Question 171356


{{{20q^2+28q-24}}} Start with the given expression



{{{4(5q^2+7q-6)}}} Factor out the GCF {{{4}}}



Now let's focus on the inner expression {{{5q^2+7q-6}}}





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Looking at {{{5q^2+7q-6}}} we can see that the first term is {{{5q^2}}} and the last term is {{{-6}}} where the coefficients are 5 and -6 respectively.


Now multiply the first coefficient 5 and the last coefficient -6 to get -30. Now what two numbers multiply to -30 and add to the  middle coefficient 7? Let's list all of the factors of -30:




Factors of -30:

1,2,3,5,6,10,15,30


-1,-2,-3,-5,-6,-10,-15,-30 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -30

(1)*(-30)

(2)*(-15)

(3)*(-10)

(5)*(-6)

(-1)*(30)

(-2)*(15)

(-3)*(10)

(-5)*(6)


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



Now which of these pairs add to 7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 7


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-30</td><td>1+(-30)=-29</td></tr><tr><td align="center">2</td><td align="center">-15</td><td>2+(-15)=-13</td></tr><tr><td align="center">3</td><td align="center">-10</td><td>3+(-10)=-7</td></tr><tr><td align="center">5</td><td align="center">-6</td><td>5+(-6)=-1</td></tr><tr><td align="center">-1</td><td align="center">30</td><td>-1+30=29</td></tr><tr><td align="center">-2</td><td align="center">15</td><td>-2+15=13</td></tr><tr><td align="center">-3</td><td align="center">10</td><td>-3+10=7</td></tr><tr><td align="center">-5</td><td align="center">6</td><td>-5+6=1</td></tr></table>



From this list we can see that -3 and 10 add up to 7 and multiply to -30



Now looking at the expression {{{5q^2+7q-6}}}, replace {{{7q}}} with {{{-3q+10q}}} (notice {{{-3q+10q}}} adds up to {{{7q}}}. So it is equivalent to {{{7q}}})


{{{5q^2+highlight(-3q+10q)+-6}}}



Now let's factor {{{5q^2-3q+10q-6}}} by grouping:



{{{(5q^2-3q)+(10q-6)}}} Group like terms



{{{q(5q-3)+2(5q-3)}}} Factor out the GCF of {{{q}}} out of the first group. Factor out the GCF of {{{2}}} out of the second group



{{{(q+2)(5q-3)}}} Since we have a common term of {{{5q-3}}}, we can combine like terms


So {{{5q^2-3q+10q-6}}} factors to {{{(q+2)(5q-3)}}}



So this also means that {{{5q^2+7q-6}}} factors to {{{(q+2)(5q-3)}}} (since {{{5q^2+7q-6}}} is equivalent to {{{5q^2-3q+10q-6}}})




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So our expression goes from {{{4(5q^2+7q-6)}}} and factors further to {{{4(q+2)(5q-3)}}}



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Answer:


So {{{20q^2+28q-24}}} factors to {{{4(q+2)(5q-3)}}}