Question 368253
{{{16(3x+2)^3+12(3x+2)^2-88(3x+2)}}} Start with the given expression.



Let {{{z=3x+2}}}



{{{16z^3+12z^2-88z}}} Replace each {{{3x+2}}} term with 'z'




Looking at the expression {{{4z^2+3z-22}}}, we can see that the first coefficient is {{{4}}}, the second coefficient is {{{3}}}, and the last term is {{{-22}}}.



Now multiply the first coefficient {{{4}}} by the last term {{{-22}}} to get {{{(4)(-22)=-88}}}.



Now the question is: what two whole numbers multiply to {{{-88}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{3}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-88}}} (the previous product).



Factors of {{{-88}}}:

1,2,4,8,11,22,44,88

-1,-2,-4,-8,-11,-22,-44,-88



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-88}}}.

1*(-88) = -88
2*(-44) = -88
4*(-22) = -88
8*(-11) = -88
(-1)*(88) = -88
(-2)*(44) = -88
(-4)*(22) = -88
(-8)*(11) = -88


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{3}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>-88</font></td><td  align="center"><font color=black>1+(-88)=-87</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-44</font></td><td  align="center"><font color=black>2+(-44)=-42</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>-22</font></td><td  align="center"><font color=black>4+(-22)=-18</font></td></tr><tr><td  align="center"><font color=black>8</font></td><td  align="center"><font color=black>-11</font></td><td  align="center"><font color=black>8+(-11)=-3</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>88</font></td><td  align="center"><font color=black>-1+88=87</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>44</font></td><td  align="center"><font color=black>-2+44=42</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>22</font></td><td  align="center"><font color=black>-4+22=18</font></td></tr><tr><td  align="center"><font color=red>-8</font></td><td  align="center"><font color=red>11</font></td><td  align="center"><font color=red>-8+11=3</font></td></tr></table>



From the table, we can see that the two numbers {{{-8}}} and {{{11}}} add to {{{3}}} (the middle coefficient).



So the two numbers {{{-8}}} and {{{11}}} both multiply to {{{-88}}} <font size=4><b>and</b></font> add to {{{3}}}



Now replace the middle term {{{3z}}} with {{{-8z+11z}}}. Remember, {{{-8}}} and {{{11}}} add to {{{3}}}. So this shows us that {{{-8z+11z=3z}}}.



{{{4z^2+highlight(-8z+11z)-22}}} Replace the second term {{{3z}}} with {{{-8z+11z}}}.



{{{(4z^2-8z)+(11z-22)}}} Group the terms into two pairs.



{{{4z(z-2)+(11z-22)}}} Factor out the GCF {{{4z}}} from the first group.



{{{4z(z-2)+11(z-2)}}} Factor out {{{11}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(4z+11)(z-2)}}} Combine like terms. Or factor out the common term {{{z-2}}}



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So {{{4z(4z^2+3z-22)}}} then factors further to {{{4z(4z+11)(z-2)}}}




So {{{16z^3+12z^2-88z}}} completely factors to {{{4z(4z+11)(z-2)}}}.



In other words, {{{16z^3+12z^2-88z=4z(4z+11)(z-2)}}}.



Now plug in {{{z=3x+2}}} to get 



{{{16(3x+2)^3+12(3x+2)^2-88(3x+2)=4(3x+2)(4(3x+2)+11)(3x+2-2)}}}



{{{16(3x+2)^3+12(3x+2)^2-88(3x+2)=4(3x+2)(12x+8+11)(3x+2-2)}}} Distribute.



{{{16(3x+2)^3+12(3x+2)^2-88(3x+2)=4(3x+2)(12x+19)(3x)}}} Combine like terms.



{{{16(3x+2)^3+12(3x+2)^2-88(3x+2)=4(3x)(3x+2)(12x+19)}}} Rearrange the terms.



{{{16(3x+2)^3+12(3x+2)^2-88(3x+2)=12x(3x+2)(12x+19)}}} Multiply



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Jim