Question 189583


Looking at {{{x^4-7x^2-8}}} we can see that the first term is {{{x^4}}} and the last term is {{{-8}}} where the coefficients are 1 and -8 respectively.


Now multiply the first coefficient 1 and the last coefficient -8 to get -8. Now what two numbers multiply to -8 and add to the  middle coefficient -7? Let's list all of the factors of -8:




Factors of -8:

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 -8

(1)*(-8)

(2)*(-4)

(-1)*(8)

(-2)*(4)


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">-8</td><td>1+(-8)=-7</td></tr><tr><td align="center">2</td><td align="center">-4</td><td>2+(-4)=-2</td></tr><tr><td align="center">-1</td><td align="center">8</td><td>-1+8=7</td></tr><tr><td align="center">-2</td><td align="center">4</td><td>-2+4=2</td></tr></table>



From this list we can see that 1 and -8 add up to -7 and multiply to -8



Now looking at the expression {{{x^4-7x^2-8}}}, replace {{{-7x^2}}} with {{{x^2-8x^2}}} (notice {{{x^2-8x^2}}} combines to {{{-7x^2}}}. So it is equivalent to {{{-7x^2}}})


{{{x^4+highlight(x^2-8x^2)+-8}}}



Now let's factor {{{x^4+x^2-8x^2-8}}} by grouping:



{{{(x^4+x^2)+(-8x^2-8)}}} Group like terms



{{{x^2(x^2+1)-8(x^2+1)}}} Factor out the GCF of {{{x^2}}} out of the first group. Factor out the GCF of {{{-8}}} out of the second group



{{{(x^2-8)(x^2+1)}}} Since we have a common term of {{{x^2+1}}}, we can combine like terms


So {{{x^4+x^2-8x^2-8}}} factors to {{{(x^2-8)(x^2+1)}}}



So this also means that {{{x^4-7x^2-8}}} factors to {{{(x^2-8)(x^2+1)}}} (since {{{x^4-7x^2-8}}} is equivalent to {{{x^4+x^2-8x^2-8}}})




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     Answer:

So {{{x^4-7x^2-8}}} factors to {{{(x^2-8)(x^2+1)}}}