Question

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

Explanation:

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

Find two numbers that multiply to \(10\) (the constant term) and add up to \(7\) (the coefficient of \(x\)). The numbers are \(2\) and \(5\).
So, \(x^{2}+7x + 10=(x + 2)(x+5)\)
To sketch the algebra - tiles:

• For \(x^{2}+7x + 10\), we have one \(x^{2}\) tile (representing \(x\times x\)), seven \(x\) tiles (representing \(x\)) and ten \(1\) - unit tiles.
• Arrange them into a rectangle. The length of the rectangle is \((x + 5)\) and the width is \((x + 2)\). The \(x^{2}\) tile is in the top - left corner. Along one side (say the length), we have \(x\) tiles and \(1\) - unit tiles. For example, if we consider the rectangle with sides \(x + 5\) and \(x + 2\), we can place \(x\) tiles and \(1\) - unit tiles to form the sides.

Step2: Factor \(2x^{2}+7x + 6\)

For \(ax^{2}+bx + c\) (here \(a = 2\), \(b = 7\), \(c = 6\)), we use the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\) or we can find two numbers \(m\) and \(n\) such that \(m\times n=2\times6 = 12\) and \(m + n=7\). The numbers are \(3\) and \(4\).
Rewrite \(2x^{2}+7x + 6\) as \(2x^{2}+3x+4x + 6\)
Group the terms: \((2x^{2}+3x)+(4x + 6)=x(2x + 3)+2(2x + 3)=(2x + 3)(x + 2)\)
To sketch the algebra - tiles:

• We have two \(x^{2}\) tiles (representing \(2x\times x\)), seven \(x\) tiles and six \(1\) - unit tiles.
• Arrange them into a rectangle. The length of the rectangle is \((2x + 3)\) and the width is \((x + 2)\). The two \(x^{2}\) tiles can be placed adjacent to each other. Then, we place \(x\) tiles and \(1\) - unit tiles to form the sides of the rectangle.

Answer:

(a) \(x^{2}+7x + 10=(x + 2)(x + 5)\)
(b) \(2x^{2}+7x + 6=(2x + 3)(x + 2)\)