Polynomial with c value

Linear, quadratic, module, parametric equations

Polynomial with c value

Postby mrlowbot » Tue Oct 16, 2018 6:09 am

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.

[tex]x^{3}-24^{2}+192x-512[/tex]
while c = 8
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Re: Polynomial with c value

Postby Guest » Tue Oct 16, 2018 8:53 pm

[tex]x^{3}[/tex]-24[tex]x^{2}[/tex]+192x-512=0

x= :?:

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

[tex](x-8)^{3}[/tex]=0

x-8=0

x=8 only one root
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Re: Polynomial with c value

Postby Guest » Sun Jun 09, 2019 8:39 am

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

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

Yes, [tex]x^3- 24x+ 192x- 512= (x- 8 )^3[/tex] so x= 8 is the only root of [tex]x^3- 24x^2+ 192x- 512= 0[/tex].
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