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Factoring is a necessary skill to have in precalculus and all math classes beyond precalculus. Factoring is a tool to simplify expressions and eases calculations. It helps us identify the roots of polynomials, making quick work of graphing without any need for computers or graphing calculators. In the same way arithmetic rules inform algebraic properties, prime factorization of real numbers informs polynomials factorization. The following real numbers have been factored into a product of prime numbers:
\(\begin{array} &36 &= 2 \cdot 2 \cdot 3 \cdot 3 \\ 40 &= 2 \cdot 2 \cdot 2 \cdot 5 \\ 45 &= 3 \cdot 3 \cdot 5 \\ 455 &= 5 \cdot 7 \cdot 13 \end{array}\)
Polynomial factorization models the same concept: break a polynomial down into the product of its prime polynomials.
This section will describe the following factoring techniques:
- Greatest Common Factor (GCF)
- Trinomial Factorization: \(x^2 − bx + c\)
- Trinomial Factorization: \(ax^2 − bx + c\) where \(a \neq 1\).
- Difference of Squares
- Perfect Square Trinomials
- Difference of Cubes
- Sum of Cubes
Greatest Common Factor (GCF)
Think of GCF factoring as reversing (or undoing) the distributive property.
\(\xrightarrow{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{multiply}} \\ a(b+c+d + ...) = a(b) + a(c) + a(d)+ ... \\ \xleftarrow[\text{factor}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;]{\text{}} \)
To find the GCF, list the factors of each term. Identify the factors in common. Then factor using the GCF method, as shown in the example below.
Example 2.3.1
Factor the trinomial \(16x^5 + 8x^3 − 4x^2\).
Solution
Break down each of the \(3\) terms of the trinomial into its factors as follows
\(\begin{array} &16x^5 &= \textcolor{red}{2} \cdot \textcolor{blue}{2} \cdot 2 \cdot 2 \cdot \textcolor{green}{x} \cdot \textcolor{pink}{x} \cdot x \cdot x \cdot x \\ 8x^3 &= \textcolor{red}{2} \cdot \textcolor{blue}{2} \cdot 2 \cdot \textcolor{green}{x} \cdot \textcolor{pink}{x} \cdot x \\ 4x^2 &= \textcolor{red}{2} \cdot \textcolor{blue}{2} \cdot \textcolor{green}{x} \cdot \textcolor{pink}{x} \end{array}\)
\(2 \cdot 2 \cdot x \cdot x = \textcolor{red}{4x^2 = \text{GCF}}\)
Next step: factor out the GCF from each of the terms.
Inside the parentheses, preserve placement of the 3 terms and the operations addition and subtraction:
\(\textcolor{red}{\text{GCF}}(\textcolor{green}{\text{Term 1}} + \textcolor{green}{\text{Term 2}} - \textcolor{green}{\text{Term 3}} )\).
Answer The factored form of the polynomial is \(4x^2(4x^3 + 2x -1)\)
Note: You can check your answer using the distributive property. This step is optional, but by multiplying out the factored form, you will internalize the concept that the product is the original polynomial.
Tip: GCF factorization is always the first technique to look for in any polynomial. Sometimes a polynomial will require several different factoring techniques to completely factor the polynomial.
Trinomial Factorization x2 + bx + c
If the leading coefficient of a \(2^{\text{nd}}\) degree trinomial is \(1\), we can factor the trinomial into the product of two binomials: \((x + m)(x + n)\). Two conditions on \(m\) and \(n\) must hold:
\(\left. \begin{array}{ll} mn = c\\ m + n =b \end{array} \right\} \text{Factoring Technique} \Longrightarrow x^2 + bx + c = (x + m)(x + n) \)
Example 2.3.2
Factor the trinomial \(x^2 −x − 12\).
Solution
Think of a pair of integers, \(m\) and \(n\), such that the two conditions hold:
\(\left. \begin{array}{ll} mn = -12\\ m + n = -1 \end{array} \right\}\)
Possible values \(m\), (n\):
| Product is \(−12\) | Success if the sum is \(−1\), |
| \((−1)(12) = −12 \\ (1)(−12) = −12\) | \( \left.\begin{array}{ll} \textcolor{red}{\times}\;\; − 1+ 12 = −11\\ \textcolor{red}{\times} \;\;12 + (−1) = 11 \end{array} \right\} \text{opposites} \) |
| \((−2)(6) = −12 \\ (2)(−6) = −12\) | \( \left.\begin{array}{ll} \textcolor{red}{\times} \;\;− 2+ 6 = 4\\ \textcolor{red}{\times} \;\;2 + (−6) = −4 \end{array} \right\} \text{opposites} \) |
| \((−3)(4) = −12 \\ (3)(−4) = −12\) | \( \left.\begin{array}{ll} \textcolor{red}{\times} \;\;−3 + 4 = 1\\ \textcolor{green}{\checkmark} \;\;3 + (−4) = −1 \end{array} \right\} \text{opposites} \) |
Answer The factored form of \(x^2 − x − 12\) is \((x + 3)(x − 4)\).
Note: You can check your answer using the FOIL method. This step is optional, but by multiplying out the factored form, you will internalize the concept that the product is the original polynomial.
Note
Some polynomials are prime, which means the polynomial cannot be factored. If every possible factoring technique fails, the polynomial is a prime polynomial.
Trinomial Factorization ax2 + bx + c where a ≠ 1
If the leading coefficient of a \(2^{\text{nd}}\) degree trinomial is not \(1\), it will require more trial-and-error to factor the polynomial. The goal, as with the case \(a=1\), is to create two binomials that, when multiplied by the FOIL method, produce the given polynomial. However, the \(a\)−value will add more attempted trials. Don’t forget that some polynomials are prime and cannot be factored. The steps below guide your approach to factoring.
Trial-and-Error Method for Factoring ax^2 + bx +c and a ≠ 1
Step 1: Is there a GCF in the trinomial? Factor out a common value if you find one.
Step 2: List the pairs of factors of \(a\) and the pairs of factors of \(c\).
Step 3: Construct the binomials. Fill in the \(4\) boxes appropriately.
Step 4: FOIL: Outer = big smiley-face. Inner = small smiley-face. Multiply to find Outer and Inner of FOIL. Find a successful combination such that Outer + Inner \(= bx =\) Target Sum.
Step 5: Keep \(a\)-factors fixed, but consider reversing \(c\)-factors if the trial fails to produce \(bx\).
Step 6: If all combinations of factors fail to produce the Target Sum \(bx\), the trinomial is prime.
Example 2.3.3
Factor \(2x^2 + 3x + 1\)
Solution
The first term of the trinomial is \(2x^2\). Since \(a \neq 1\), let’s walk through each step:
Step 1: GCF \(= 1\), so no common factors.
Step 2: List pairs of factors of \(a\) and of \(c\):
| \(a = 2\) | \(c = 1\) |
| \(\textcolor{red}{2 \cdot 1}\) | \(\textcolor{blue}{1 \cdot 1}\) |
Step 3: \((\textcolor{red}{2}x + \textcolor{blue}{1})(\textcolor{red}{1}x + \textcolor{blue}{1}) \)
Step 4: Let’s check Outer and Inner to make sure their sum is \(bx = 3x\).
\(\text{Outer } + \text{ Inner } = 2x + x = 3x\)
Success! We found the correct factorization!
Answer The factored form of the trinomial \(2x^2 + 3x + 1 = (2x+1)(x+1)\)
Example 2.3.4
Factor \(6t^2 − 28t + 16\)
Solution
Step 1: GCF \(=2\). Factor out the common factor: \(2(3t^2 − 14t + 8)\)
Step 2: List pairs of factors of \(a\) and of \(c\):
Notice the Target Sum is negative.
\(bx = −14t\)
However, \(c\) is positive. The \(c\)-factors must both be negative (see the table below)
| \(a=3\) | \(c=8\) |
| \(\textcolor{red}{3 \cdot 1}\) | \(\textcolor{blue}{-8 \cdot -1}\) |
| \(\textcolor{blue}{-4 \cdot -2}\) |
Step 3: Place the \(a\)-factors, then choose a pair of \(c\)-factors. Since the target sum is \(−14t\), it’s more likely that the smaller numbers will be successful.
Trial 1: \(2(\textcolor{red}{3}t − \textcolor{blue}{4})(\textcolor{red}{1}t − \textcolor{blue}{2})\)
Step 4: Let’s check the Outer and Inner.
Trial 1 fails. \(−4t − 6t \neq −14t\)
Trial 2: \(2(\textcolor{red}{3}t − \textcolor{blue}{2})(\textcolor{red}{1}t − \textcolor{blue}{4})\)
Step 5: Let’s check the Outer and Inner
Trial 2 is a success! \(−12t − 2t = −14t\)
Answer \(6t^2 − 28t + 16 = 2(3t-2)(t-4)\). The trinomial is completely factored.
Difference of Squares
Below are three examples that demonstrate a pattern after using the FOIL method.
| Multiply the Binomials | FOIL Method | Simplified Result | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| \((x + 3)(x − 3)\) | \(=\) | \(x^2 + \underbrace{−3x + 3x}_{\text{O + I cancels}} −3 \cdot 3\) | \(=\) | \(x^2 − 9\) | ||||||
| \((y-10)(y+10)\) | \(=\) | \(y^2 + \underbrace{+10y - 10y}_{\text{O + I cancels}} −10 \cdot 10\) | \(=\) | \(y^2 - 100\) | ||||||
| \((n^2 + 4)(n^2 − 4)\) | \(=\) | \(n^4 + \underbrace{-4n^2 + 4n^2}_{\text{O + I cancels}} −4 \cdot 4\) | \(=\) | \(n^4-16\) | ||||||
Two binomials are multiplied. One is the conjugate of the other. That is, in one binomial \(x\) and \(a\) are added, and in the other \(x\) and \(a\) are subtracted. Notice that the outer and inner terms always cancel each other under this pattern.
Special Product = Difference of Squares: \((A+B)(A-B) = A^2 - B^2\)
Factoring is the reverse of multiplying. In order to factor, we have to unsquare the perfect squares, creating a product:
\(\xrightarrow{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{multiply}} \\ (x+a)(x-a) = \underbrace{x^2-a^2}_{\text{difference of squares}} \\ \xleftarrow[\text{factor}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;]{\text{}} \)
Example 2.3.5
Factor: \(n^2 − 64\)
Solution
Both terms are perfect squares. We’ll use the Special Product formula in reverse: \(A^2 − B^2 = (A + B)(A − B)\). In order to factor, we need the unsquare of each term!
Here, \(A=n\) and \(B=8\). The formula guides our factoring:
\(\begin{array} &&(A+B)(A-B) \\ &(n+8)(n-8) \end{array}\)
Answer \(n^2 − 64 = (n+8)(n-8)\)
Example 2.3.6
Factor completely. \(4x^2y − 81x\)
Solution
When factoring, the first thing to look for is a Greatest Common Factor (GCF). Both terms of the polynomial share the factor \(y\). After factoring out the GCF, we factor the parentheses as a difference of squares.
\(4x^2y - 81y\)
GCF \(= y\) Factor out GCF
\(=y(4x^2-81)\)
Difference of Squares \(A=2x\) and \(B=9\)
\(=y(2x+9)(2x-9)\)
Note
The word “difference” translates to “subtraction.” The Difference of Squares requires subtraction between the two squared terms: \(A^2 − B^2\). Aside from common factors, the sum of squares \(A^2 + B^2\) is not factorable.
Example 2.3.7
Factor \(n^2 + 25\).
Solution
Although both terms are perfect squares, this is not a difference of squares. The operation between the squares is not subtraction. The GCF \(=1\). Since we cannot factor out a common factor, we conclude that \(n^2 + 25\) is a prime polynomial.
Perfect Square Trinomials
Another pattern emerges when two identical binomials are multiplied. Below are three examples that demonstrate a pattern after using the FOIL method.
| Multiply the Binomials | FOIL Method | Simplified Result | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| \((t + 5)(t + 5)\) | \(=\) | \(t^2 + \underbrace{−5t + 5t}_{\text{O + I} = 2(5t)} + 5^2\) | \(=\) | \(t^2 + 10t + 25\) | ||||||
| \((2y − 9)(2y − 9)\) | \(=\) | \((2y)^2 - \underbrace{18y - 18y}_{\text{O + I} =2(2y)(−10)} + 9^2\) | \(=\) | \(4y^2 − 36y + 81\) | ||||||
| \((pq + 7)(pq + 7)\) | \(=\) | \((pq)^2 + \underbrace{7pq + 7pq}_{\text{O + I} =2(7pq)} + 7^2\) | \(=\) | \(p^2q^2 + 14pq + 49\) | ||||||
Special Product = Perfect Square Trinomial
\( \begin{array} &&(A + B)^2 = (A + B)(A + B) = A^2 + 2AB + B^2 \\ &(A − B)^2 = (A − B)(A − B) = A^2 − 2AB + B^2 \end{array} \)
Factoring is the reverse of multiplying. In order to factor, identify the unsquare of the first and last term. That is, find \(A\) and \(B\), then follow the formulas above.
Example 2.3.8
Factor \(25n^2 − 90n + 81\)
Solution
The trinomial has GCF \(= 1\). However, the first and last terms are perfect squares, and the polynomial is a trinomial. Let’s see if we were given a Perfect Square Trinomial.
- Match the formula with the given trinomial.
- Determine the value of \(A\) and \(B\).
- Check that \(2AB\) also matches.
- The minus symbol is a major clue telling us which special product formula we’ll use.
- \(A^2 − 2AB + B^2 = (A − B)^2\)
Be sure to verify that \(2AB\) also matches the given middle term. In Example \(2.3.8\), we were given a Perfect Square Trinomial because \(2AB\) matched the middle term, \(90n\). Therefore, we can factor this trinomial using the formula for factoring Perfect Square Trinomials.
Answer: \(25n^2 − 90n + 81 = (5n-9)^2\)
Note
Caution: If a polynomial cannot be factored using one factoring technique, it doesn’t necessarily mean the polynomial is prime. Don’t give up! Exhaust all possible factoring techniques before concluding that the polynomial is prime.
Example 2.3.9
Factor \(9y^2 + 15y + 4\)
Solution
The GCF \(=1\). However, the first and last terms are perfect squares, and the polynomial is a trinomial. Let’s see if we were given a Perfect Square Trinomial.
By matching the terms, we determine \(A = 3y\) and \(B = 2\). However, \(2AB = 12y\) does not match the given middle term \(15y\). The trinomial is not a Perfect Square Trinomial.
The trinomial is not prime, however. It can be factored using the trial-and-error method.
Step 1: GCF \(=1\). There is no common factor.
Step 2: List pairs of factors of \(a\) and of \(c\): (see the chart below).
| \(a=9\) | \(c=4\) |
| \(\textcolor{red}{9 \cdot 1}\) | \(\textcolor{blue}{4 \cdot 1}\) |
| \(\textcolor{red}{3 \cdot 3}\) | \(\textcolor{blue}{2 \cdot 2}\) |
\(bx = 15y = \text{ Target Sum}\)
Step 3: Place the a-factors, then choose a pair of \(c\)-factors. We already know that \((3y + 2)(3y + 2)\) does not work. Let’s try: \((3y + 4)(3y + 1)\)
Step 4: Let’s check the Outer and Inner
This trial is a success! \(12y + 3y = 15y\)
Other trials that fail:
| \((3y + 2)(3y + 2)\) | \(6y + 6y= 12y\) | \(\neq 15y\) |
| \((9y + 4)(y + 1)\) | \(4y + 9y = 13y\) | \(\neq 15y\) |
| \((9y + 1)(y + 4)\) | \(y+36y=37y\) | \(\neq 15y\) |
Answer \(9y^2 + 15y + 4 = (3y+4)(3y+1)\).
Factor Sum of Cubes and Difference of Cubes
In order to identify the correct strategy for factoring, it’s a good idea to have at least the first five cubes memorized. The formulas shown below guide our factoring.
\(\begin{array} &&1^3 = 1 \\ &2^3 = 8 \\ &3^3 = 27 \\&4^3 = 84 \\&5^3 = 125 \end{array}\)
Factoring Sums and Differences of Cubes
\(\begin{array} &&A^3 + B^3 = (A + B)(A^2 − AB + B^2 ) \\ &A^3 − B^3 = (A − B)(A^2 + AB + B^2 ) \end{array}\)
Example 2.3.10
Factor \(27 − 8m^3\)
Solution
The binomial is a difference of cubes.
Use the formula with \(A = 3\) and \(B = 2m\).
\(27 − 8m^3 = (3 − 2m)(9 + 6m + 4m^2)\)
Try It! (Exercises)
Factor #1-3 following using GCF factorization:
- \(12u − 6u^2\)
- \(10x^3y + 15xy^2\)
- \(12z^3 − 9z^2 − 3z\)
Factor #4-6 using trinomial factorization:
- \(t^2 + 2t - 15\)
- \(h^2 − 12h + 20\)
- \(r^2 + 14r + 24\)
Factor #7-12 using both GCF and trinomial factorization:
- \(3b^2 + 12b + 9\)
- \(4q^2 − 4q − 80\)
- \(x^3 − 11x^2 + 10x\)
- \(2a^2b + 24ab − 56b\)
- \(7x^4 + 49x^3 + 70x^2\)
- \(3n^2m − 12mn − 63m\)
Factor #13-18 using trinomial factorization where the leading coefficient \(\neq 1\).
- \(2d^2 − 9d − 18\)
- \(3h^2 − 29h + 18\)
- \(7t^2 + 17t − 12\)
- \(4y^2 − 3y − 7\)
- \(10n^2 − 11n − 8\)
- \(8x^2 + 6x − 9\)
Factor #19-24 using difference of squares factorization:
- \(r^2 − 100\)
- \(81 − h^2\)
- \(4t^2 − 1\)
- \(25 − n^4\)
- \(v^4 − 36\)
- \(16p^2 − 25q^2\)
Factor #25-27 using sum of cubes formula:
- \(c^3 + 125\)
- \(27 + 64n^3\)
- \(1 + a^6\)
Factor #28-30 using difference of cubes formula:
- \(q^3 − 8\)
- \(125d^3 − 27\)
- \(64u^3 − v^3\)
Factor #31-36 using any appropriate method. Factor completely.
- \(30m^2 − 5m − 5\)
- \(2c^2 − 8\)
- \(h^4 − 81\)
- \(6p^3q + 6q\)
- \(15k^3 − 2k^2 − k\)
- \(a^2b^2 − a^2\)
- What is a prime polynomial? Give \(3\) examples of prime trinomials.
- Give an example of a sum of squares. Is it factorable? Explain.
- The number \(64\) is both a perfect square and a cube. What other numbers are both a square and a cube? Do you think there are there only a few or infinitely many?
