Question

Divide the following polynomial
https://media.glynlyon.com/g_alg02_ccss_2013/4/q21230.gif

Answer 1

Okay, what is the first term of the numerator divided by the first term of the denominator?

Answer 2

2x^2

Answer 3

Very good. Multiply 2x^2 by the denominator and carefully subtract that from the starting numerator. What do you get?

Answer 4

I don't know

Answer 5

What is \[6x^3+11x^2-4x-4 - 2x^2(3x-2)\]

Answer 6

Expand and simplify

Answer 7

\[6x^3+11x^2-4x-4-6x^3-(-2)2x^2 = \cancel{6x^3}+11x^2-4x-4\cancel{-6x^3}+4x^2 \]\[\qquad= 15x^2-4x-4\]

What is the first term of that divided by the first term of the denominator? We repeat this process until what we have left is either 0 or smaller than the denominator.

Answer 8

1.8?

Answer 9

Huh? How do you figure that \[15x^2/x=1.8\]?!?

Answer 10

Sorry, \[15x^2/3x=1.8\]

No, the next term of our quotient is \(5x\) because \(15x^2/3x = 5x\)

Answer 11

So we multiply \(5x(3x-2)=15x^2-10x\) and subtract that from our polynomial:
\[15x^2-4x-4-(15x^2-10x) = 15x^2-4x-4-15x^2+10x = 6x-4\]
Can you divide \(6x-4\) by \(3x-2\)?

Answer 12

2x^2 - 2

Answer 13

No, you forgot the \(5x\) term., and the last term has the wrong sign. (6x-4) = 2(3x-2).

All together, our quotient is \(2x^2+5x+2\)

Checking the result:
\[(3x-2)(2x^2+5x+2) = 3x*2x^2+3x*5x+3x*2 -2*2x^2-2*5x-2*2\]\[\qquad=6x^3+15x^2+6x-4x^2-10x-4\]\[\qquad=6x^3+11x^2-4x-4\checkmark\]