# How do you factor #4x^4-19x^3+16x^2-19x+12#?

##### 1 Answer

Use the rational root theorem, divide by the first factor found, then factor by grouping to find:

#4x^4-19x^3+16x^2-19x+12 = (4x-3)(x-4)(x^2+1)#

#### Explanation:

Let

By the rational root theorem, any rational roots of

Also, since the signs of the terms are alternating, all the Real roots of

That means that the only possible rational roots are:

#1/4# ,#1/2# ,#3/4# ,#1# ,#3/2# ,#2# ,#3# ,#4# ,#6# ,#12#

Let's try a few:

#f(1/4) = 4/256-19/64+16/16-19/4+12 = (1-19+64-304+768)/64 = 446/64#

#f(1/2) = 4/16-19/8+16/4-19/2+12 = (2-19+32-76+96)/8 = 35/8#

So

Divide by

#4x^4-19x^3+16x^2-19x+12 = (4x-3)(x^3-4x^2+x-4)#

The remaining cubic factor factors nicely by grouping:

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

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

So putting this together, we have:

#4x^4-19x^3+16x^2-19x+12 = (4x-3)(x-4)(x^2+1)#

The remaining quadratic factor

If you really want to factor it further:

#x^2+1 = (x-i)(x+i)#

where

Giving us:

#4x^4-19x^3+16x^2-19x+12 = (4x-3)(x-4)(x-i)(x+i)#