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Factor the polynomial completely.
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\[7x ^{3}-4x ^{2}+8x\]
how did you get that? and aren't you supposed to find the 2 numbers after factoring (using the table or box method)
ok
\[7x^3-4x^2+8x=x(7x^2-4x+8)\] 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 \[7x^3-4x^2+8x=x(x- \frac27 - \frac{\sqrt{52}}7 i)(x- \frac27 + \frac{\sqrt{52}}7 i)\]
Which is quite ugly. Stick to the real numbers. It is neater
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so what would be the final answer?
Have you worked with complex factorising before?
If not, use the first line of my answer
ok. and no i'm in algebra 2. thanks!
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