Question 133352
Do you want to factor?


# 1


{{{64x^3+343}}} Start with the given expression.



{{{(4x)^3+(7)^3}}} Rewrite {{{64x^3}}} as {{{(4x)^3}}}. Rewrite {{{343}}} as {{{(7)^3}}}.



{{{(4x+7)((4x)^2-(4x)(7)+(7)^2)}}} Now factor by using the sum of cubes formula. Remember the <a href="http://www.purplemath.com/modules/specfact2.htm">sum of cubes formula</a> is {{{A^3+B^3=(A+B)(A^2-AB+B^2)}}}



{{{(4x+7)(16x^2-28x+49)}}} Multiply


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Answer:


So {{{64x^3+343}}} factors to {{{(4x+7)(16x^2-28x+49)}}}.


In other words, {{{64x^3+343=(4x+7)(16x^2-28x+49)}}}




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# 2



Looking at {{{1x^4+8x^2-9}}} we can see that the first term is {{{1x^4}}} and the last term is {{{-9}}} where the coefficients are 1 and -9 respectively.


Now multiply the first coefficient 1 and the last coefficient -9 to get -9. Now what two numbers multiply to -9 and add to the  middle coefficient 8? Let's list all of the factors of -9:




Factors of -9:

1,3


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


These factors pair up and multiply to -9

(1)*(-9)

(-1)*(9)


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



Now which of these pairs add to 8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 8


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-9</td><td>1+(-9)=-8</td></tr><tr><td align="center">-1</td><td align="center">9</td><td>-1+9=8</td></tr></table>



From this list we can see that -1 and 9 add up to 8 and multiply to -9



Now looking at the expression {{{1x^4+8x^2-9}}}, replace {{{8x^2}}} with {{{-1x^2+9x^2}}} (notice {{{-1x^2+9x^2}}} adds up to {{{8x^2}}}. So it is equivalent to {{{8x^2}}})


{{{1x^4+highlight(-1x^2+9x^2)+-9}}}



Now let's factor {{{1x^4-1x^2+9x^2-9}}} by grouping:



{{{(1x^4-1x^2)+(9x^2-9)}}} Group like terms



{{{x^2(x^2-1)+9(x^2-1)}}} Factor out the GCF of {{{x^2}}} out of the first group. Factor out the GCF of {{{9}}} out of the second group



{{{(x^2+9)(x^2-1)}}} Since we have a common term of {{{x^2-1}}}, we can combine like terms


So {{{1x^4-1x^2+9x^2-9}}} factors to {{{(x^2+9)(x^2-1)}}}



So this also means that {{{1x^4+8x^2-9}}} factors to {{{(x^2+9)(x^2-1)}}} (since {{{1x^4+8x^2-9}}} is equivalent to {{{1x^4-1x^2+9x^2-9}}})






So {{{x^4+8x^2-9}}} factors to {{{(x^2+9)(x^2-1)}}}



{{{(x^2+9)(x+1)(x-1)}}} Now factor {{{x^2-1}}} by using the difference of squares

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Answer:



So {{{x^4+8x^2-9}}} factors to {{{(x^2+9)(x+1)(x-1)}}}