# How do you find all the zeros of #f(x)=12x^4+14x^3-8x^2-14x-4#?

##### 1 Answer

Use some observations of sums of coefficients, division and AC Method factoring to find zeros:

#### Explanation:

First note that the sum of the coefficients of

#12+14-8-14-4 = 0#

So

Also all of the terms are divisible by

#12x^4+14x^3-8x^2-14x-4#

#=2(6x^4+7x^3-4x^2-7x-2)#

#=2(x-1)(6x^3+13x^2+9x+2)#

The remaining cubic factor has all positive coefficients, so no zeros for positive values of

#-6+13-9+2 = 0#

So

#=2(x-1)(x+1)(6x^2+7x+2)#

The remaining quadratic factor can be factored using an AC method: Find a pair of factors of

#=2(x-1)(x+1)(6x^2+3x+4x+2)#

#=2(x-1)(x+1)(3x(2x+1)+2(2x+1))#

#=2(x-1)(x+1)(3x+2)(2x+1)#

So the zeros of