x^4 + 2x^3 + x^2 + 8x - 12
Lucky guess, try x = 1, see if it makes the problem = 0.
1^4 + 2(1^3) + 1^2 + 8(1) - 12
1 + 2 + 1 + 8 - 12 = 0
It works. x = 1 is a zero of x^4 + 2x^3 + x^2 + 8x - 12. If x = 1, then x-1 = 0, so x-1 should be a factor of x^4 + 2x^3 + x^2 + 8x - 12
You can do synthetic division to factor x-1 out of x^4 + 2x^3 + 8x -12. You should get:
(x-1)(x^3 + 3x^2 + 4x + 12)
.
Can we factor x^3 + 3x^2 + 4x + 12 easily?
Let's find us some zeros.
If f(x) = x^3 + 3x^2 + 4x + 12, then we want to choose an x so that f(x) = 0.
f(-1) = -1 + 3 - 4 + 12 = 10. Not a zero.
f(-2) = -8 + 12 - 8 + 12 = 8. Not a zero.
f(-3) = -27 + 27 - 12 + 12 = 0.
x = -3 is a zero of x^3 + 3x^2 + 4x + 12, so (x+3) should be a factor of x^3 + 3x^2 + 4x + 12
I'll leave it to you to do synthetic division or whatever method you want. The result is (x + 3)(x^2 + 4) = x^3 + 3x^2 + 4x + 12
(x - 1)(x + 3)(x^2 + 4) = x^4 + 2x^3 + x^2 + 8x - 12
x^2 + 4 doesn't factor
Answer: (x - 1)(x + 3)(x^2 + 4)