What is the missing constant term in the perfect square that starts with x2+2x

Answer 1

First, remember F-O-I-L. (ax+b)(cx+d) = acx² + adx + bcx + bd

Now, make that a perfect square by making the two factors the same:

(ax+b)(ax+b) = a²x² + abx + abx + b²

(ax+b)(ax+b) = a²x² + 2abx + b²

This is especially easy if a=1: (x+b)(x+b) = x² +2bx + b²

The problem gives: x² + 2x + _____

For the coefficients to match, 2b=2, so b=1

That makes, b² = 1

(x+1)(x+1) = x² + 2x + 1

Answer 2

The constant is 1, as in (x+1)^2 = x^2 + 2x + 1

Answer 3

Just take half the coefficient of the linear term, x, and square it.

The coefficient of the linear term is 2. Half is 2/2 = 1. square it, it's still 1.

That's the constant term to make x² + 2x into a perfect square: x² + 2x + 1

You can do this if the squared term has a coefficient of 1.

Let's say your started with x²+8x. To get the constant term to make a perfect square, just take half of 8 = 4 and square it to get 16. 16 is the constant term that makes it into a perfect square (x+4)²