Question

Find the zeros of the following polynomial. 3x3 + 9x2 - 12x

Answer 1

x = 4
x = -1
x = 0

Equation at the end of step 1 :

((3 • (x3)) - 32x2) - 12x = 0
Step 2 :

Equation at the end of step 2 :

(3x3 - 32x2) - 12x = 0
Step 3 :

Step 4 :

Pulling out like terms :

4.1 Pull out like factors :

3x3 - 9x2 - 12x = 3x • (x2 - 3x - 4)

Trying to factor by splitting the middle term

4.2 Factoring x2 - 3x - 4

The first term is, x2 its coefficient is 1 .
The middle term is, -3x its coefficient is -3 .
The last term, "the constant", is -4

Step-1 : Multiply the coefficient of the first term by the constant 1 • -4 = -4

Step-2 : Find two factors of -4 whose sum equals the coefficient of the middle term, which is -3 .

-4 + 1 = -3 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and 1
x2 - 4x + 1x - 4

Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-4)
Add up the last 2 terms, pulling out common factors :
1 • (x-4)
Step-5 : Add up the four terms of step 4 :
(x+1) • (x-4)
Which is the desired factorization

Answer 2

If you factor 3x from the expression, you have

$3x^3+9x^2-12x=3x(x^2+3x-4)$

So, we have

$3x(x^2+3x-4)=0 \iff 3x=0\lor x^2+3x-4=0$

We easily have

$3x=0\iff x=0$

So, one solution is x=0.

The other solutions depend on the quadratic equation:

$x^2+3x-4=0 \iff x=-4 \lor x=1$

So, the solutions are $x=-4,\ 0,\ 1$