Question

What is the missing constant term in the perfect square that starts with x^2-16x? Pretty stumped on this one. Thanks in advance!

Answer 1

Perfect squares are the result of an integer being multiplied by itself.

The trinomial is given by the following formula:


`ax^2 + bx + c`

We are missing c from this trinomial, so we must solve for c.

An equation that we can use to solve for c in this equation is the following formula:


`c = ((b)/(2))^2`

Plug b into the equation:


`b = -16`

`(-16)/(2) = -8`

`-8^2 = 64`


`c = 64`

The missing constant term is 64.

Answer 2

You'll have to find a number, in which, when factored, the factors would add up to -16

x^2 - 16x + 64 (for example)
x -8
x -8

(x - 8) (x - 8). If you use FOIL, you will get the trinomial again.

64 can be your answer

hope this helps