```Question 540990
I'm assuming you're wondering how to factor {{{b^2-7b+10}}} to get {{{(b-5)(b-2)}}}

Looking at the expression {{{b^2-7b+10}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-7}}}, and the last term is {{{10}}}.

Now multiply the first coefficient {{{1}}} by the last term {{{10}}} to get {{{(1)(10)=10}}}.

Now the question is: what two whole numbers multiply to {{{10}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-7}}}?

To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{10}}} (the previous product).

Factors of {{{10}}}:

1,2,5,10

-1,-2,-5,-10

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

These factors pair up and multiply to {{{10}}}.

1*10 = 10
2*5 = 10
(-1)*(-10) = 10
(-2)*(-5) = 10

Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-7}}}:

<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>10</font></td><td  align="center"><font color=black>1+10=11</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>2+5=7</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>-1+(-10)=-11</font></td></tr><tr><td  align="center"><font color=red>-2</font></td><td  align="center"><font color=red>-5</font></td><td  align="center"><font color=red>-2+(-5)=-7</font></td></tr></table>

From the table, we can see that the two numbers {{{-2}}} and {{{-5}}} add to {{{-7}}} (the middle coefficient).

So the two numbers {{{-2}}} and {{{-5}}} both multiply to {{{10}}} <font size=4><b>and</b></font> add to {{{-7}}}

Now replace the middle term {{{-7b}}} with {{{-2b-5b}}}. Remember, {{{-2}}} and {{{-5}}} add to {{{-7}}}. So this shows us that {{{-2b-5b=-7b}}}.

{{{b^2+highlight(-2b-5b)+10}}} Replace the second term {{{-7b}}} with {{{-2b-5b}}}.

{{{(b^2-2b)+(-5b+10)}}} Group the terms into two pairs.

{{{b(b-2)+(-5b+10)}}} Factor out the GCF {{{b}}} from the first group.

{{{b(b-2)-5(b-2)}}} 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.

{{{(b-5)(b-2)}}} Combine like terms. Or factor out the common term {{{b-2}}}

===============================================================

So {{{b^2-7b+10}}} factors to {{{(b-5)(b-2)}}}.

In other words, {{{b^2-7b+10=(b-5)(b-2)}}}.

Note: you can check the answer by expanding {{{(b-5)(b-2)}}} to get {{{b^2-7b+10}}} 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>

Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you

Jim
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