Question

(x^2-9)/(2x^2+7x+3) * (2x^2+11x+5)/(x^2-3x)

Answer 1

is this 2 problems

Answer 2

No it is all one problem. The left one multiplied by the right one.

Answer 3

oh sorry I see the * is multiply

Answer 4

$$\huge \frac{x^2-9}{2x^2+7x+3} * \frac{2x^2+11x+5}{x^2-3x}$$

Answer 5

Yea.

Answer 6

Looks like you have some factoring to do :)

Answer 7

x^2-9 is factored into (x+3) (x-3)

Answer 8

I'm lost after this?

Answer 9

factor an x out of $$x^2 -3x$$

Answer 10

(x-3)?

Answer 11

somebody?

Answer 12

so far...\[\huge \frac{(x+3)(x-3)}{2x^2+7x+3} * \frac{2x^2+11x+5}{x(x-3)}\]

Answer 13

How do I factor 2x^2+7x+3?

Answer 14

2x^2+7x+3

multiply the first and last numbers: 2*3 = 6

now find two numbers that multiply to equal 6, and add up to the middle number (7).

Answer 15

6*1. So (x+6) (x+1)?

Answer 16

It's not that simple this time. break up the 7x like this:

\[\Large 2x^2+7x+3\]

\[\Large 2x^2+6x+1x+3\]

\[\Large 2x(x+3)+1(x+3)\]

Answer 17

Ok, so how do I break up the 10 and one for the next part?

Answer 18

Do you know how to finish factoring this \[\Large 2x(x+3)+1(x+3)\]

Answer 19

x(x+3)?

Answer 20

Take out the common factor (x+3), what gets left behind?

2x(x+3)+1(x+3)

(x+3)(2x+1)

Answer 21

Ok, I got that.

Answer 22

\[\huge \frac{(x+3)(x-3)}{(x+3)(2x+1)} * \frac{2x^2+11x+5}{x(x-3)}\]
try factoring the 2x^2+11x+5 using a similar method to above.