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.
Method
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.
Example
Factor the trinomial
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Exercise
Factor the trinomial completely.
x3+2x2−3x
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.
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.
x3+2x2−3x=x(x2+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 x2+2x−3 can be written as the product (x+3)(x−1). Thus, x3+2x2−3x can be factored into
x(x+3)(x−1).
Method
Factoring by Grouping
Sometimes it's possible to factor polynomials even if its terms do not have a common factor. The polynomial
y=2x3−x2+6x−3
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.
y=2x3−x2+6x−3=(2x3−x2)+(6x−3)
2
Factor out the GCF in each pair
In the first pair, factor out the GCF. Here, the GCF is x2.(2x3−x2)+(6x−3)x2(2x−1)+(6x−3)
For the second pair, the GCF is 3.x2(2x−1)+(6x−3)x2(2x−1)+3(2x−1)
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 (2x−1).x2(2x−1)+3(2x−1)(2x−1)(x2+3)
This means that y=2x3−x2+6x−3 can be written as
y=(2x−1)(x2+3).
Example
Solve the equation by factoring by grouping
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Exercise
Solve the equation by factoring.
2x3+3x2−18x−27=0
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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 x2, 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.
x2(2x+3)−9(2x+3)=(x2−9)(2x+3)=0
Using the Zero Product Property, we can separate this equation into two new equations, that can be solved individually.