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Question 737332: how do you factor x cubed+4xsquared -4x-16
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! All odd powered polynomials have at least one real root. You have a cubic
(1) x^3 + 4x^2 - 4x -16 which is odd, so it has at least one real root. Big deal! How do we find the roots?
First we set (1) equal to zero,
(2) x^3 + 4x^2 - 4x -16 = 0
I start by setting only the last two terms of (2) equal to zero to get an estimate of the value of the real root (a root is the value of x that makes the whole equation equal to zero). This step gives us
(3) -4x -16 = 0 or
(4) 4x = -16 or
(5) x = -4
Now see if this is a root by substituting x = -4 into (2),
Is ( (-4)^3 + 4(-4)^2 - 4(-4) - 16 = 0)?
Is ( -64 + 64 + 16 - 16 = 0)?
Is (0 = 0)? Yes
We got lucky, our first step yielded a real root!!!
Now since the root is -4 we write
(6) x = -4 or
(7) (x + 4) = 0,
Equation (7) is a factor of (1), and we use long division to find the other factor (a quadratic - one order less than the cubic). I can't type this division, but it is given by
(8) (x^3 + 4x^2 - 4x -16)/(x + 4) = (x^2 - 4)
To check (8) FOIL
(9) (x + 4)*(x^2 - 4) = x^3 + 4x^2 - 4x - 16
The second factor is the difference of two perfect squares and factors into
(10) (x^2 - 4) = (x + 2)*(x - 2)
Setting the two factors of (10) to zero gives us two more roots
(11) x + 2 = 0 or
(12) x = -2 and
(13) x - 2 = 0 or
(14) x = 2
We check (12) and (14) in (2)
Is ( (-2)^3 + 4(-2)^2 - 4(-2) - 16 = 0)?
Is ( -8 + 16 + 8 - 16 = 0)?
Is (0 = 0)? Yes
Is ( (2)^3 + 4(2)^2 - 4(2) - 16 = 0)?
Is ( 8 + 16 -8 - 16 = 0)?
Is (0 = 0)? Yes
Answer: (x^3 + 4x^2 - 4x - 16) factors into (x+4)*(x+2)*(x-2)
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