Practice Factoring Polynomials by Grouping
| Site: | Saylor Academy |
| Course: | MA007: Algebra |
| Book: | Practice Factoring Polynomials by Grouping |
| Printed by: | Guest user |
| Date: | Tuesday, 15 July 2025, 7:37 AM |
Description
buttons.Factoring Common Binomials
The first step in the factoring process is often factoring the common monomials from a polynomial. Sometimes polynomials have common terms that are binomials. For example, consider the following expression.
\(\begin{align*}x(3x+2)-5(3x+2)\end{align*}\)
You can see that the term \(\begin{align*}(3x+2)\end{align*}\) appears in both terms of the polynomial. This common term can be factored by writing it in front of a set of parentheses. Inside the parentheses, we write all the terms that are left over when we divide them by the common factor.
\(\begin{align*}(3x+2)(x-5)\end{align*}\)
This expression is now completely factored. Let's look at some examples.
Example 2
Factor \(\begin{align*}3x(x-1)+4(x-1)\end{align*}\).
Solution:
\(\begin{align*}3x(x-1)+4(x-1)\end{align*}\) has a common binomial of \(\begin{align*}(x-1)\end{align*}\).
When we factor the common binomial, we get \(\begin{align*}(x-1)(3x+4)\end{align*}\).
Factoring Polynomials Completely
We say that a polynomial is factored completely when we factor as much as we can and we are unable to factor any more. Here are some suggestions that you should follow to make sure that you factor completely.
- Factor all common monomials first.
- Identify special products such as difference of squares or the square of a binomial. Factor according to their formulas.
- If there are no special products, factor using the methods we learned in the previous sections.
- Look at each factor and see if any of these can be factored further.
Example 1
Factor the following polynomials completely.
(a) \(\begin{align*}2x^2-8\end{align*}\)
(b) \(\begin{align*}x^3+6x^2+9x\end{align*}\)
Solution:
(a) Look for the common monomial factor. \(\begin{align*}2x^2-8=2(x^2-4)\end{align*}\). Recognize \(\begin{align*}x^2-4\end{align*}\) as a difference of squares. We factor \(\begin{align*}2(x^2-4)=2(x+2)(x-2)\end{align*}\). If we look at each factor we see that we can't factor anything else. The answer is \(\begin{align*}2(x+2)(x-2)\end{align*}\).
(b) Recognize this as a perfect square and factor as \(\begin{align*}x(x+3)^2\end{align*}\). If we look at each factor we see that we can't factor anything else. The answer is \(\begin{align*}x(x+3)^2\end{align*}\).
Source: cK-12, https://www.ck12.org/book/basic-algebra/section/9.7/
This work is licensed under a Creative Commons Attribution-NonCommercial 3.0 License.
Factoring by Grouping
It may be possible to factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factoring by grouping. The following example illustrates how this process works.
Example 3
Factor \(\begin{align*}2x+2y+ax+ay\end{align*}\).
Solution
There isn't a common factor for all four terms in this example. However, there is a factor of 2 that is common to the first two terms and there is a factor of \(\begin{align*}a\end{align*}\) that is common to the last two terms. Factor 2 from the first two terms and factor \(\begin{align*}a\end{align*}\) from the last two terms.
\(\begin{align*}2x+2y+ax+ay = 2(x+y) +a(x+y)\end{align*}\)
Now we notice that the binomial \(\begin{align*}(x+y)\end{align*}\) is common to both terms. We factor the common binomial and get.
\(\begin{align*}(x+y)(2+a)\end{align*}\)
Our polynomial is now factored completely.
We know how to factor Quadratic Trinomials \(\begin{align*}(ax^2+bx+c)\end{align*}\) where \(\begin{align*}a \neq 1\end{align*}\) using methods we have previously learned. To factor a quadratic polynomial where \(\begin{align*}a \neq 1\end{align*}\), we follow the following steps.
- We find the product \(\begin{align*}ac\end{align*}\).
- We look for two numbers that multiply to give \(\begin{align*}ac\end{align*}\) and add to give \(\begin{align*}b\end{align*}\).
- We rewrite the middle term using the two numbers we just found.
- We factor the expression by grouping.
Let's apply this method to the following examples.
Example 4
Factor \(\begin{align*}3x^2+8x+4\end{align*}\) by grouping.
Solution:
Follow the steps outlined above.
\(\begin{align*}ac=3 \cdot 4=12\end{align*}\)
The number 12 can be written as a product of two numbers in any of these ways:
\(\begin{align*}12&=1 \times 12 && and && 1+12=13\\ 12 & =2 \times 6 && and && 2+6=8 \qquad \text{This is the correct choice}.\end{align*}\)
Rewrite the middle term as: \(\begin{align*}8x=2x+6x\end{align*}\), so the problem becomes the following.
\(\begin{align*}3x^2+8x+4 = 3x^2+2x+6x+4\end{align*}\)
Factor an \(\begin{align*}x\end{align*}\) from the first two terms and 2 from the last two terms.
\(\begin{align*}x(3x+2)+2(3x+2)\end{align*}\)
Now factor the common binomial \(\begin{align*}(3x+2)\end{align*}\).
\(\begin{align*}(3x+2)(x+2)\end{align*}\)
Our answer is \(\begin{align*}(3x+2)(x+2)\end{align*}\).
In this example, all the coefficients are positive. What happens if the \(\begin{align*}b\end{align*}\) is negative?
Example 5:
Factor \(\begin{align*}6x^2-11x+4\end{align*}\) by grouping.
Solution:
\(\begin{align*}ac=6 \cdot 4 = 24\end{align*}\)
The number 24 can be written as a product of two numbers in any of these ways.
\(\begin{align*}24&=1\times 24 && and && 1+24=25\\ 24&=(-1) \times (-24) && and && (-1)+(-24)=-25\\ 24&=2 \times 12 && and && 2+12=14\\ 24&=(-2) \times (-12) && and && (-2)+(-12)=-14\\ 24&=3 \times 8 && and && 3+8=11\\ 24&=(-3) \times (-8) && and && (-3)+(-8)=-11 \quad \text{This is the correct choice}.\end{align*}\)
Rewrite the middle term as \(\begin{align*}-11x=-3x-8x\end{align*}\), so the problem becomes:
\(\begin{align*}6x^2-11x+4=6x^2-3x-8x+4\end{align*}\)
Factor by grouping. Factor a \(\begin{align*}3x\end{align*}\) from the first two terms and factor –4 from the last two terms.
\(\begin{align*}3x(2x-1)-4(2x-1)\end{align*}\)
Now factor the common binomial \(\begin{align*}(2x-1)\end{align*}\).
Our answer is \(\begin{align*}(2x-1)(3x-4)\end{align*}\).
Solving Real-World Problems Using Polynomial Equations
Now that we know most of the factoring strategies for quadratic polynomials, we can see how these methods apply to solving real-world problems.
Example 6:
The product of two positive numbers is 60. Find the two numbers if one of the numbers is 4 more than the other.
Solution:
\(\begin{align*}x=\end{align*}\) one of the numbers and \(\begin{align*}x+4\end{align*}\) equals the other number. The product of these two numbers equals 60. We can write the equation.
\(\begin{align*}x(x+4)=60\end{align*}\)
Write the polynomial in standard form.
\(\begin{align*}x^2+4x&=60\\ x^2+4x-60&=0\end{align*}\)
Factor: \(\begin{align*}-60=6 \times(-10)\end{align*}\) and \(\begin{align*}6+(-10)=-4\end{align*}\)
\(\begin{align*}-60=-6 \times 10\end{align*}\) and \(\begin{align*}-6+10=4\end{align*}\) This is the correct choice.
The expression factors as \(\begin{align*}(x+10)(x-6)=0\end{align*}\).
Solve:
\(\begin{align*}x+10=0 && x-6& =0\\ \text{or} \\ x=-10 && x& =6\end{align*}\)
Since we are looking for positive numbers, the answer must be positive.
\(\begin{align*}x=6\end{align*}\) for one number, and \(\begin{align*}x+4=10\end{align*}\) for the other number.
Check: \(\begin{align*}6 \cdot 10=60\end{align*}\) so the answer checks.
Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
Factor by Grouping and Factoring Completely
Factor by grouping.
- \(\begin{align*}6x^2-9x+10x-15\end{align*}\)
- \(\begin{align*}5x^2-35x+x-7\end{align*}\)
- \(\begin{align*}9x^2-9x-x+1\end{align*}\)
- \(\begin{align*}4x^2+32x-5x-40\end{align*}\)
- \(\begin{align*}12x^3-14x^2+42x-49\end{align*}\)
- \(\begin{align*}4x^2+25x-21\end{align*}\)
- \(\begin{align*}24b^3+32b^2-3b-4\end{align*}\)
- \(\begin{align*}2m^3+3m^2+4m+6\end{align*}\)
- \(\begin{align*}4xy+32x+20y+160\end{align*}\)
- \(\begin{align*}10ab+40a+6b+24\end{align*}\)
- \(\begin{align*}9mn+12m+3n+4\end{align*}\)
- \(\begin{align*}4jk-8j^2+5k-10j\end{align*}\)
- \(\begin{align*}24ab+64a-21b-56\end{align*}\)