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factor completely: x^7-14x^6-51x^5

Mathematics
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Find the common factor and put that outside the bracket.
If I'm correct

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Other answers:

You are correct. @pokemon23
x^5(x^2-14x-51x) ?
\[x^5(x^2-14x-51)\]
thanks parth nice to meet you I'ma long time user but I departed OS for three months and I'm read to rock and roll!
Oops, you got the 51x incorrect @pokemon23
\(\Large \color{purple}{\rightarrow x^5(51x) = 51x^6 }\)
\[x ^{5}\times(x-3)\times(x+17)\]
so what's the full equation?
an amir. sat has something totally different
can someone1 clarify for me plz
@endiia Amir is correct, he has actually grouped the quadratic expression.
x^2−14x−51=(x-3)*(x+17)
so when u factor it the answer is (x-3)(x+17)
so x is 3 & 17
Yes, a better and cleared answer is: \(\LARGE \color{purple}{\rightarrow x^5[(x - 3)(x + 17)] }\)
sorry it's x^2−14x−51=(x+3)*(x-17) so answer is x^5×(x+3)×(x-17)
and x is 0 & -3 & 17

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