Find all real zeros of x^4-6x^3+7x^2-6x+6

Question

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Answer 1

x^4-6x^3+7x^2-6x+6

By the rational root therom

Since the leading coefficient is 1 and constant is 6.

So the possible real zeros are ± 1,±2,±3,±6

Let check the possbilities.

P(1) = (1)^4-6(1)^3+7(1)^2-6(1)+6 = 1-6+7-6+6 = 2

P(-1) = (-1)^4-6(-1)^3+7(-1)^2-6(-1)+6 = 1+6+7+6+6 = 26

P(2) = (2)^4-6(2)^3+7(2)^2-6(2)+6 = 16-48+28-12+6 = -10

P(-2) = (-2)^4-6(-2)^3+7(-2)^2-6(-2)+6 = 16+48+28+12+6 = 110

P(3) = (3)^4-6(3)^3+7(3)^2-6(3)+6 = 81-162+63-18+6 = -30

P(-3) = (-3)^4-6(-3)^3+7(-3)^2-6(-3)+6 = 81+162+63+18+6 = 330

P(6) = (6)^4-6(6)^3+7(6)^2-6(6)+6 = 1296-1296+252-36+6 = 222

P(-6) = (-6)^4-6(-6)^3+7(-6)^2-6(-6)+6 = 1296+1296+252+36+6 = 2878

There are no real zeros of given function.

 

Answer 2

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