endiia 3 years ago factor completely: x^7-14x^6-51x^5

1. endiia

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2. ParthKohli

Find the common factor and put that outside the bracket.

3. pokemon23

If I'm correct

4. ParthKohli

You are correct. @pokemon23

5. endiia

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

6. optiquest

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

7. pokemon23

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!

8. ParthKohli

Oops, you got the 51x incorrect @pokemon23

9. ParthKohli

$$\Large \color{purple}{\rightarrow x^5(51x) = 51x^6 }$$

10. amir.sat

$x ^{5}\times(x-3)\times(x+17)$

11. endiia

so what's the full equation?

12. endiia

an amir. sat has something totally different

13. endiia

can someone1 clarify for me plz

14. ParthKohli

@endiia Amir is correct, he has actually grouped the quadratic expression.

15. amir.sat

x^2−14x−51=(x-3)*(x+17)

16. endiia

so when u factor it the answer is (x-3)(x+17)

17. endiia

so x is 3 & 17

18. ParthKohli

Yes, a better and cleared answer is: $$\LARGE \color{purple}{\rightarrow x^5[(x - 3)(x + 17)] }$$

19. amir.sat

sorry it's x^2−14x−51=(x+3)*(x-17) so answer is x^5×(x+3)×(x-17)

20. amir.sat

and x is 0 & -3 & 17