Question 173192


{{{4x^3+4x^2-80x}}} Start with the given expression



{{{4x(x^2+x-20)}}} Factor out the GCF {{{4x}}}



Now let's focus on the inner expression {{{x^2+x-20}}}





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Looking at {{{1x^2+1x-20}}} we can see that the first term is {{{1x^2}}} and the last term is {{{-20}}} where the coefficients are 1 and -20 respectively.


Now multiply the first coefficient 1 and the last coefficient -20 to get -20. Now what two numbers multiply to -20 and add to the  middle coefficient 1? Let's list all of the factors of -20:




Factors of -20:

1,2,4,5,10,20


-1,-2,-4,-5,-10,-20 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -20

(1)*(-20)

(2)*(-10)

(4)*(-5)

(-1)*(20)

(-2)*(10)

(-4)*(5)


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



Now which of these pairs add to 1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 1


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-20</td><td>1+(-20)=-19</td></tr><tr><td align="center">2</td><td align="center">-10</td><td>2+(-10)=-8</td></tr><tr><td align="center">4</td><td align="center">-5</td><td>4+(-5)=-1</td></tr><tr><td align="center">-1</td><td align="center">20</td><td>-1+20=19</td></tr><tr><td align="center">-2</td><td align="center">10</td><td>-2+10=8</td></tr><tr><td align="center">-4</td><td align="center">5</td><td>-4+5=1</td></tr></table>



From this list we can see that -4 and 5 add up to 1 and multiply to -20



Now looking at the expression {{{1x^2+1x-20}}}, replace {{{1x}}} with {{{-4x+5x}}} (notice {{{-4x+5x}}} adds up to {{{1x}}}. So it is equivalent to {{{1x}}})


{{{1x^2+highlight(-4x+5x)+-20}}}



Now let's factor {{{1x^2-4x+5x-20}}} by grouping:



{{{(1x^2-4x)+(5x-20)}}} Group like terms



{{{x(x-4)+5(x-4)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(x+5)(x-4)}}} Since we have a common term of {{{x-4}}}, we can combine like terms


So {{{1x^2-4x+5x-20}}} factors to {{{(x+5)(x-4)}}}



So this also means that {{{1x^2+1x-20}}} factors to {{{(x+5)(x-4)}}} (since {{{1x^2+1x-20}}} is equivalent to {{{1x^2-4x+5x-20}}})




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So our expression goes from {{{4x(x^2+x-20)}}} and factors further to {{{4x(x+5)(x-4)}}}



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Answer:


So {{{4x^3+4x^2-80x}}} factors to {{{4x(x+5)(x-4)}}}