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vekshin1
3 years ago
12

Find the zeroes of 6x^4+x^3+2x^2-4x+1

Mathematics
1 answer:
Masja [62]3 years ago
8 0
By the rational root theorem, you have the following candidates for roots:

\pm\dfrac16,\pm\dfrac13,\pm\dfrac12

Plugging in each of these will tell you which one is actually a zero. You'll find that both x=\dfrac12 and x=\dfrac13 both work, which means x-\dfrac12 and x-\dfrac13 are linear factors to the quartic.

To find the remaining factor(s), divide the quartic by the known factors:

\dfrac{6x^4+x^3+2x^2-4x+1}{x-\dfrac12}=6x^3+4x^2+4x-2
\dfrac{6x^3+4x^2+4x-2}{x-\dfrac13}=6x^2+6x+6

Since 6x^2+6x+6=6(x^2+x+1)=0 has no real roots, you are left with

6x^4+x^3+2x^2-4x+1=\left(x-\dfrac12\right)\left(x-\dfrac13\right)(6x^2+6x+6)=(2x-1)(3x-1)(x^2+x+1)=0

which has two real zeros at x=\dfrac12, x=\dfrac13. It also has two complex roots at x=-\dfrac{1\pm\sqrt3i}2.
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Step-by-step explanation:

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