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Solving quadratic equations in the form $ax_{2}+bx+c=0$ can be done in various ways. One way is by *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 $12$ can be factored in several different ways:
$12=12=12= 2⋅63⋅42⋅2⋅3 $
Similarly, multiplicative algebraic expressions, such as $4x_{3},$ can be rewritten by rewriting their coefficients and variables as a product of their factors.

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 $x_{2}+2x.$
Notice that each term contains $x.$ **Factoring out** $x$ results in the product $x(x+2).$ Notice that if we use the Distributive Property on the factors, what results is the original expression.
$x_{2}x⋅xx(x +2x+2⋅x+2) $

$4x+2y$

$3a_{2}−9a$

$7ab+4b$

$6x+60$

$-4x+4$

$12x+24y$

Find and factor the GCF in the following expressions. $4a+2and6x_{2}−2x$

Show Solution

To find the GCF between terms, it's necessary to analyze the number and variable part of each term. Notice that $4a$ means $4⋅a.$ Thus, $4$ and $a$ are both factors of this term. Additionally, $4$ can be written in terms of its factors. This gives $4a↔2⋅2a.$ The second term of the original expression, $2,$ does not have any variable part. Thus, it can be written in terms of its factors as $2↔2⋅1.$ Notice that $4a$ and $2$ both share a factor of $2.$ Thus, $2$ is the GCF. Factoring $2$ out of the expression gives $2(2a+1).$

Following the same process as above, the terms in this expression can be written in terms of their factors. $6x_{2}2⋅3⋅x⋅x −2x−2⋅1⋅x $ Notice that each term contains a $2$ and an $x$ in its factors. Thus, $2x$ is the expression's GCF. Factoring $2x$ gives $2x(3x−1).$

An expression written in factored form and set equal to $0$ can be solved using the Zero Product Property. When the product of two or more factors is $0,$ at least one of the factors must equal $0.$ Consider the following equation. $(3x−9)(x+5)=0 $

Set each factor equal to $0$

Solve the equations

**both** solve the original equation. In this case, that means that $x=3$ and $x=-5$ solve
$(3x−9)(x+5)=0. $

When factoring an expression in the form $ax_{2}+bx+c$ it can be difficult to ### 1

The first step is to find all possible pairs of integers that multiply to $a⋅c.$ In this case, $a=4$ and $c=3.$ Thus, their product is $3⋅4=12.$ To find all factor pairs, start with the pair where one factor is $1.$ The other factor must then be $12.$ Then continue with the pair where one of the factors is $2,$ and so forth. In this case, there are three pairs. $ 1and122and63and4 $

### 2

If the given expression is factorable, one of the factor pairs will add to equal $b.$ In this case, $b=13.$ $1+122+63+4 =13=8=7 $ Here, $1$ and $12$ is the only factor pair that adds to $13.$

### 3

Now, use the factor pair to rewrite the $x$-term of the original expression as a sum. Since the factor pair is $1$ and $12,$ the middle term can be written as $13x=12x+x.$ This gives the following equivalent expression. $4x_{2}4x_{2} +13x+12x+x +3+3 $ Notice that the expression hasn't been changed. Rather, it has been rewritten.

### 4

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, $4x_{2}$ and $12x.$ Notice how each can be written as a product of its factors. $4x_{2}4⋅x⋅x +12x+4⋅3⋅x $ The GCF is $4x.$ Therefore, it's possible to factor out $4x.$ $4x_{2}+12x+x+34x(x+3)+x+3 $

### 5

Next, repeat the same process with the last two terms. In this case, $x$ and $3$ do not have any common factors, but it's always possible to write expressions as a product of $1$ and itself. $4x(x+3)4x(x+3) +x+3+1(x+3) $

### 6

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 $(x+3)$ out of both terms gives $4x(x+3)+(x+3)(4 1(x+3)x+1). $ This means that $4x_{2}+13x+3$ can be written in factored form as $(x+3)(4x+1).$

seeits factors. $4x_{2}+13x+3 $ This expression can be factored by finding a pair of integers whose product is $a⋅c,$ which here is $4⋅3,$ and whose sum is $b,$ which in the example is $13.$

List all pairs of integers whose product is equal to $a⋅c$

Find the pair whose sum is $b$

Write $bx$ as a sum

Factor out the GCF in the first two terms

Factor out the GCF in the last two terms

Factor out the common factor

Solve the following equation by factoring. $x_{2}−2x−3=0$

Show Solution

We'll focus on the left-hand side of the equation before solving for $x.$ Notice that $1$ and $-3$ are the factors that multiply to equal $-3$ and add to $-2.$
Notice that, in factored form, the numbers in the parentheses with $x$ are the numbers from the chosen factor pair. This is because the coefficient of $x_{2}$ is $1.$ 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−3=0 ↔x=-1↔x=-3 $
Thus, the solutions to the equation $x_{2}−2x−3=0$ are $x=-1$ and $x=3.$

$x_{2}−2x−3=0$

$x_{2}+x−3x−3=0$

FactorOutFactor out $x$

$x(x+1)−3x−3=0$

FactorOutFactor out $-3$

$x(x+1)−3(x+1)=0$

FactorOutFactor out $(x+1)$

$(x+1)(x−3)=0$

The area of a rectangle is $12$ square meters and can described by $2x_{2}+5x+9.$ Write an equation that represents this area. Solve the equation to find $x.$

Show Solution

To begin, we can create an equation from the given information. Equating the area of the rectangle, $12$ square meters, with the given rule, we have $2x_{2}+5x+9=12.$
Factoring and solving this equation will allow us to determine $x.$ Before we factor, we must set the equation equal to $0.$
$2x_{2}+5x+9=12⇔2x_{2}+5x−3=0$
In factored form, we have $(x+3)(2x−1)=0.$ We'll use the Zero Product Property to create two indvidual equations. $x+3=02x−1=0 ⇔x=-3⇔x=0.5 $
Thus, the values of $x$ that make the area equation true are $x=-3$ and $x=0.5.$

$2x_{2}+5x−3=0$

$2x_{2}+6x−x−3=0$

FactorOutFactor out $2x$

$2x(x+3)−x−3=0$

FactorOutFactor out $-1$

$2x(x+3)−1(x+3)=0$

FactorOutFactor out $x+3$

$(x+3)(2x−1)=0$

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