I am a bit confused about conjugates in algebra. I am supposed to multiply 8/(the square root of two + 4) by the square root of 2 -4. My math book then goes on to simplify this to (8 times the square root of two minus 32) divided by (2-4 times the square root of two plus 4 times the square root of 2 minus 16). I do not see how the bottom part came to be. Could someone offer an explanation, please?

8/(sqrt2 + 4)

multiply top and bottom by (sqrt 2 -4) to get rid of the sqrt on the bottom because remember:
(a-b)(a+b) = a^2 - b^2
thereby squaring any square roots that might have been in b
8/(sqrt 2 + 4) * (sqrt 2-4)/(sqrt 2-4)

= 8 (sqrt 2 - 4) / [ (sqrt 2)^2 - 16 ]

= (8 sqrt 2 - 32) / [2-16]

= (8 sqrt 2 - 32) / -14

= (16 - 4 sqrt 2 / 7

By the way your math book did the

(a-b)(a+b) by FOIL

a^2 + ab -ab - b^2 which is a^2-b^2
but I know and you should know that
(a-b)(a+b) = a^2 - b^2
without going through the FOIL

Squareroot of A* square root of AB ^2

Sure! I can help you understand how the denominator of your expression was simplified.

When dealing with expressions involving radicals, it is often helpful to rationalize the denominator. This means eliminating the radical from the denominator. In this case, the denominator is given as (2 - 4√2 + 4√2 - 16).

To rationalize this expression, we need to eliminate the square root terms. One way to do this is by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of a binomial is obtained by changing the sign between the terms.

The conjugate of (2 - 4√2 + 4√2 - 16) is (2 - 4√2 + 4√2 + 16).

To simplify the denominator, multiply the numerator and denominator by the conjugate:

[(8 × √2 - 32) / (2 - 4√2 + 4√2 - 16)] × [(2 - 4√2 + 4√2 + 16) / (2 - 4√2 + 4√2 + 16)]

Now, let's simplify the denominator:

(2 - 4√2 + 4√2 - 16) = (2 - 16) = -14

So, the denominator simplifies to -14.

Now, you can simplify the entire expression:

(8 × √2 - 32) / (2 - 4√2 + 4√2 - 16) = (8 × √2 - 32) / (-14)

And that's how the denominator was obtained in the final expression.