Is it correct?

Question

Hi, i factorized this thing, and i would like to know if i did correctly and if there is any way to do it.

A=(3-2x)^2-(7x+2)^2
=[3-2x)+(7x+2)][(3-2x)-(7x+2)]
=[3-2x+7x+2][3-2x-7x+2)
=(5x+5)(9x-5)

Thanks


Thanks

Answer 1

Hi,
You certainly took the right approach. This is the difference of two squares as you apparently percieved. So, we know that:
(a²-b²) = (a-b)(a+b)
You did great until you got to the last line. There you reversed the signs in the last expression. You have this:
=[3-2x+7x+2][3-2x-7x+2)
Notice that in the expression [3-2x-7x+2) you have two positive-signed numbers and two negative-signed variables. So, when you combine like terms you would have (-9x+5) to give a final answer of
(5x+5)(-9x+5)
rather than
=(5x+5)(9x-5).
Great job except for a slight bobble near the end.

FE

Answer 2

OK I think it's close

let 3-2x =l; let 7x+2 =m

A = l^2 - m^2
A = (l+m)(l-m)
A=(5x+5)(-9x+5)

I could be wrong - but I think the signs are not right on your second term.

Hope this helps.

Answer 3

Uh-uh.

(3-2x)^2-(7x+2)^2
(9 - 12x + 4x^2) - (49x^2 + 28x + 4)

-45x^2 - 40x + 5

-5*(9x^2 + 8x - 1)

-5*(x + 1)*(9x - 1)

Ah, that's better.

Answer 4

(5x+5)(1-9x)

3rd line is wrong
should be =[3-2x+7x+2][3-2x-7x-2)

Answer 5

(3-2x)(3-2x)-(7x+2)(7x+2)
9-12x+4x^2 - (49x^2+28x+4)
5-40x-45x^2
5(1-8x-9x^2)

i'm and idiot but this is what I get from the structure of the original equation

Answer 6

First, the word is factored, not factorized. 2nd, you did not do it correctly. From your 2nd to 3rd step, you did not distribute your negative sign on the (7x+2) properly. Then, from the third to the fourth step, you did not add your "x" terms correctly. Review it again.

I am not going to tell you how to do it, I'm just pointing out your mistakes.



*EDIT*

The only person that has it correct is the first poster, MrMath.

Answer 7

the last step is wrong it is

=(5x+5)(-9x+5)