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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.

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
$x(x+3)(x−1).$

$x_{2}+2x−3$

Rewrite

Rewrite $2x$ as $3x−x$

$x_{2}+3x−x−3$

FactorOut

Factor out $x$

$x(x+3)−x−3$

FactorOut

Factor out $-1$

$x(x+3)−(x+3)$

FactorOut

Factor 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

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