Question

3x^3+19x^2-72x=0

Answer 1

Divide both sides by \(x\).\[3x^2 + 19x - 72 = 0 \]Now, solve this quadratic equation.

Answer 2

ok

Answer 3

Hint: x+9 is a factor of the quadratic.

Answer 4

is the other (3x-8)

Answer 5

Yes.

Answer 6

ok but when i add -8+9 it does not equal 19

Answer 7

why is that?

Answer 8

\[(x+9) (3 x-8)=3x^2+27x-8x-72=3 x^2+19 x-72 \]

Answer 9

ok...so then how would i get the 3x^3 and 19x^2 back??

Answer 10

\[x*\left(3 x^2+19 x-72\right)=3 x^3+19 x^2-72 x \]

Answer 11

ok thanks!!

Answer 12

You're welcome.