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Find the common factor and put that outside the bracket.

If I'm correct

You are correct. @pokemon23

x^5(x^2-14x-51x) ?

\[x^5(x^2-14x-51)\]

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