Question

(2x+3)(9x+6)

Answer 1

did you try and work it out?? the way i told you

Answer 2

yess

Answer 3

what did you get??

Answer 4

=2x+11x+8
=2x+27x16

thats my answer

Answer 5

(2x+3)(9x+6)is the product of two binomials and will thus the product will result in "the square of x" plus two other terms. it is very important that you express "the square of x" correctly mathematically. Do this by typing

x^2 or \[x^2\]

Answer 6

I don't see either of these in your final answer.

(2x+3)(9x+6) => 2x^2 +12x +27x +18.

Simplify this by combining like terms.

Answer 7

tnz i have so much learn

Answer 8

Yes, that's a lot to learn!

The "like terms" are 12x and 27x. Combine them, to get ... what?

Answer 9

12x
+27x
-----
?

Answer 10

39

Answer 11

39x, to be exact.

So, your final answer is 2x^2 + 39x + ?

Answer 12

i dont know

Answer 13

I'm sure you've understood some of this. Try to put your finger on (identify) what you don't know.

Answer 14

no i taught you how to foil. (2x+3)(9x+6)
first you multiply 2x * 9x = 18x^2 and 2x * 6 = 12x (18x^2 + 12x)
then you multiply 3 * 9x = 27x and 3 * 6 = 18 then add 18x^2 + 12x + 27x + 18
which will give you 18x^2 + 39x + 18

Answer 15

Yes, FriedRice caught our mistake; the first term is 18x^2, not 2x^2.

If you multiply the last terms of our two binomials (2x+3)(9x+6), you get (or should get) 18. Try it, please.

Again, what part of this problem is clear for you, and which part is not?

Answer 16

dont get mad each other plzz

Answer 17

I'm not the least bit "mad" at FriedRice; I was acknowledging that you (danica) and I were wrong and that FriedRice found our mistake on the first term.

Please, could we go back and finish this problem?

What remains unclear for y ou?

Answer 18

math never gets mad

Answer 19

Oh, yes he does! But he tries not to get mad here on OpenStudy! :)

Answer 20

Danica: We must multiply (2x+3(9x+6).

Please go over the work already done (above) and come up with the correct answer (which we call a "product"). Thanks. Write your result here.