Question

how could brent use a rectangle to model the factors of x² - 7x + 6? he could draw a diagram of a rectangle with dimensions x - 3 and x - 4 and then show the area is equivalent to the sum of x²,-3x,-4x, and half of 12. he could draw a diagram of a rectangle with dimensions x + 7 and x - 1 and then show the area is equivalent to the sum of x²,7x,-x, and 6. he could draw a diagram of a rectangle with dimensions x - 1 and x - 6 and then show the area is equivalent to the sum of x²,-x,-6x, and 6. he could draw a diagram of a rectangle with dimensions x - 4 and x + 3 and then show the area is equivalent to the sum of x²,-4x,3x, and half of -12.

Explanation:

Step1: Factor the quadratic expression

We use the formula for factoring \(ax^{2}+bx + c\) (\(a = 1\), \(b=-7\), \(c = 6\)). We need to find two numbers \(m\) and \(n\) such that \(m\times n=ac=6\) and \(m + n=b=-7\). The numbers are \(-1\) and \(-6\) since \((-1)\times(-6)=6\) and \(-1+( - 6)=-7\). So \(x^{2}-7x + 6=(x - 1)(x - 6)\).

Step2: Recall the area formula for a rectangle

The area of a rectangle is \(A = l\times w\), where \(l\) is the length and \(w\) is the width. If the length \(l=x - 1\) and the width \(w=x - 6\), then \(A=(x - 1)(x - 6)=x^{2}-6x-x + 6=x^{2}-7x + 6\).

Answer:

He could draw a diagram of a rectangle with dimensions \(x - 1\) and \(x - 6\) and then show the area is equivalent to the sum of \(x^{2},-x,-6x\), and \(6\).