11 x + 5 = (2 x + 1) (x + 5)$

Solution: ${3 x}^{2} + 5 x - 12 = (3 x - 4) (x + 3)$