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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!
Sure
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