Question

Find the product of the following by using methods of special products.

1. (2a^2+7)(2a^2-7)
2. (3x^2-yz^3)^2
3. (3x-5y)(5x+3y)
4. (3a^2b-7)^3
5. (2x-5)(4x^2-10x+25)

Help again? Lol I'm so sorry i'm just not good at maaaaaath

Answer 1

(2a^2+7)(2a^2-7)

2a^2 * 2a^2 = ___
2a^2 * -7 = ____

7 * 2a^2 = ___
7 * -7 = ____

Answer 2

just multiply

Answer 3

why not give it a try @bottledthoughts

Answer 4

Or you could use the method of grouping. \[\large (2a^2+7)(2a^2-7)\]\[\large \color{green}{2a^2}(2a^2-7) +\color{green}7(2a^2-7)\]

Answer 5

Or you could see that there's a special product called square difference that is
(x+y)(x-y)=x^2-y^2
so, you just replace 2a^2=x and 7=y and voila...

Answer 6

Yeah, difference of squares is the concept being requested here.

Answer 7

(for the first and third ones)

Answer 8

For number two you have a binomial square
(x-y)^2=x^2-2xy+y^2
4 is a binomial cube
(x-y)^3=x^3-3x^2y+3xy^2-y^3
And number five, factorize the second term and you will see magic happening (Remember about the perfect binomial square)