Question

1. sketch the algebra tiles for each trinomial arranged as a rectangle to show how they can be factored. (a) ( x^{2}+7x + 10 ) ( x^{2}+7x + 10=(x + 2)(x + 5) ) (b) ( 2x^{2}+7x + 6 ) (c) ( x^{2}-x - 20 )

Explanation:

Step1: Factor the trinomial \(x^{2}+7x + 10\)

We need to find two numbers \(m\) and \(n\) such that \(m + n=7\) and \(m\times n = 10\).
The numbers \(m = 2\) and \(n=5\) satisfy \(2 + 5=7\) and \(2\times5 = 10\).
So, \(x^{2}+7x + 10=(x + 2)(x + 5)\)

Step2: Sketch the algebra - tiles

• For the term \(x^{2}\), we have one large square (representing \(x\times x\)).
• For the term \(7x\), we have seven rectangles (where the length is \(x\) and the width is \(1\)).
• For the term \(10\), we have ten small squares (representing \(1\times1\)).

We arrange them into a rectangle. The length of the rectangle is \((x + 5)\) and the width is \((x + 2)\)

Answer:

To sketch the algebra - tiles for \(x^{2}+7x + 10\):

• Place one \(x^{2}\) tile (large square).
• Place seven \(x\) - tiles (rectangles with side - lengths \(x\) and \(1\)).
• Place ten \(1\times1\) tiles (small squares).

Arrange them into a rectangle with dimensions \((x + 2)\) and \((x + 5)\) (where the number of \(x\) - tiles and \(1\times1\) tiles are distributed along the sides of the rectangle formed by the \(x^{2}\) tile).