```Question 527931

{{{3x(y^2-2y-15)}}} Factor out the GCF {{{3x}}}.

Now let's try to factor the inner expression {{{y^2-2y-15}}}

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Looking at the expression {{{y^2-2y-15}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-2}}}, and the last term is {{{-15}}}.

Now multiply the first coefficient {{{1}}} by the last term {{{-15}}} to get {{{(1)(-15)=-15}}}.

Now the question is: what two whole numbers multiply to {{{-15}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-2}}}?

To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-15}}} (the previous product).

Factors of {{{-15}}}:

1,3,5,15

-1,-3,-5,-15

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

These factors pair up and multiply to {{{-15}}}.

1*(-15) = -15
3*(-5) = -15
(-1)*(15) = -15
(-3)*(5) = -15

Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-2}}}:

<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>-15</font></td><td  align="center"><font color=black>1+(-15)=-14</font></td></tr><tr><td  align="center"><font color=red>3</font></td><td  align="center"><font color=red>-5</font></td><td  align="center"><font color=red>3+(-5)=-2</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>15</font></td><td  align="center"><font color=black>-1+15=14</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>-3+5=2</font></td></tr></table>

From the table, we can see that the two numbers {{{3}}} and {{{-5}}} add to {{{-2}}} (the middle coefficient).

So the two numbers {{{3}}} and {{{-5}}} both multiply to {{{-15}}} <font size=4><b>and</b></font> add to {{{-2}}}

Now replace the middle term {{{-2y}}} with {{{3y-5y}}}. Remember, {{{3}}} and {{{-5}}} add to {{{-2}}}. So this shows us that {{{3y-5y=-2y}}}.

{{{y^2+highlight(3y-5y)-15}}} Replace the second term {{{-2y}}} with {{{3y-5y}}}.

{{{(y^2+3y)+(-5y-15)}}} Group the terms into two pairs.

{{{y(y+3)+(-5y-15)}}} Factor out the GCF {{{y}}} from the first group.

{{{y(y+3)-5(y+3)}}} Factor out {{{5}}} 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.

{{{(y-5)(y+3)}}} Combine like terms. Or factor out the common term {{{y+3}}}

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So {{{3x(y^2-2y-15)}}} then factors further to {{{3x(y-5)(y+3)}}}

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So {{{3xy^2-6xy-45x}}} completely factors to {{{3x(y-5)(y+3)}}}.

In other words, {{{3xy^2-6xy-45x=3x(y-5)(y+3)}}}.

Note: you can check the answer by expanding {{{3x(y-5)(y+3)}}} to get {{{3xy^2-6xy-45x}}} or by graphing the original expression and the answer (the two graphs should be identical).

If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com">jim_thompson5910@hotmail.com</a>

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Jim```