What is the missing constant term in the perfect square that starts with x^2-16x

Answer 1

This problem is about finding "c" in the quadratic expression: ax^2 +bx +c=0

In order to find "c" we need to use the following formula C= (-b/2)^2.

If you notice in the above expression a=1, b=-16 and c=? is unknown.

Let apply the formula: C= (-(-16/2))^2 = (8)^2=64

Now we are going to substitute C to complete the expressions ax^2+bx+c=0

ax^2 +bx +c=0 , x^2 -16x +64=0, so the constant term is c=64

To check the answer , we can see that the new expression is a perfect trinomial.

x^2 -16x+64=0

Factoring (x-8)(x-8)=0 . or (x-8)^2 . therefore the answer is c=64.

Answer 2

Divide the negative 16 by 2 =- 8

Then square -8.

So the missing constant is 64

The perfect square would then be

(X-8)²