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Polynomials are usually written in standard form. However, depending on what they describe and what information is needed, it's sometimes useful to write them in factored form. By rewriting polynomials as products of their factors, it's possible to solve polynomial equations.

There are many ways to factor an expression, such as factoring by GCF or factoring a quadratic trinomial. If a polynomial expression has a degree higher than $2,$ it can be helpful to combine these methods.

Factor the trinomial completely. $x_{3}+2x_{2}−3x$

Show Solution

To factor the trinomial, let's start by determining if the terms have any common factor. To do this, we'll write each term as a product of its factors.
$x⋅x⋅x+2⋅x⋅x−3⋅x$
It can be seen that each term has an $x.$ Thus, we can factor the GCF of $x$ out of the entire expression.
$x_{3}+2x_{2}−3x=x(x_{2}+2x−3)$
The parentheses now contain a quadratic trinomial, which can be factored. Notice that $3$ and $-1$ multiply to equal $-3$ and add to equal $2.$
The expression $x_{2}+2x−3$ can be written as the product $(x+3)(x−1).$ Thus, $x_{3}+2x_{2}−3x$ can be factored into
$x(x+3)(x−1).$

$x_{2}+2x−3$

$x_{2}+3x−x−3$

FactorOutFactor out $x$

$x(x+3)−x−3$

FactorOutFactor out $-1$

$x(x+3)−(x+3)$

FactorOutFactor out $(x+3)$

$(x+3)(x−1)$

Sometimes it's possible to factor polynomials even if its terms do not have a common factor. The polynomial
$y=2x_{3}−x_{2}+6x−3$
can be factored by *grouping* the terms.
### 1

To begin, the first two terms and the last two terms are grouped. This can be done using parentheses. $y=2x_{3}−x_{2}+6x−3=(2x_{3}−x_{2})+(6x−3)$

### 2

In the first pair, factor out the GCF. Here, the GCF is $x_{2}.$ $(2x_{3}−x_{2})+(6x−3)x_{2}(2x−1)+(6x−3) $ For the second pair, the GCF is $3.$ $ x_{2}(2x−1)+(6x−3)x_{2}(2x−1)+3(2x−1) $

### 3

If the polynomial is factorable, this should lead to a sum of two terms with a common factor. In this case, that factor is $(2x−1).$ $x_{2} (2x−1)+3(2x−1)(2x−1)(x_{2}+3) $ This means that $y=2x_{3}−x_{2}+6x−3$ can be written as $y=(2x−1)(x_{2}+3).$

Group the terms in pairs

Factor out the GCF in each pair

Factor out the GCF in the resulting sum

Solve the equation by factoring. $2x_{3}+3x_{2}−18x−27=0$

Show Solution

To begin, we must determine which type of factoring we can use to solve the given polynomial equation. Notice that there is no common factor amongst all four terms.
$(2⋅x⋅x⋅x)+(3⋅x⋅x)−(2⋅3⋅3⋅x)−(3⋅3⋅3)$
The first three terms all have an $x$ in common, while the last three terms have a $3$ in common. Thus, we must factor by grouping the first two terms and the last two terms. In the first pair, the GCF is $x_{2},$ and in the second it is $-9.$ Let's factor out the GCFs separately.
Now, both terms have the factor $2x+3,$ which can be factored out.
$x_{2}(2x+3)−9(2x+3)=(x_{2}−9)(2x+3)=0$
Using the Zero Product Property, we can separate this equation into two new equations, that can be solved individually.
Thus, the original equation has three solutions: $x=-3,$ $x=-1.5,$ and $x=3.$

$2x_{3}+3x_{2}−18x−27=0$

FactorOutFactor out $x_{2}$

$x_{2}(2x+3)−18x−27=0$

FactorOutFactor out $-9$

$x_{2}(2x+3)−9(2x+3)=0$

$(x_{2}−9)(2x+3)=0$

ZeroProdPropUse the Zero Product Property

$x_{2}−9=02x+3=0 (I)(II) $

$x_{2}=92x+3=0 $

$x=±9 2x+3=0 $

CalcRoot$(I):$ Calculate root

$x=±32x+3=0 $

$x=±32x=-3 $

$x=±3x=-1.5 $

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