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# factor by grouping

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8x^4+8x^3+27x+27

#1
+1829
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Just like in the previous problem you posted, factor out the GCF of each term in the parentheses.

 $$(8x^4+8x^3)+(27x+27)$$ Now, factor out the GCF. $$8x^3(x+1)+27(x+1)$$ (x+1) is common to both factors, so we can rewrite this. $$(8x^3+27)(x+1)$$ (8x^3+27) happens to be a sum of cubes and must be dealth with as such. $$a=\sqrt[3]{8x^3}=\sqrt[3]{(2x)^3}=2x\\ b=\sqrt[3]{27}=3$$ Now that a and b have been identified expand the sum of cubes. $$(2x+3)((2x)^2-(3)(2x)+3^2)(x+1)$$ Now, simplify. $$(2x+3)(4x^2-6x+9)(x+1)$$ 4x^2-6x+9 is an irreducible trinomial and nothing else can be reduced further.
TheXSquaredFactor  Nov 28, 2017
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#1
+1829
+1
 $$(8x^4+8x^3)+(27x+27)$$ Now, factor out the GCF. $$8x^3(x+1)+27(x+1)$$ (x+1) is common to both factors, so we can rewrite this. $$(8x^3+27)(x+1)$$ (8x^3+27) happens to be a sum of cubes and must be dealth with as such. $$a=\sqrt[3]{8x^3}=\sqrt[3]{(2x)^3}=2x\\ b=\sqrt[3]{27}=3$$ Now that a and b have been identified expand the sum of cubes. $$(2x+3)((2x)^2-(3)(2x)+3^2)(x+1)$$ Now, simplify. $$(2x+3)(4x^2-6x+9)(x+1)$$ 4x^2-6x+9 is an irreducible trinomial and nothing else can be reduced further.