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Algebraic expressions Simplifying expressions Equivalent expressions Linear equations Polynomials Quadratic graphsHere you will learn how to factor polynomials with a strategy called factor by grouping.
Students first learn how to factor by grouping in high school.
Factor by grouping is writing the polynomial as a product of its factors. It is the inverse process of multiplying algebraic expressions using the distributive property.
There are several strategies for factoring polynomials. This page will overview the strategy factor by grouping for polynomial equations.
For example,
Factor this four-term polynomial by grouping:
x^2+x+3x+3Group the first two terms together and the second two terms together.
Factor out the greatest common factor (GCF) in the first group and the second group.
Notice how both sets of parentheses are the same. When factoring by grouping, you want the binomials to be the same after factoring out the GCF.
Looking at the expression, the common binomial is (x+1) so like the GCF, it can be factored out.
x^2+x+3 x+3=(x+1)(x+3)How does this relate to high school math?
Use this worksheet to check your high school studentsβ understanding of factoring by grouping. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREEUse this worksheet to check your high school studentsβ understanding of factoring by grouping. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREEIn order to factor by grouping:
Factor the expression: x^2-8 x-2 x+16
Group the terms that have common factors. In this case, it is the first term with the second term and the third term with the fourth term. Use parentheses to show the groupings.
\left(x^2-8x\right)+(-2x+16)2Factor out the \textbf{GCF} from each binomial.
In x^2-8x , both terms have a factor of x, so x^2-8x=x(x-8).
In -2x+16, both terms have a factor of 2, so -2x+16=2(-x+8).
Note here that factoring out the positive coefficient creates the binomial -x+8 but as the binomial from the previous two terms is x-8 so factor the negative coefficient to match the binomials.
Factor -2 from -2x+16, so -2x+16=-2(x-8).
3Factor out the common binomial.
\begin{aligned}&x^2-8x-2x+16\\\\&=\left(x^2-8x\right)+(-2x+16)\\\\&=x(x-8)+-2(x-8)\\\\&=(x-8)(x-2)\end{aligned}Notice what happens when you use the commutative property before factoring a quadratic equation:
\begin{aligned}& x^2-8 x-2 x+16 \\\\& =x^2-2 x-8 x+16 \\\\& =x(x-2)-8(x-2) \\\\& =(x-8)(x-2)\end{aligned}This shows us that we can group the x^2 term with either x term and the other x term with the constant.
Factor the expression: 3x^2+6 x+12 x+24
Group the first two terms together and the last two terms together.
Factor out the \textbf{GCF} from each binomial.
The GCF of 3x^2+6x is 3x, so 3x^2+6 x=3 x(x+2).
The GCF of 12x+24 is 12, so 12x+24=12(x+2).
Factor out the common binomial.
Factor the expression: 2x^2-10 x+3 x-15
Group the first two terms together and the last two terms together.
Factor out the \textbf{GCF} from each binomial.
The GCF of 2x^2-10x is 2x, so 2x^2-10 x=2 x(x-5).
The GCF of 3x-15 is 3, so 3x-15=3(x-5).
Factor out the common binomial.
Factor the expression: x^2+\cfrac{1}{2} x+2x+1
Group the first two terms together and the last two terms together.
Factor out the \textbf{GCF} from each binomial.
In x^2+\cfrac{1}{2}x, both terms have a factor of x, so x^2+\cfrac{1}{2} x=x\left(x+\cfrac{1}{2}\right).
In 2x+1, both terms have a factor of 2, so 2x+1=2\left(x+\cfrac{1}{2}\right).
Note here that \cfrac{1}{2} is not the GCF. If the GCF does not lead to a common factor, you try other rational numbers.
Factor out the common binomial.
Factor the expression: x^2-7 x-30
Group the first two terms together and the last two terms together.
In order to use factor by grouping, you need to split the middle term. Split it in a way that creates common factors.
\begin{aligned}& x^2-7 x-30 \\\\& =x^2(-10 x+3 x)-30 \\\\& =\left(x^2-10 x\right)+(3 x-30)\end{aligned}
Factor out the \textbf{GCF} from each binomial.
In x^2-10x, both terms have a factor of x, so x^2-10x=x(x-10).
In 3x-30, both terms have a factor of 3, so 3x -30=3(x-10).
Factor out the common binomial.
Factor the expression: 4x^3-x^2+32x^2-8x
Group the first two terms together and the last two terms together.
Notice, the terms of the polynomial are different from a quadratic function. A cubic function is a polynomial with an exponent to the power of 3. Given four terms, you can still factor by grouping:
\left(4 x^3-x^2\right)+\left(32 x^2-8 x\right)
Factor out the \textbf{GCF} from each binomial.
In 4x^3-x^2, both terms have a factor of x^2, so 4x^3-x^2=x^2(4 x-1).
In 32x^2-8x, both terms have a factor of 8x, so 32x^2-8x=8x(4x-1).
Factor out the common binomial.
1. Which is the factored form of x^2-6x+7x-42?
Group the terms that have common factors. \left(x^2-6 x\right)+(7 x-42)
Factor out the GCF from each binomial.
In x^2-6 x, both terms have a factor of x, so x^2-6 x=x(x-6).
In 7x-42, both terms have a factor of 7, so 7x-42=7(x-6).
So,
\begin{aligned} & x^2-6 x+7 x-42 \\\\ & =\left(x^2-6 x\right)+(7 x-42) \\\\ & =x(x-6)+7(x-6) \\\\ & =(x+7)(x-6) \end{aligned}
2. Which is the factored form of 5x^2-45 x+3 x-27?
Group the terms that have common factors. \left(5x^2-45 x\right)+(3 x-27)
Factor out the GCF from each binomial.
In 5x^2-45 x, both terms have a factor of 5x, so 5x^2-45 x=5 x(x-9).
In 3x-27, both terms have a factor of 3, so 3x-27=3(x-9).
So,
\begin{aligned} & 5 x^2-45 x+3 x-27 \\\\ & =\left(5 x^2-45 x\right)+(3 x-27) \\\\ & =5 x(x-9)+3(x-9) \\\\ & =(5 x+3)(x-9) \end{aligned}
3. Which is the factored form of -2 x^2-4 x+4 x+8?
Group the terms that have common factors. \left(-2 x^2-4x\right)+(4 x+8)
Factor out the GCF from each binomial.
In -2x^2-4x, both terms have a factor of -2x, so -2x^2-4x=-2x(x+2).
In 4x+8, both terms have a factor of 4, so 4x+8=4(x+2).
So,
\begin{aligned} & -2 x^2-4 x+4 x+8 \\\\& =\left(-2 x^2-4 x\right)+(4 x+8) \\\\& =-2 x(x+2)+4(x+2) \\\\& =(-2 x+4)(x+2)\end{aligned}
4. Which is the factored form of x^2+36 x+\cfrac{1}{3} x+12?
Group the terms that have common factors. \left(x^2+36 x\right)+\left(\cfrac{1}{3} x+12\right)
Factor each binomial to find a common factor.
In x^2+36x, both terms have a factor of x, so x^2+36 x=x(x+36).
In \cfrac{1}{3} x+12, both terms have a factor of \cfrac{1}{3}, so \cfrac{1}{3} x+12=\cfrac{1}{3}(x+36).
So,
\begin{aligned} & x^2+36 x+\cfrac{1}{3} x+12 \\\\ & =\left(x^2+36 x\right)+\left(\cfrac{1}{3} x+12\right) & =x(x+36)+\cfrac{1}{3}(x+36) \\\\& =\left(x+\cfrac{1}{3}\right)(x+36) \end{aligned}
5. Which is the factored form of x^2+3 x-88?
In order to use factor by grouping, you need to split the middle term. Split it in a way that creates common factors.
\begin{aligned} & x^2+3 x-88 \\\\ & =x^2+(-8 x+11 x)-88 \\\\ & =\left(x^2-8 x\right)+(11 x-88) \end{aligned}
Factor out the GCF from each binomial.
In x^2-8x, both terms have a factor of x, so x^2-8x=x(x-8).
In 11x-88, both terms have a factor of 11, so 11x-88=11(x-8).
So,
\begin{aligned} & x^2+3 x-88 \\\\ & =x^2+(-8 x+11 x)-88 \\\\ & =\left(x^2-8 x\right)+(11 x-88) \\\\ & =x(x-8)+11(x-8) \\\\ & =(x+11)(x-8) \end{aligned}
6. Which is the factored form of x^3+10 x^2-3 x-30?
Group the terms that have common factors. \left(x^3+10 x^2\right)+(-3 x-30)
Factor out the GCF from each binomial.
In x^3+10 x^2, both terms have a factor of x^2, so x^3+10 x^2=x^2(x+10).
In 3x+30, both terms have a factor of 4, so 3x+30=3(x+10).
So,
\begin{aligned} & x^3+10 x^2-3 x-30 \\\\ & =\left(x^3+10 x^2\right)-(3 x-30) \\\\ & =x^2(x+10)-3(x+10) \\\\ & =\left(x^2-3\right)(x+10) \end{aligned}
This is typically covered in Algebra 1 or Integrated Math II.
Yes, for example 8x^2 can be shown as the factors 2x(4x), among other factors.
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