Question

factor to find the dimensions.
1 the area of a rectangle is $4x^2 + 4x + 1$. factor to find the dimensions of the shape.
2 the area of a rectangle is $121x^2 - 36y^2$. factor to find the width and length.

Explanation:

Problem 1

Step1: Identify the form

The quadratic \(4x^{2}+4x + 1\) looks like a perfect square trinomial. The form of a perfect square trinomial is \(a^{2}+2ab + b^{2}=(a + b)^{2}\).
Here, \(a^{2}=4x^{2}\), so \(a = 2x\) (since \((2x)^{2}=4x^{2}\)), and \(b^{2}=1\), so \(b = 1\) (since \(1^{2}=1\)). Now check the middle term: \(2ab=2\times(2x)\times1 = 4x\), which matches the middle term of the given quadratic.

Step2: Factor the quadratic

Using the perfect square trinomial formula, \(4x^{2}+4x + 1=(2x + 1)^{2}\). Since the area of a rectangle is length times width, and this factors to a square, the dimensions of the rectangle are both \(2x + 1\) (it is a square, a special case of a rectangle).

Step1: Identify the form

The expression \(121x^{2}-36y^{2}\) is a difference of squares. The formula for factoring a difference of squares is \(a^{2}-b^{2}=(a + b)(a - b)\).
Here, \(a^{2}=121x^{2}\), so \(a = 11x\) (since \((11x)^{2}=121x^{2}\)), and \(b^{2}=36y^{2}\), so \(b = 6y\) (since \((6y)^{2}=36y^{2}\)).

Step2: Factor the expression

Using the difference of squares formula, \(121x^{2}-36y^{2}=(11x + 6y)(11x - 6y)\). Since the area of a rectangle is length times width, the length and width of the rectangle are \(11x + 6y\) and \(11x - 6y\) (or vice versa).

Answer:

The dimensions of the rectangle are \(2x + 1\) and \(2x + 1\) (or \((2x + 1)\) by \((2x + 1)\)).

Problem 2