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OpenStudy (anonymous):

Factor the polynomial completely.

OpenStudy (anonymous):

\[7x ^{3}-4x ^{2}+8x\]

OpenStudy (anonymous):

how did you get that? and aren't you supposed to find the 2 numbers after factoring (using the table or box method)

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

\[7x^3-4x^2+8x=x(7x^2-4x+8)\] 7x^2-4x+8 has no real roots so, to continue factorising, we must use complex numbers. We solve 7x^2-4x+8 using the quadratic formula to get x = 2/7 +- root(52)/7 i Thus \[7x^3-4x^2+8x=x(x- \frac27 - \frac{\sqrt{52}}7 i)(x- \frac27 + \frac{\sqrt{52}}7 i)\]

OpenStudy (anonymous):

Which is quite ugly. Stick to the real numbers. It is neater

OpenStudy (anonymous):

so what would be the final answer?

OpenStudy (anonymous):

Have you worked with complex factorising before?

OpenStudy (anonymous):

If not, use the first line of my answer

OpenStudy (anonymous):

ok. and no i'm in algebra 2. thanks!

OpenStudy (anonymous):

Sure

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