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

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

Answer 2

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:

$c = (\frac{b}{2})^2$

Plug b into the equation:

$b = -16$
$\frac{-16}{2} = -8$
$-8^2 = 64$

$c = 64$

The missing constant term is 64.