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

User Twisted
by
7.5k points

2 Answers

1 vote

Answer:

2x² + 3x -2

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

User Klompenrunner
by
8.0k points
4 votes

Answer:
2x^2+3x-2

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
(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
(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
(-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

User Sidharth Mudgal
by
8.2k points

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