The area of a quadrilateral is 4x^2+20x+25. If the dimensions of the quadrilateral are the factors of 4x^2+20x+25 could the rectangle be a square? EXPLAIN

Question

VERY URGENT DUE TOMORROW!!!

Answer 1

You list two different values for the area of the quadrilateral - one with 20 and one with 20x. I assume you are working with the second trinomial.

You are given the area of the figure and told to determine if the figure is a square. The formula for the area of a square is A = s². To find s, you need to do the opposite of what's done to it in the equation. Here, it is squared, so to find s, find the square root of both sides of the equation. In short, you need to find out if the original trinomial for the area is a perfect square. To do this, you have to be able to factor the trinomial into its square roots.

The model for factoring a trinomial into its square roots is a² + 2ab + b² = (a + b)². Notice, the first term (a²) and the third term (b²) of the trinomial are both squares. Their square roots are a and b. The middle term is 2 times the product of those roots (2ab).

The trinomial for this problem is 4x² + 20x + 25. Can you factor this so that you have the same binomial squared (a + b)²? Do the following:

1. First, find square roots for 4x² and 25.
2. Next, see if the middle term (20x) is the product of 2 times the square root of the first and third terms. If you find a binomial which works here, you have a perfect square.
3. If you have a perfect square, then the answer to the original question is yes.

Answer 2

Yes, 4x² + 20x + 25 = (2x + 5)²