How do you find the zeroes of $f(x)=6x^4-5x^3-12x^2+5x+6$ ?

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Answer 1 · 2016-06-24T17:51:00

$6x^4-5x^3-12x^2+5x+6 = (x-1)(x+1)(3x+2)(2x-3)$

$f(x) = 6x^4-5x^3-12x^2+5x+6$

First note that the sum of the coefficients is $0$ .

That is:

$6-5-12+5+6=0$

Hence $x=1$ is a zero and $(x-1)$ a factor:

$6x^4-5x^3-12x^2+5x+6 = (x-1)(6x^3+x^2-11x-6)$

If you reverse the signs of the terms of odd degree on the remaining cubic factor, then the sum is $0$ .

That is:

$-6+1+11-6 = 0$

Hence $x=-1$ is a zero and $(x+1)$ a factor:

$6x^3+x^2-11x-6=(x+1)(6x^2-5x-6)$

To factor the remaining quadratic expression, use an AC method:

Find a pair of factors of $AC = 6*6 = 36$ which differ by $B=5$ .

The pair $9, 4$ works.

Use this pair to split the middle term and factor by grouping:

$6x^2-5x-6$

$=6x^2-9x+4x-6$

$=(6x^2-9x)+(4x-6)$

$=3x(2x-3)+2(2x-3)$

$=(3x+2)(2x-3)$

Putting it all together:

$6x^4-5x^3-12x^2+5x+6 = (x-1)(x+1)(3x+2)(2x-3)$

How do you find the zeroes of f(x)=6x^4-5x^3-12x^2+5x+6 f(x)=6x^4-5x^3-12x^2+5x+6?