Question

If `(8z-9)(8z+9) = az^2-b`, what is the value of `a`?

Answer 1

`a = 64`

Explanation:

difference of two squares identity:

`(a+b)(a-b) = a^2-b^2`

this means that `(8z-9)(8z+9) = (8z)^2 - 9^2`

`= 64z^2 - 81`

this can be shown by expanding brackets (FOIL) rule:

F (firsts): `8z` and `8z`
`8z * 8z = 64z^2`

O (outsides): `8z` and `9`
`8z * 9 = 72z`

I (insides): `-9` and `8z`
`-9 * 8z = -72z`

L (lasts): `-9` and `9`
`-9 * 9 = -81`

#(8z-9)(8z+9) is found by adding these products together:

`64z^2 + 72z - 72z - 81`

`72z` and `-72z` cancel each other out (add to `0`).

the result is `64z^2 - 81`.

in the question, `az^2 - b = 64z^2 - 81`

so `a = 64`.