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 factor \(x^{2}-7x + 6\) using the formula for factoring \(ax^{2}+bx + c\) (here \(a = 1\), \(b=-7\), \(c = 6\)). We need two numbers that multiply to \(ac=6\) and add up to \(b=-7\). 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: Find the area of the rectangle

The area of a rectangle with length \(l=x - 1\) and width \(w=x - 6\) is \(A=l\times w=(x - 1)(x - 6)\). Using the FOIL method, \((x - 1)(x - 6)=x\times x+x\times(-6)+(-1)\times x+(-1)\times(-6)=x^{2}-6x - x+6\).

Answer:

C. 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 \(6\).