3.02 Factoring polynomials
Concept summary
Factoring polynomials is a process of expressing polynomials from a sum or difference of terms to a product of its factors. In other words, it is the inverse process of multiplying polynomials.
A greatest common factor from each term of a polynomial is factored out
ax+ay+\ldots=a(x+y+\ldots)
A method for factoring an expression containing at least four terms, by grouping the terms in pairs and taking out common factors
ax + ay + bx + by = a\left(x + y\right) + b\left(x + y\right) = \left(x+y\right)\left(a+b\right)
A trinomial that can be expressed as the product of two binomials
ax^{2} + bx + c= \left(x + p\right)\left(x + q\right) where pq=c and p+q=b
A trinomial that can be expressed as the product of two binomials
ax^{2} + bx + c= \left(mx + p\right)\left(nx + q\right) where mn=a,pq=c and np+mq=b
We have a few identities that can help us.
A trinomial that is formed by multiplying a binomial by itself
a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2} \text{ or } a^{2} - 2 a b + b^{2} = \left(a - b\right)^{2}
The result of a perfect square being subtracted from another perfect square
a^{2} - b^{2} = \left(a+b\right)\left(a-b\right)
Two perfect cube expressions being added
a^{3} + b^{3} = \left(a+b\right)\left(a^2-ab+b^2\right)
The result of a perfect cube being subtracted from another perfect cube
a^{3} - b^{3} = \left(a-b\right)\left(a^2+ab+b^2\right)
Worked examples
Example 1
Factor 6 x^{4} - 10x^{3} - 24x^{2}.
Approach
First, we find any common factors for all three terms and factor it out. Then, from the simpler trinomial, we find the value of r and s, that multiply to ac and add up to b. After finding r and s, we write the trinomial in the form ax^{2} + rx + sx + c and factor it by grouping.
Solution
The common factors of the three terms is 2x^{2}. We factor out 2x^{2} and get 2x^{2}\left(3 x^{2} - 5x - 12\right).
For 3 x^{2} - 5x - 12, we find two numbers that have a product of ac = \left(3\right)\left(-12\right) = -36 and sum of b = -5.
The factors of -36 include \pm1, \pm2, \pm3,\pm4,\pm6,\pm9,\pm18 and \pm36. Among these factors are 1 and -6, 4 and -9 that add up to -5. We choose r=4 and s=-9.
We write 3 x^{2} - 5x - 12 in the form ax^{2} + rx + sx + c. Substituting the values for r and s, we get
3x^{2} +4x -9 x - 12
Now, we factor the expression by grouping
| \displaystyle 3x^{2} +4x -9x -12 | \displaystyle = | \displaystyle (3x^{2} +4x) +(-9x -12) | Split based on common factors |
| \displaystyle = | \displaystyle x\left(3x + 4\right) - 3\left(3x + 4\right) | Factor out the GCF: x and -3 | |
| \displaystyle = | \displaystyle \left(x-3\right)\left(3x+4\right) | Factor out the common binomial factor |
Therefore, 6 x^{4} - 10x^{3} - 24x^{2}=2x^{2}\left(x-3\right)\left(3x+4\right).
Reflection
We can check the answer by multiplying the factored form 2x^{2}\left(x-3\right)\left(3x+4\right).
| \displaystyle 2x^{2}\left(x-3\right)\left(3x+4\right) | \displaystyle = | \displaystyle 2x^{2}\left(x\left(3x+4\right)-3\left(3x+4\right)\right) | Distribute 3x + 4 |
| \displaystyle = | \displaystyle 2x^{2}\left(3x^{2} +4x -9x - 12\right) | Distribute x and -3 | |
| \displaystyle = | \displaystyle 6 x^{4} + 8x^{3} -18x^{3} - 24x^{2} | Distribute 2x^{2} | |
| \displaystyle = | \displaystyle 6 x^{4} - 10x^{3} - 24x^{2} | Combine like terms |
Example 2
Factor 18m^{3}-2mn^2+9m^2n-n^3.
Approach
We arrange the terms first, grouping those with common factor. We factor out the GCF on each pair and the common binomial factor afterwards.
Solution
| \displaystyle 18m^{3}-2mn^2+9m^2n-n^3 | \displaystyle = | \displaystyle \left(18m^{3}-2mn^2\right)+\left(9m^2n-n^3\right) | Split based on common factors |
| \displaystyle = | \displaystyle 2m\left(9m^{2}-n^{2}\right) + n\left(9m^{2}-n^{2}\right) | Factor out the GCF: 2m and n | |
| \displaystyle = | \displaystyle \left(2m+n\right)\left(9m^{2}-n^{2}\right) | Factor out the common binomial factor |
Observe that 9m^{2}-n^{2}=\left(3m\right)^{2}-\left(n\right)^{2}. This means that 9m^{2}-n^{2} is a difference of two squares, so we use the formula in factoring:
| \displaystyle a^{2}-b^{2} | \displaystyle = | \displaystyle (a-b)(a+b) | Formula of difference of two squares |
| \displaystyle \left(3m\right)^{2}-\left(n\right)^{2} | \displaystyle = | \displaystyle (3m-n)(3m+n) | Substitue a=3m and b=n |
If we substitute 9m^{2}-n^{2}=(3m+n)(3m-n), we get 18m^{3}-2mn^2+9m^2n-n^3=\left(2m+n\right)(3m-n)(3m+n)
Reflection
Alternatively, we can group 8m^{3} \text{ and } 9m^{2}n and -2mn^{2} \text{ and } -n^{3} together and get the same answer.
\displaystyle 18m^{3}-2mn^2+9m^2n-n^3 \displaystyle = \displaystyle \left(18m^{3}+9m^2n \right)+ \left(-2mn^2-n^3\right) Split based on common factors \displaystyle = \displaystyle 9m^{2}\left(2m+n\right) + \left(-n^{2}\right)\left(2m+n\right) Factor out the GCF: 9m^2 and -n^2 \displaystyle = \displaystyle \left(9m^{2} - n^{2}\right)\left(2m + n\right) Factor out the common binomial factor \displaystyle = \displaystyle \left(3m +n\right)\left(3m -n\right)\left(2m + n\right) Apply the formula for difference of two squares We can check the answer by multiplying the factored form \left(3m +n\right)\left(3m -n\right)\left(2m + n\right).
\displaystyle \left(3m +n\right)\left(3m -n\right)\left(2m + n\right) \displaystyle = \displaystyle \left(9m^{2} - n^{2}\right)\left(2m + n\right) Apply the formula for difference of two squares \displaystyle = \displaystyle 2m\left(9m^{2} - n^{2}\right)+n\left(9m^{2} - n^{2}\right) Distribute \\9m^{2} - n^{2} \displaystyle = \displaystyle 18m^{3}-2mn^2+9m^2n-n^3 Simplify
Example 3
Factor 64x^{3}+125y^{3}.
Approach
We check first whether 64x^{3}+125y^{3} is a special product and identify its type.
Solution
Observe that 64x^{3}+125y^{3}=\left(4x\right)^{3}+\left(5y\right)^{3}. This means that 64x^{3}+125y^{3} is a sum of two cubes, so we use the formula in factoring:
| \displaystyle a^{3} + b^{3} | \displaystyle = | \displaystyle \left(a+b\right)\left(a^2-ab+b^2\right) | Formula of sum of two cubes |
| \displaystyle \left(4x\right)^{3}+\left(5y\right)^{3} | \displaystyle = | \displaystyle \left(4x+5y\right)\left((4x)^{2}+\left(4x\right)\left(5y\right)+\left(5y\right)^{2}\right) | Substitute a=4x and b=5y |
| \displaystyle = | \displaystyle \left(4x+5y\right)\left(16x^{2}+20xy+25y^{2}\right) | Simplify |
Therefore, 64x^{3}+125y^{3}=\left(4x+5y\right)\left(16x^{2}+20xy+25y^{2}\right).
Reflection
We can check the answer by multiplying the factored form \left(4x+5y\right)\left(16x^{2}+20xy+25y^{2}\right).
Observe that \left(4x+5y\right)\left(16x^{2}+20xy+25y^{2}\right)=\left(4x+5y\right)\left((4x)^{2}+\left(4x\right)\left(5y\right)+\left(5y\right)^{2}\right).
Now,
| \displaystyle \left(4x+5y\right)\left(16x^{2}+20xy+25y^{2}\right) | \displaystyle = | \displaystyle \left(4x\right)^{3}+\left(5y\right)^{3} | Formula of sum of two cubes |
| \displaystyle = | \displaystyle 64x^{3}+125y^{3} | Simplify |
Example 4
Factor 27p^{3}-\dfrac{1}{8}.
Approach
We check first whether 27p^{3}-\dfrac{1}{8} is a special product and identify its type.
Solution
Observe that 27p^{3}-\dfrac{1}{8}=\left(3p\right)^{3}-\left(\dfrac{1}{2}\right)^{3}. This means that 27p^{3}-\dfrac{1}{8} is a difference of two cubes, so we use the formula in factoring:
| \displaystyle a^{3} - b^{3} | \displaystyle = | \displaystyle \left(a-b\right)\left(a^2+ab+b^2\right) | Formula of difference of two cubes |
| \displaystyle (3p)^{3}-\left(\dfrac{1}{2}\right)^{3} | \displaystyle = | \displaystyle \left(3p-\dfrac{1}{2}\right)\left((3p)^{2}+(3p)\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^{2}\right) | Substitute a=3p and b=\dfrac{1}{2} |
| \displaystyle = | \displaystyle \left(3p-\dfrac{1}{2}\right)\left(9p^{2}+\dfrac{3}{2}p+\dfrac{1}{4}\right) | Simplify |
Therefore, 27p^{3}-\dfrac{1}{8}=\left(3p-\dfrac{1}{2}\right)\left(9p^{2}+\dfrac{3}{2}p+\dfrac{1}{4}\right).
Reflection
We can check the answer by multiplying the factored form \left(3p-\dfrac{1}{2}\right)\left(9p^{2}+\dfrac{3}{2}p+\dfrac{1}{4}\right).
Observe that \left(3p-\dfrac{1}{2}\right)\left(9p^{2}+\dfrac{3}{2}p+\dfrac{1}{4}\right)=\left(3p-\dfrac{1}{2}\right)\left((3p)^{2}+(3p)\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^{2}\right).
Now,
| \displaystyle \left(3p-\dfrac{1}{2}\right)\left(9p^{2}+\dfrac{3}{2}p+\dfrac{1}{4}\right) | \displaystyle = | \displaystyle (3p)^{3}-\left(\dfrac{1}{2}\right)^{3} | Formula of difference of two cubes |
| \displaystyle = | \displaystyle 27p^{3}-\dfrac{1}{8} | Simplify |
Outcomes
MA.912.AR.1.8
Rewrite a polynomial expression as a product of polynomials over the real or complex number system.