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

##### 1 Answer

#### Answer:

#### Explanation:

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

First note that the sum of the coefficients is

That is:

#6-5-12+5+6=0#

Hence

#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

That is:

#-6+1+11-6 = 0#

Hence

#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

The pair

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)#