Question

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

Answer 1

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

By the rational root theorem, any rational roots of `f(x) = 0` will be expressible in lowest terms as `p/q`, where `p, q in ZZ`, `q != 0`, `p` a divisor of the constant term `12` and `q` a divisor of the coefficient `4` of the leading term.

Also, since the signs of the terms are alternating, all the Real roots of `f(x) = 0` are positive.

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`

`f(3/4) = 81/64 - 513/64 + 9 - 57/4 + 12 = (81-513+576-912+768)/64 = 0`

So `x = 3/4` is a root and `(4x-3)` is a factor.

Divide by `(4x-3)` to find:

`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 `x^2 + 1` has no simpler linear factors with Real coefficients since `x^2 + 1 >= 1 > 0` for all `x in RR`.

If you really want to factor it further:

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

where `i` is the imaginary unit.

Giving us:

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