# Polynomial with c value

### Polynomial with c value

So i was given this polynomial with its c value, this is from precalculas!

The question is "you are given a polynomial and one of its zeros. Use the techniques in this section to ﬁnd the rest of the real zeros and factor the polynomial".

But i've never done that using a c value with it. So i don't really know how i should use it.

$$x^{3}-24^{2}+192x-512$$
while c = 8
mrlowbot

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### Re: Polynomial with c value

$$x^{3}$$-24$$x^{2}$$+192x-512=0

x=

$$x^{3}$$-3$$x^{2}$$8+3x$$8^{2}$$-$$8^{3}$$=0 [ formula ]

$$(x-8)^{3}$$=0

x-8=0

x=8 only one root
Guest

### Re: Polynomial with c value

Equivalently, if x= 8 is a root of $$x^3- 24x^2+ 192x- 512= 0$$ then x- 8 evenly divides $$x^3- 24x^2+ 192x- 512$$. x- 8 divides into $$x^3- 24x^2$$ $$x^2$$ times with remainder $$-16x^2+192x- 512$$. And x- 8 divides into that -16x times with remainder $$16x^2+ 192x- 512- 16x(x- 8 )= 64x- 512$$. x- 8 divides into that 64 times with remainder $$64x- 512- 64(x- 8 )= 0$$. That is, $$x^3- 24x^2+ 192x- 512= (x- 8 )(x^2- 16x+ 64)$$.

You can then either recognize $$x^2- 16x+ 64$$ as $$(x- 8 )^2$$ or, using the "rational root theorem" see that any rational root must be an integer factor of 64. Trying $$\pm 1$$, $$\pm 2$$, and $$\pm 4$$ does not work but $$8^2- 16(8 )+ 64= 64- 128+ 64= 0$$ so x= 8 is a root of $$x^2- 16x+ 64$$. Then x- 8 divides into $$x^2- 16x$$ x times with remainder $$x^2- 16x+ 64- x(x- 8 )= x^2- 16x+ 64- x^2+ 8x= -8x- 64$$ and $$x- 8$$ divides into that $$8$$ times. $$x^2- 16x+ 64= (x- 8 )(x- 8 )= (x- 8 )^2$$

Yes, $$x^3- 24x+ 192x- 512= (x- 8 )^3$$ so x= 8 is the only root of $$x^3- 24x^2+ 192x- 512= 0$$.
Guest