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{{ printedBook.courseTrack.name }} {{ printedBook.name }} Factoring Polynomials

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.

Factoring Trinomials

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.
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Exercise
Factor the trinomial completely.
x3+2x23x
Show Solution
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.
xxx+2xx3x
It can be seen that each term has an x. Thus, we can factor the GCF of x out of the entire expression.
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.
x2+2x3
x2+3xx3
x(x+3)x3
x(x+3)(x+3)
(x+3)(x1)
The expression x2+2x3 can be written as the product (x+3)(x1). Thus, x3+2x23x can be factored into
x(x+3)(x1).

Factoring by Grouping

Sometimes it's possible to factor polynomials even if its terms do not have a common factor. The polynomial
y=2x3x2+6x3
can be factored by grouping the terms.

1

Group the terms in pairs
To begin, the first two terms and the last two terms are grouped. This can be done using parentheses.

2

Factor out the GCF in each pair
In the first pair, factor out the GCF. Here, the GCF is x2.
For the second pair, the GCF is 3.

3

Factor out the GCF in the resulting sum
If the polynomial is factorable, this should lead to a sum of two terms with a common factor. In this case, that factor is (2x1).
This means that y=2x3x2+6x3 can be written as
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Exercise
Solve the equation by factoring.
2x3+3x218x27=0
Show Solution
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.
(2xxx)+(3xx)(233x)(333)
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 x2, and in the second it is -9. Let's factor out the GCFs separately.
2x3+3x218x27=0
x2(2x+3)18x27=0
x2(2x+3)9(2x+3)=0
Now, both terms have the factor 2x+3, which can be factored out.
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.