SOLUTION: Can anyone help me? I did the first step to this question and am lost on how to complete it factor to find all the real zeros of 3x^3+7x^2-6x. What am I to do??
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: Can anyone help me? I did the first step to this question and am lost on how to complete it factor to find all the real zeros of 3x^3+7x^2-6x. What am I to do??
Log On
Question 107109: Can anyone help me? I did the first step to this question and am lost on how to complete it factor to find all the real zeros of 3x^3+7x^2-6x. What am I to do?? Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! 3x^3+7x^2-6x
=x(3x^2+7x-6)
Let's try to factor.
3*-6=-18 multiply the 1st and last numbers.
what are 2 factors of -18 that add up to 7? Answer +9 and -2
(3x+9)(3x-2)=9x^2+21x-18 Use the coordinate of x^2 for the 1st number in both parentheses. We are 3 times too high on both sides.
(x+3)(3x-2)=3x^2+7x-6 Divide (3x+9) by 3.
x(x+3)(3x-2)
Ed
