Question 125884


Looking at {{{1x^2+6x-16}}} we can see that the first term is {{{1x^2}}} and the last term is {{{-16}}} where the coefficients are 1 and -16 respectively.


Now multiply the first coefficient 1 and the last coefficient -16 to get -16. Now what two numbers multiply to -16 and add to the  middle coefficient 6? Let's list all of the factors of -16:




Factors of -16:

1,2,4,8


-1,-2,-4,-8 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -16

(1)*(-16)

(2)*(-8)

(-1)*(16)

(-2)*(8)


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



Now which of these pairs add to 6? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 6


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-16</td><td>1+(-16)=-15</td></tr><tr><td align="center">2</td><td align="center">-8</td><td>2+(-8)=-6</td></tr><tr><td align="center">-1</td><td align="center">16</td><td>-1+16=15</td></tr><tr><td align="center">-2</td><td align="center">8</td><td>-2+8=6</td></tr></table>



From this list we can see that -2 and 8 add up to 6 and multiply to -16



Now looking at the expression {{{1x^2+6x-16}}}, replace {{{6x}}} with {{{-2x+8x}}} (notice {{{-2x+8x}}} adds up to {{{6x}}}. So it is equivalent to {{{6x}}})


{{{1x^2+highlight(-2x+8x)+-16}}}



Now let's factor {{{1x^2-2x+8x-16}}} by grouping:



{{{(1x^2-2x)+(8x-16)}}} Group like terms



{{{x(x-2)+8(x-2)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{8}}} out of the second group



{{{(x+8)(x-2)}}} Since we have a common term of {{{x-2}}}, we can combine like terms


So {{{1x^2-2x+8x-16}}} factors to {{{(x+8)(x-2)}}}



So this also means that {{{1x^2+6x-16}}} factors to {{{(x+8)(x-2)}}} (since {{{1x^2+6x-16}}} is equivalent to {{{1x^2-2x+8x-16}}})




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     Answer:

So {{{x^2+6x-16}}} factors to {{{(x+8)(x-2)}}}