```Question 278244
Good job. Now let's factor {{{r^2+3r-40}}}

Looking at the expression {{{r^2+3r-40}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{3}}}, and the last term is {{{-40}}}.

Now multiply the first coefficient {{{1}}} by the last term {{{-40}}} to get {{{(1)(-40)=-40}}}.

Now the question is: what two whole numbers multiply to {{{-40}}} (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 {{{-40}}} (the previous product).

Factors of {{{-40}}}:

1,2,4,5,8,10,20,40

-1,-2,-4,-5,-8,-10,-20,-40

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

These factors pair up and multiply to {{{-40}}}.

1*(-40) = -40
2*(-20) = -40
4*(-10) = -40
5*(-8) = -40
(-1)*(40) = -40
(-2)*(20) = -40
(-4)*(10) = -40
(-5)*(8) = -40

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>-40</font></td><td  align="center"><font color=black>1+(-40)=-39</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-20</font></td><td  align="center"><font color=black>2+(-20)=-18</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>4+(-10)=-6</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>-8</font></td><td  align="center"><font color=black>5+(-8)=-3</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>40</font></td><td  align="center"><font color=black>-1+40=39</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>20</font></td><td  align="center"><font color=black>-2+20=18</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>-4+10=6</font></td></tr><tr><td  align="center"><font color=red>-5</font></td><td  align="center"><font color=red>8</font></td><td  align="center"><font color=red>-5+8=3</font></td></tr></table>

From the table, we can see that the two numbers {{{-5}}} and {{{8}}} add to {{{3}}} (the middle coefficient).

So the two numbers {{{-5}}} and {{{8}}} both multiply to {{{-40}}} <font size=4><b>and</b></font> add to {{{3}}}

Now replace the middle term {{{3r}}} with {{{-5r+8r}}}. Remember, {{{-5}}} and {{{8}}} add to {{{3}}}. So this shows us that {{{-5r+8r=3r}}}.

{{{r^2+highlight(-5r+8r)-40}}} Replace the second term {{{3r}}} with {{{-5r+8r}}}.

{{{(r^2-5r)+(8r-40)}}} Group the terms into two pairs.

{{{r(r-5)+(8r-40)}}} Factor out the GCF {{{r}}} from the first group.

{{{r(r-5)+8(r-5)}}} Factor out {{{8}}} 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.

{{{(r+8)(r-5)}}} Combine like terms. Or factor out the common term {{{r-5}}}

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Answer:

So {{{2r^2+6r-80}}} completely factors to {{{2(r+8)(r-5)}}}.

In other words, {{{2r^2+6r-80=2(r+8)(r-5)}}}.

I'll let you continue on to solve for 'r'.```