How do I factorise (2x + 2)² - 9?

Question

i think i have to change the 9 to 3²
but that's all I know

Answer 1

yes it's a special binomial (the difference of two squares)

=(2x + 2)² - 9

=(2x + 2)² - 3²

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

=(2x+5)(2x-1)

x= -5/2 or 1/2

Answer 2

(2x + 2)^2 - 9
(2x + 2)(2x + 2) - 9
4x^2 +4x +4x +4 - 9
4x^2 +8x -5 < --- answer

This can be refactored into (2x+5)(2x-1)
If you need to solve for x, set each factor = 0

(2x+5) = 0
2x = -5
x = -5/2

(2x-1) = 0
2x = 1
x=1/2

Answer 3

i cant do the squared number 2 in small case, but hope this helps anyway...
4x2 +4x +4x +4 - 9=0
4x2 + 8x -5 =0
(2x +5)(2x -1)=0
take either side

2x + 5=0
2x =-5
x = -5/2

or....
2x -1=0
2x =1
x = 1/2

Answer 4

if it's saying simplify, do the solutions above..

but if it says factorize,

then the answer is..

[(2x+2)+3][(2x+2)-3]
(2x+5)(2x-1)
or, give the statement "the factors of (2x+2)^2-9 is (2x+5) and (2x-1)"

Answer 5

Well I think it's asking you to do this:
(2x + 2)^2 - 9
(2x + 2)(2x + 2) - 9
4x^2 +4x +4x +4 - 9
4x^2 +8x -5 < --- answer

Answer 6

You have to write out (2x+2)^2....(^2 means squared.)

So, (2x+2) * (2x+2) -9
After multiplying that, you get 4x^2 + 4x + 4x + 4 - 9
Then just combine like integers 4x^2 + 8x -5
And if you factor that out, you get (2x-1) (2x+5)
If you need to get an exact number for X, then set both equal to 0 and solve.

2x-1=0
2x=1
X=1/2

2x+5=0
2x=-5
X=-5/2

Hope this helps.

Answer 7

(2x + 2)² =

(2x + 2) (2x + 2) = (4x² + 8x + 4)

(4x² + 8x + 4) - 9 =

4x² + 8x - 5

Hope that's right but it's been years since I did maths.

Answer 8

you have to square the first expression
and you get 4x^2 + 8x +4 -9
that's 4x^2 + 8x - 5
That factors out to (2x +5)(2x - 1)

Answer 9

multiply out the firt part and get

(4x^2+*x+4)-9
that gives you

4x^2+8-5
that factors to

(2x+5)(2x-1)

good luck

Answer 10

(2x+2)^2 -9=(2x-2)^2-3^2
= (2x-2-3)(2x-2+3)
=(2x-5)(2x+1 )
that's easy