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Factoring Quadratics

Solving quadratic equations in the form ax2+bx+c=0ax^2+bx+c=0 can be done in various ways. One way is by factoring.
Concept

Factoring

Given the product of a multiplicative expression, factoring is the process of breaking a number down into its smaller factor components. For example, the integer 1212 can be factored in several different ways: 12= 2612= 3412= 223\begin{aligned} 12=&\ 2\cdot6\\ 12=&\ 3\cdot4\\ 12=&\ 2\cdot2\cdot3 \end{aligned} Similarly, multiplicative algebraic expressions, such as 4x3,4x^3, can be rewritten by rewriting their coefficients and variables as a product of their factors.

4x3=22xxx\begin{gathered} 4x^3 = 2\cdot2\cdot x\cdot x\cdot x \end{gathered}
Method

Factor by GCF

When all terms in an expression contain a common factor, the expression can be rewritten as a product of its factors. Each term in the expression is divided by the common factor. It is written in front of a parentheses which contains the quotient of the terms. Consider the expression x2+2x. x^2 + 2x. Notice that each term contains x.x. Factoring out xx results in the product x(x+2).x(x + 2). Notice that if we use the Distributive Property on the factors, what results is the original expression. x2+2xxx+2xx(x+2)\begin{aligned} x^2 &+2x \\ x \cdot x &+ 2 \cdot x \\ x(x &+ 2) \end{aligned}

4x+2y4x+2y

3a29a3a^2-9a

7ab+4b7ab+4b

6x+606x+60

-4x+4\text{-} 4x+4

12x+24y12x+24y

Sometimes, it's possible to factor out more than one factor. Because factoring allows expressions to be simplified, factoring out the greatest common factor, GCF, is preferred.
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Exercise

Find and factor the GCF in the following expressions. 4a+2and6x22x 4a+2 \quad \text{and} \quad 6x^2-2x

Show Solution
Solution
Example

4a+24a+2

To find the GCF between terms, it's necessary to analyze the number and variable part of each term. Notice that 4a4a means 4a.4 \cdot a. Thus, 44 and aa are both factors of this term. Additionally, 44 can be written in terms of its factors. This gives 4a22 a. 4a \quad \leftrightarrow \quad {\color{#0000FF}{2}} \cdot 2 \ a. The second term of the original expression, 2,2, does not have any variable part. Thus, it can be written in terms of its factors as 221. 2 \quad \leftrightarrow \quad {\color{#0000FF}{2}} \cdot 1. Notice that 4a4a and 22 both share a factor of 2.2. Thus, 22 is the GCF. Factoring 22 out of the expression gives 2(2a+1). 2(2a+1).

Example

6x22x6x^2-2x

Following the same process as above, the terms in this expression can be written in terms of their factors. 6x22x23xx21x\begin{aligned} 6x^2 &- 2x \\ {\color{#0000FF}{2}} \cdot 3 \cdot {\color{#009600}{x}} \cdot x &- {\color{#0000FF}{2}} \cdot 1 \cdot {\color{#009600}{x}} \end{aligned} Notice that each term contains a 22 and an xx in its factors. Thus, 2x2x is the expression's GCF. Factoring 2x2x gives 2x(3x1). 2x(3x-1).

Method

Solving an Equation using the Zero Product Property

An expression written in factored form and set equal to 00 can be solved using the Zero Product Property. When the product of two or more factors is 0,0, at least one of the factors must equal 0.0. Consider the following equation. (3x9)(x+5)=0\begin{aligned} (3x-9)(x+5)=0 \end{aligned}

1

Set each factor equal to 00

Since one of the factors must equal 0,0, set them each equal to 00 and solve for the variable. Notice that new equations are created. 3x9=0andx+5=0\begin{aligned} 3x-9=0 \quad \text{and} \quad x+5=0 \end{aligned}

2

Solve the equations

Now, use inverse operations to solve the equations. 3x9=0x=3x+5=0x=-5\begin{aligned} 3x-9=0\quad&\Leftrightarrow\quad x=3\\ x+5=0\quad&\Leftrightarrow\quad x=\text{-}5 \end{aligned} The solutions to the new equations both solve the original equation. In this case, that means that x=3x=3 and x=-5x=\text{-}5 solve (3x9)(x+5)=0.\begin{aligned} (3x-9)(x+5)=0. \end{aligned}

Method

Factoring a Quadratic Trinomial

When factoring an expression in the form ax2+bx+cax^2+bx+c it can be difficult to see its factors. 4x2+13x+3\begin{gathered} 4x^2+13x+3 \end{gathered} This expression can be factored by finding a pair of integers whose product is ac,a\cdot c, which here is 43,4\cdot 3, and whose sum is b,b, which in the example is 13.13.

1

List all pairs of integers whose product is equal to aca\cdot c

The first step is to find all possible pairs of integers that multiply to ac.a\cdot c. In this case, a=4a=4 and c=3.c=3. Thus, their product is 34=12. 3\cdot 4 = 12. To find all factor pairs, start with the pair where one factor is 1.1. The other factor must then be 12.12. Then continue with the pair where one of the factors is 2,2, and so forth. In this case, there are three pairs. 1and122and63and4\begin{aligned} &1 \quad \text{and} \quad 12\\ &2 \quad \text{and} \quad 6\\ &3 \quad \text{and} \quad 4 \end{aligned}

2

Find the pair whose sum is bb

If the given expression is factorable, one of the factor pairs will add to equal b.b. In this case, b=13.b=13. 1+12=132+6=83+4=7\begin{aligned} 1+12&=13 \\ 2+6&=8 \\ 3+4&=7 \end{aligned} Here, 11 and 1212 is the only factor pair that adds to 13.13.

3

Write bxbx as a sum

Now, use the factor pair to rewrite the xx-term of the original expression as a sum. Since the factor pair is 11 and 12,12, the middle term can be written as 13x=12x+x. 13x = 12x + x. This gives the following equivalent expression. 4x2+  13x +34x2+   12x+x+3\begin{aligned} 4x^2 &+ \ \ \quad {\color{#0000FF}{13x}}\ &&+ 3 \\ 4x^2 &+ \ \ \ {\color{#0000FF}{12x +x}} &&+ 3 \end{aligned} Notice that the expression hasn't been changed. Rather, it has been rewritten.

4

Factor out the GCF in the first two terms

Now, the expression has four terms, which can be grouped into the first two terms and the last two terms. Then, the GCF of each group can be factored out. Here, begin with the first group of terms, 4x24x^2 and 12x.12x. Notice how each can be written as a product of its factors. 4x2 + 12x4xx + 43x\begin{aligned} 4x^2 \ &+ \ 12x \\ 4 \cdot x \cdot x \ &+ \ 4 \cdot 3 \cdot x \end{aligned} The GCF is 4x.4x. Therefore, it's possible to factor out 4x.4x. 4x2+12x+x+34x(x+3)+x+3\begin{aligned} 4x^2 +12x \textcolor{lightgray}{+}\textcolor{lightgray}{x} \textcolor{lightgray}{+} \textcolor{lightgray}{3} \\ 4x(x+3)\textcolor{lightgray}{+} \textcolor{lightgray}{x}\textcolor{lightgray}{+}\textcolor{lightgray}{3} \end{aligned}

5

Factor out the GCF in the last two terms

Next, repeat the same process with the last two terms. In this case, xx and 33 do not have any common factors, but it's always possible to write expressions as a product of 11 and itself. 4x(x+3)+x+34x(x+3)+1(x+3)\begin{aligned} \textcolor{lightgray}{4x(x+3)} & \textcolor{lightgray}{+} x+3 \\ \textcolor{lightgray}{4x(x+3)} & \textcolor{lightgray}{+} 1(x+3) \end{aligned}

6

Factor out the common factor

If all previous steps have been performed correctly, there should now be two terms with a common factor, which can be seen as a repeated parentheses. Factoring (x+3)(x+3) out of both terms gives 4x(x+3)+1(x+3)(x+3)(4x+1).\begin{aligned} {\color{#0000FF}{4x}}(x+3) + & {\color{#0000FF}{1}} (x+3) \\ (x+3) ({\color{#0000FF}{4}}&{\color{#0000FF}{x}}+{\color{#0000FF}{1}}). \end{aligned} This means that 4x2+13x+34x^2+13x+3 can be written in factored form as (x+3)(4x+1).(x+3)(4x+1).

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Exercise

Solve the following equation by factoring. x22x3=0 x^2-2x-3=0

Show Solution
Solution
We'll focus on the left-hand side of the equation before solving for x.x. Notice that 11 and -3\text{-} 3 are the factors that multiply to equal -3\text{-} 3 and add to -2.\text{-} 2.
x22x3=0x^2 - 2x-3 = 0
x2+x3x3=0x^2+x - 3x-3 = 0
x(x+1)3x3=0x(x+1) - 3x-3 = 0
x(x+1)3(x+1)=0x(x+1) - 3(x+1) = 0
(x+1)(x3)=0(x+1)(x-3) = 0
Notice that, in factored form, the numbers in the parentheses with xx are the numbers from the chosen factor pair. This is because the coefficient of x2x^2 is 1.1. We could have written the expression in factored form immediately after finding the factor pair. Lastly, we can use the zero product property to solve the equation. x+1=0x=-1x3=0x=-3\begin{aligned} x+1=0 \quad &\leftrightarrow \quad x=\text{-} 1\\ x-3=0 \quad &\leftrightarrow \quad x= \phantom{\text{-}} 3 \end{aligned} Thus, the solutions to the equation x22x3=0x^2-2x-3=0 are x=-1x=\text{-} 1 and x=3.x=3.
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Exercise

The area of a rectangle is 1212 square meters and can described by 2x2+5x+9.2x^2+5x+9. Write an equation that represents this area. Solve the equation to find x.x.

Show Solution
Solution
To begin, we can create an equation from the given information. Equating the area of the rectangle, 1212 square meters, with the given rule, we have 2x2+5x+9=12. 2x^2+5x+9=12. Factoring and solving this equation will allow us to determine x.x. Before we factor, we must set the equation equal to 0.0. 2x2+5x+9=122x2+5x3=0 2x^2+5x+9=12 \quad \Leftrightarrow \quad 2x^2+5x-3=0
2x2+5x3=02x^2+5x-3=0
2x2+6xx3=02x^2+6x-x-3=0
2x(x+3)x3=02x(x+3)-x-3=0
2x(x+3)1(x+3)=02x(x+3)-1(x+3)=0
(x+3)(2x1)=0(x+3)(2x-1)=0
In factored form, we have (x+3)(2x1)=0.(x+3)(2x-1)=0. We'll use the Zero Product Property to create two indvidual equations. x+3=0x=-32x1=0x=0.5\begin{aligned} x+3 =0 &\Leftrightarrow x=\text{-} 3 \\ 2x-1 =0 &\Leftrightarrow x=0.5 \end{aligned} Thus, the values of xx that make the area equation true are x=-3x=\text{-} 3 and x=0.5.x=0.5.
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