Question

2x^3-3x^2-11x+6 divide by x-3

Answer 1

Answer: $2x^2+3x-2$

Step-by-step explanation:

You can do long division, which is very very hard to show with typing on a keyboard. You essentially want to divide the leading coefficient for each term. Ill try my best to explain it.

Do $\frac{2x^3}{x}=2x^2$. Write 2x^2 down. Now multiply (x - 3) by it. Then subtract it from the trinomial.

$2x^2*(x-3)=2x^3 -6x^2\\(2x^3 -3x^2-11x+6)-(2x^3-6x^2) = 3x^2-11x+6$

Now do $\frac{3x^2}{x} =3x$. Write that down next to your 2x^2. Multiply 3x by (x - 3) to get:

$3x(x-3)=3x^2-9x\\(3x^2-11x+6)-(3x^2-9x)=-2x+6$

Your final step is to do $\frac{-2x}{x} =-2$. Write this -2 next to your other two parts

Multiply -2 by (x - 3) to get:

$-2(x-3)=-2x+6\\(-2x+6)-(-2x+6)=0$

Our remainder is 0 so that means (x - 3) goes into that trinomial exactly:

$2x^2+3x-2$ times

Answer 2

Answer:

2x² + 3x -2

Step-by-step explanation:

2x³ - 3x² - 11x + 6 : (x - 3)

2x³ - 6x² from (x - 3) * 2x²

-------------------------- —

3x² - 11x + 6

3x² - 9x from (x - 3) * 3x

-------------------------- —

- 2x + 6

- 2x + 6 from (x - 3) * (-2)

-------------------------- —

0

so 2x³ - 3x² - 11x + 6 : (x - 3) = 2x² + 3x -2