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

gadzooks1753  Nov 28, 2017

Best Answer 

 #1
avatar+2266 
+1

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
 #1
avatar+2266 
+1
Best Answer

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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