how do i multiply 2x^2-6x all over x^2+18x+81 times 9x+81 all over x^2-9?

Question

how do i go about solving this problem

Answer 1

You must factor numerator and denominator then cross cancel matching sets
2x(x-3) ........... 9(x+9)
-------------- x ---------------
(x+9)(x+9) ..(x+3)(x-3)

When you cancel the matching sets you are left with
18x
-------------
(x+9)(x+3)

Answer 2

Work with one equation at a time:

2x^2-6x can be reduced to 2x(x-3)

x^2+18x+81 can be reduced to (x+9)(x-9)

9x+81 = 9(x+9)

x^2-9 = (x+3)(x-3)

Cross multiply and you will cancel the (x-3) and (x+9) on both sides. Combine and you get:

18x/(x-9)(x+3)

Answer 3

(2x^2-6x/x^2+18x+81)*(9x+81/x^2-9)

First: factor 2x^2-6x > the least common factor is "2x"

2x^2 - 6x = 2x(x-3)

You have [2x(x-3)/x^2+18x+81]*(9x+81/x^2-9)

SEc: factor x^2+18x+81. multiply the 1st & 3rd term to get 81. find two numbers that give you 81 when multiplied & 18 when added/subtracted. the numbers are (9 & 9). you should get...

x^2+18x+81 = (x+9)(x+9)

You have [2x(x-3)/(x+9)(x+9)]*(9x+81/x^2-9)

Third: factor 9x+81 which is, 9(x+9)

You have [2x(x-3)/(x+9)(x+9)]*[9(x+9)/x^2-9]

Fourth: factor x^2 -9 which, is the difference of squares.

x^2 - 9 = (x+3)(x-3)

You have [2x(x-3)/(x+9)(x+9)]*[9(x+9)/(x+3)(x-3)]

Fifth: cross cancel "like" terms & combine the rest.

[2x/(x+9)]*[9/(x+3)]
(2x*9)/(x+9)(x+3)
= 18x/(x+9)(x+3)

Answer 4

I won't do this for you, but I'll tell you how. You can assume with these questions that they don't actually expect you to multiply every term by every other term. That doesn't really teach you anything. What they want is for you to be able to break longer equations into their roots so as to be able to cancel out parts. So you need to look for ways to do this.

In this equation all of the partswith x^2 can be can be broken down into roots or can factor out an x. If you are good at factoring do it that way or use the quadratic equation if you can't see what the roots would be. Once you have each part broken down into single x terms you can cancel anything that is the same on the top and bottom. Then multiple the remaining parts back together. You should end up with a similar type of equation to the one you started with just less complex.

Good Luck

Answer 5

2x(x-3) * 9(x+9)
---------------------------
(x+9)(x+9) (x-3)(x+3)

the 2 x-3 and x+9 cancel out

2x*9/ (x+9)(x+3)

18x/ (x+9)(x+3)

Answer 6

2x(x-3)/(x+9)(x+9) .....x ..... 9(x+9)/(x-3)(x+3)=

2x/x+9 ......x...... 9/(x+3) =

18x/(x+9)(x+3)

Answer 7

2x^2-6x/(x^2+18x+81)*9x+81/(x^2-9)
2x(x-3)(x+9))^2*9(x+9)/(x-3)(x+3
=2x(x+9)^3*9/x+3
18x(x+9)^3/(x+3). answer