Question

Question #ec66c

Answer 1

On `R` you can't.

Explanation:

If you can write it as product of 1 degree factors it should be
`P(x)=a(x-x_1)(x-x_2)`
where `x_1` and `x_2` are roots of `P(x)`, but if you try to solve the second degrre equation you will find
`Delta/4=8^2-3*24=-8<0`
so there are no roots on `R`.

Answer 2

Complete the square or use the quadratic formula to find the roots, then find the factors.

Explanation:

`3x^2-16x+24 = 0` has solutions

`x = (8+-2sqrt2i)/3`

So

`x-(8+2sqrt2i)/3`, and `x-(8-2sqrt2i)/3` are factors of the quadratic.

In order to get leading coefficient `3`, we need a constant factor of `3`.

So

`3x^2-16x+24 = 3(x-(8+2sqrt2i)/3)(x-(8-2sqrt2i)/3)`

If you want strict standard for for the imaginaries, write

`3x^2-16x+24 = 3(x-(8/3+(2sqrt2)/3i))(x-(8/3-(2sqrt2)/3i))`