SOLUTION: The polynomial g(x)= -3x^3-15x^2-6x+24 has a zero at x=-2. Find all the other zeros of g(x) using polynomial long division.

Question 804507: The polynomial g(x)= -3x^3-15x^2-6x+24 has a zero at x=-2. Find all the other zeros of g(x) using polynomial long division.
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Given:
(1) g(x) = -3x^3 -15x^2 -6x +24
We are told that a root (zero) of this trinomial is
(2) x = -2
Therefore
(3) x + 2 = 0
which makes (x + 2) a factor of g(x). That is, if you divide g(x) by (x + 2) you will obtain a quadratic
(4) ax^2 +bx +c
Long division is the easy way to do the division which I'm sure you've had in class, but I can't type it here. It's just like dividing numbers. I will you show you how to get the answer using multiplication, ok?
Before we begin, remember we are deriving the zeroes of g(x), which means we will set g(x) = 0. If we set (1) equal to zero we get
(5) -3x^3 -15x^2 -6x +24 = 0
Note that a common factor of all terms is -3, so we have
(6) -3*(x^3 +5x^2 +2x -8) = 0 or equivalently
(7) x^3 +5x^2 +2x -8 = 0, much friendlier!
We know that (7) factors into
(8) (x+2)*(ax^2+bx+c) = x^3+5x^2+2x-8
As stated above (and by the question) we can find a,b, and c by dividing both sides of (8) by the known factor (x+2), but I can't do that here.
Alternatively, multiply out the left side of (8) to get
(9) a*x^3+b*x^2+c*x+2ax^2+2bx+2*c = x^3+5x^2+2x-8
Since the left side equals the right side of (9), equate like terms to get
(10) a*x^3 = x^3
(11) (b + 2*a)*x^2 = 5*x^2
(12) (c + 2*b)*x = 2*x and
(13) 2*c = -8
We can solve (10) for a
(14) a = 1
and (13) for c
(15) c = -4
In (11), set the coefficients equal to get
(16) b + 2*a = 5 and use a = 1 to get
(16) b + 2*1 = 5 or
(17) b = 5 - 2 or
(18) b = 3
We don't need (12), but we can verify a,b,and c.
In (12), set the coefficients equal to get
(19) c + 2*b = 2 and use c = -4 and b = 3 to get
Is (-4 + 2*3 = 2)?
Is (-4 + 6 = 2)?
Is (2 = 2)? Yes
The quadratic factor of (4) is
(20) x^2 +3*x -4 or
(21) g(x) = -3*(x+2)*(x^2+3x-4)
Now we can further factor the quadratic of (20) to get
(22) x^2 +3*x -4 = (x+4)*(x-1)
which when set equal to zero gives
(23) x + 4 = 0 or
(24) x = -4
and
(25) x -1 = 0 or
(26) x = 1
Answer: The zeroes of g(x) are {-4,-2,1}

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