Question 200536

{{{a^2b+2ab-15b}}} Start with the given expression



{{{b(a^2+2a-15)}}} Factor out the GCF {{{b}}}



Now let's focus on the inner expression {{{a^2+2a-15}}}





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Looking at {{{a^2+2a-15}}} we can see that the first term is {{{a^2}}} and the last term is {{{-15}}} where the coefficients are 1 and -15 respectively.


Now multiply the first coefficient 1 and the last coefficient -15 to get -15. Now what two numbers multiply to -15 and add to the  middle coefficient 2? Let's list all of the factors of -15:




Factors of -15:

1,3,5,15


-1,-3,-5,-15 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -15

(1)*(-15)

(3)*(-5)

(-1)*(15)

(-3)*(5)


note: remember, the product of a negative and a positive number is a negative number



Now which of these pairs add to 2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 2


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-15</td><td>1+(-15)=-14</td></tr><tr><td align="center">3</td><td align="center">-5</td><td>3+(-5)=-2</td></tr><tr><td align="center">-1</td><td align="center">15</td><td>-1+15=14</td></tr><tr><td align="center">-3</td><td align="center">5</td><td>-3+5=2</td></tr></table>



From this list we can see that -3 and 5 add up to 2 and multiply to -15



Now looking at the expression {{{a^2+2a-15}}}, replace {{{2a}}} with {{{-3a+5a}}} (notice {{{-3a+5a}}} adds up to {{{2a}}}. So it is equivalent to {{{2a}}})


{{{a^2+highlight(-3a+5a)+-15}}}



Now let's factor {{{a^2-3a+5a-15}}} by grouping:



{{{(a^2-3a)+(5a-15)}}} Group like terms



{{{a(a-3)+5(a-3)}}} Factor out the GCF of {{{a}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(a+5)(a-3)}}} Since we have a common term of {{{a-3}}}, we can combine like terms


So {{{a^2-3a+5a-15}}} factors to {{{(a+5)(a-3)}}}



So this also means that {{{a^2+2a-15}}} factors to {{{(a+5)(a-3)}}} (since {{{a^2+2a-15}}} is equivalent to {{{a^2-3a+5a-15}}})




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So our expression goes from {{{b(a^2+2a-15)}}} and factors further to {{{b(a+5)(a-3)}}}



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Answer:


So {{{a^2b+2ab-15b}}} completely factors to {{{b(a+5)(a-3)}}}

    


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