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Chapter 6: Problem 76

Factor by grouping. $$x^{3}+6 x^{2}-2 x-12$$

Short Answer

Expert verified
The factored form of the polynomial \(x^{3}+6 x^{2}-2 x-12\) by grouping is \((x^{2}-2)(x+6)\)

Step by step solution

01

Group the terms

Group the first two terms together and the last two terms together in the given polynomial: \\((x^{3}+6 x^{2})+(-2 x-12)\)
02

Factor out the GCF from each group

Factor out the GCF from each group. In the first group, \(x^{2}\) can be factored out, and in the second group, \(2\) can be factored out: \\(x^{2}(x+6) - 2(x+6)\)
03

Check if the results from step 2 are the same

The expressions in the parentheses are both \(x+6\). So, \(x+6\) is the common binomial and it can be factored out.
04

Factor out the binomial

Factor out the common binomial \(x+6\) to write the original expression in its factored form: \ ((x^{2}-2)(x+6))

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Most popular questions from this chapter

Use the \([\mathrm{GRAPH}]\) or \([\mathrm { TABLE }]\) feature of a graphing utility to determine if the polynomial on the left side of each equation has been correctly factored. If not, factor the polynomial correctly and then use your graphing utility to verify the factorization. $$4 x^{2}-12 x+9=(4 x-3)^{2} ;[-5,5,1] \text { by }[0,20,1]$$

Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$16 a^{2}-32 a b+12 b^{2}$$

Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$4 a y^{3}-12 a y^{2}+9 a y$$

Factor completely. $$4 x^{4}-9 x^{2}+5$$

Factor completely. $$(x-7)^{2}-4 a^{2}$$
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