Maths help needed please. Completing the square.?

Provided you have 1 x^2 , a will be half the number in front of the x term, with the same sign. So here (x-4)^2 now when you expand this it gives x^2 - 8x (which you want) but + 16 (which is not right) so you need to add another 5 --> (x-4)^2 +5 and that is called completing the square.

(x + a)^2 = x^2 + 2ax + a^2 - front squared plus twice product plus back squared. Match this up to what you've got, and always get a first (worry about b afterwards):

(x + a)^2 = x^2 + 2ax + a^2

= x^2 - 8x + 21

so 2a = - 8, and a = - 4.

Plug this back in:

(x - 4)^2 = x^2 - 8x + 16

but you need x^2 - 8x + 21 - you are 5 short, so add five to both sides:

(x - 4)^2 + 5 = x^2 - 8x + 21

a = 4 and b = 5.

What number could "a" be to produce -8 as the middle coefficient of x?

4, right?

So what is (x - 4)^2 equal to?

What, then, is the value of "b" to get the expression back to the original value?

a=12

b=3