Factor the polynomial completely. A zero has been provided

f(x) = 48x^3 - 80x^2 + 41x - 6 = 0
Zero: x = 2/3

Any help would be greatly appreciated! I have a test tomorrow and do not understand this content :(

Since you've been given a zero, (x - 2/3) or (3x - 2) is a factor of f(x). Divide f(x)/(3x-2) using your favourite method (say, long division or synthetic division) to get:

f(x) = (3x-2)(16x² - 16x + 3)

Now factorise the quadratic using your favourite method.

Factor theorem tells us that, because x = 2/3 is a root of f(x), x - 2/3 will be a factor of f(x). We could simply divide out x - 2/3, but let's divide out 3x - 2 instead. It's simply 3 times x - 2/3, and we get to work with integers. Dividing:

....................... 16x^2 - 16x + 3
........ ____________________
3x - 2) 48x^3 - 80x^2 + 41x - 6
.......... 48x^3 - 32x^2
...................... -48x^2 + 41x - 6
...................... -48x^2 + 32x
...................................... 9x - 6
...................................... 9x - 6
......................................… 0

So we therefore have:

f(x) = (3x - 2)(16x^2 - 16x + 3)

We have a quadratic factor that we need to factor, if possible. You can try traditional factoring methods, if you want (they do work). You can try to discover the roots of 16x^2 - 16x + 3 using the quadratic formula, or completing the square, and using the factor theorem to discover the linear factors that way. I'm going to factorise using the completing the square method:

f(x) = (3x - 2)(16x^2 - 16x + 4 - 1)
f(x) = (3x - 2)((4x - 2)^2 - 1^2)
f(x) = (3x - 2)(4x - 2 - 1)(4x - 2 + 1)
f(x) = (3x - 2)(4x - 3)(4x - 1)

There's the answer!