6.7: Factoring by Grouping
The volume of a rectangular prism is \begin{align*}3x^5 - 27x^4 - 2x^2 + 18x\end{align*}. What are the lengths of the prism's sides?
Watch This
James Sousa: Factor By Grouping
Guidance
In the Factoring when the Leading Coefficient Doesn't Equal 1 concept (in the previous chapter), we introduced factoring by grouping. We will expand this idea to other polynomials here.
Example A
Factor \begin{align*}x^4+7x^3-8x-56\end{align*} by grouping.
Solution: First, group the first two and last two terms together. Pull out any common factors.
\begin{align*}\underbrace{x^4+7x^3}_{x^3{\color{red}(x+7)}}\underbrace{-8x-56}_{-8{\color{red}(x+7)}}\end{align*}
Notice what is inside the parenthesis is the same. This should always happen when factoring by grouping. Pull out this common factor.
\begin{align*}& x^3(x+7)-8(x+7)\\ & (x+7)(x^3-8)\end{align*}
Look at the factors. Can they be factored any further? Yes. The second factor is a difference of cubes. Use the formula.
\begin{align*}& (x+7)(x^3-8)\\ & (x+7)(x-2)(x^2+2x+4)\end{align*}
Example B
Factor \begin{align*}x^3+5x^2-x-5\end{align*} by grouping.
Solution: Follow the steps from above.
\begin{align*}& x^3+5x^2-x-5\\ & x^2(x+5)-1(x+5)\\ & (x+5)(x^2-1)\end{align*}
Look to see if we can factor either factor further. Yes, the second factor is a difference of squares.
\begin{align*}& (x+5)(x^2-1)\\ &(x+5)(x-1)(x+1)\end{align*}
Example C
Find all real-number solutions of \begin{align*}2x^3-3x^2+8x-12 = 0\end{align*}.
Solution: Follow the steps from Example A.
\begin{align*}2x^3-3x^2+8x-12 &= 0\\ x^2(2x-3)+4(2x-3) &= 0\\ (2x-3)(x^2+4) &= 0\end{align*}
Now, determine if you can factor further. No, \begin{align*}x^2+4\end{align*} is a sum of squares and not factorable. Setting the first factor equal to zero, we get \begin{align*}x = \frac{3}{2}\end{align*}.
Intro Problem Revisit
We need to factor \begin{align*}3x^5 - 27x^4 - 2x^2 + 18x\end{align*} to find the lengths of the prism's sides.
First, pull out the common factor. \begin{align*}x(3x^4 - 27x^3 - 2x + 18)\end{align*}
Next, factor \begin{align*}(3x^4 - 27x^3 - 2x + 18)\end{align*} by grouping the first two and last two terms together.
\begin{align*}(3x^4 - 27x^3 - 2x + 18)\\ = (3x^4 - 27x^3) + (- 2x + 18)\\ = 3x^3(x - 9) - 2(x - 9)\end{align*}
Now pull out the common factor.
\begin{align*}3x^3(x - 9) - 2(x - 9)\\ = (3x^3 - 2)(x - 9)\end{align*}
The expression can't be factored further, so \begin{align*}3x^5 - 27x^4 - 2x^2 + 18x = x(3x^3 - 2)(x - 9)\end{align*} and the lengths of the sies of the rectangular prism are \begin{align*}x\end{align*}, \begin{align*}3x^3 - 2\end{align*}, and \begin{align*}x - 9\end{align*}.
Guided Practice
Factor the following polynomials by grouping.
1. \begin{align*}x^3+7x^2-2x-14\end{align*}
2. \begin{align*}2x^4-5x^3+2x-5\end{align*}
3. Find all the real-number solutions of \begin{align*}4x^3-8x^2-x+2 = 0\end{align*}.
Answers
Each of these problems is done in the same way: Group the first two and last two terms together, pull out any common factors, what is inside the parenthesis is the same, factor it out, then determine if either factor can be factored further.
1. \begin{align*}& x^3+7x^2-2x-14\\ & x^2(x+7)-2(x+7)\\ & (x+7)(x^2-2)\end{align*}
\begin{align*}x^2-2\end{align*} is not a difference of squares because 2 is not a square number. Therefore, this cannot be factored further.
2. \begin{align*}& 2x^4-5x^3+2x-5\\ & x^3(2x-5)+1(2x-5)\\ & (2x-5)(x^3+1) \quad \ \ \text{Sum of cubes, factor further}.\\ & (2x-5)(x+1)(x^2+x+1)\end{align*}
3. Factor by grouping.
\begin{align*}4x^3-8x^2-x+2 &= 0\\ 4x^2(x-2)-1(x-2) &= 0\\ (x-2)(4x^2-1) &= 0\\ (x-2)(2x-1)(2x+1) &= 0\\ x & = 2, \frac{1}{2}, -\frac{1}{2}\end{align*}
Practice
Factor the following polynomials using factoring by grouping. Factor each polynomial completely.
- \begin{align*}x^3-4x^2+3x-12\end{align*}
- \begin{align*}x^3+6x^2-9x-54\end{align*}
- \begin{align*}3x^3-4x^2+15x-20\end{align*}
- \begin{align*} 2x^4-3x^3-16x+24\end{align*}
- \begin{align*}4x^3+4x^2-25x-25\end{align*}
- \begin{align*}4x^3+18x^2-10x-45\end{align*}
- \begin{align*}24x^4-40x^3+81x-135\end{align*}
- \begin{align*}15x^3+6x^2-10x-4\end{align*}
- \begin{align*}4x^3+5x^2-100x-125\end{align*}
- \begin{align*}3x^3-2x^2+12x-8\end{align*}
Find all the real-number solutions of the polynomials below.
- \begin{align*}9x^3-54x^2-4x+24 = 0\end{align*}
- \begin{align*}x^4+3x^3-27x-81 = 0\end{align*}
- \begin{align*}x^3-2x^2-4x+8 = 0\end{align*}
- Challenge Find ALL the solutions of \begin{align*}x^6-9x^4-x^2+9 = 0\end{align*}.
- Challenge Find ALL the solutions of \begin{align*}x^3+3x^2+16x+48 = 0\end{align*}.
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
---|---|---|---|
Please Sign In to create your own Highlights / Notes | |||
Show More |
Factor by Grouping
Factoring by grouping is a method of factoring a polynomial by factoring common monomials from groups of terms.Grouping Symbols
Grouping symbols are parentheses or brackets used to group numbers and operations.Volume
Volume is the amount of space inside the bounds of a three-dimensional object.Image Attributions
Here you'll learn how to factor and solve certain polynomials by grouping.