How to rewrite y = x^2 – 20x + 111 into four terms in which the first three terms form a perfect square trinomial

x^2 - 20x + 100 + 11 = (x - 10)^2 + 11

Step-by-step explanation:

We're basically completing the square here.

We have the equation x^2 - 20x + 11. Let's just focus on the first two terms for now:

x^2 - 20x

In order to make this a perfect square trinomial, we need to find the constant term. We use the formula (b/2)^2, where b is the coefficient of the x term. Here, b = -20, so the constant term will be: (-20/2)^2 = (-10)^2 = 100.

We now have: x^2 - 20x + 100

However, we can't forget about the 111 from the original equation. We can't do this: x^2 - 20x + 100 + 111 because then we're left with an equation that is NOT the same as the original. Instead, we need to subtract 100 from 11 because we're already adding it in our perfect square trinomial:

x^2 - 20x + 100 + 111 - 100

x^2 - 20x + 100 + 11

The equation can then be written as: (x - 10)^2 + 11

Hope this helps!

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i