How to factor 3x^4+3x^3-3x-3

How to factor 3x^4+3x^3-3x-3

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^{4} =
3x^{3} · x

3x^{3} =
3x^{3} · 1

a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})

~~~~~~~~~

3x^{4} + 3x^{3} - 3x -3 =

(3x^{4} + 3x^{3}) - (3x + 3) =

3x^{3}(x +1) - 3(x + 1) =

(x +1)(3x^{3} - 3) =

3(x +1)(x^{3} - 1) =

**3(x + 1)(x - 1)(x**^{2}** + x + 1)**

Hey Daisy!

Looks like x= 1, -1 both work => (x+1)(x-1)= (x^2-1) can come out & bring 3 out:

3(x^4+x^3-x-1) / (x^2-1) = x^2 + x + 1

(x^4-x^2) <<<<<<<<<<<<\/ \/ \/

0+x^3+x^2 \/ \/

(x^3-x)<<<<<<<<<<<<<<<<\/ \/

0+x^2-1 <<<<<<<<<<<<<<<<<<\/

Answer ==> 3(x^2+x+1)(x+1)(x-1) ... I like getting started by trying x= 1,-1, etc ... Best wishes

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## Comments

I agree. First, look for anything that can be factored out from every term. Add the coefficients to see if 1 is a solution. If it is, you can use synthetic division to pull out (x-1) which reduces the degree and makes it easier to factor from there. If you check -1, you can do synthetic division again to pull out (x+1). Or you can identify that (x

^{2}-1) is a factor and do long division.I like your explanation, Brad; I just wanted to clarify that step.