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How to factor completely?

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2 Answers

3x4 = 3x3 · x
3x3 = 3x3 · 1
a3 - b3 = (a - b)(a2 + ab + b2)
~~~~~~~~~ 
3x4 + 3x3 - 3x -3 =
(3x4 + 3x3) - (3x + 3) =
3x3(x +1) - 3(x + 1) =
(x +1)(3x3 - 3) =
3(x +1)(x3 - 1) =
3(x + 1)(x - 1)(x2 + 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

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 (x2-1) is a factor and do long division.

I like your explanation, Brad; I just wanted to clarify that step.

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