Question

how could brent use a rectangle to model the factors of $x^{2}-7x + 6$?
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 $x^{2},-3x,-4x$, 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 $x^{2},7x,-x$, 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 $x^{2},-x,-6x$, 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 $x^{2},-4x,3x$, and half of -12.

Explanation:

Step1: Expand the product of factors

For a rectangle with dimensions \( (x - a)(x - b)=x^{2}-(a + b)x+ab \).
For \(x^{2}-7x + 6\), we need to find \(a\) and \(b\) such that \(a + b=7\) and \(ab = 6\). The values \(a = 1\) and \(b=6\) satisfy these equations.
The product \((x - 1)(x - 6)\) using the distributive property (FOIL method):
\((x-1)(x - 6)=x\times x+x\times(-6)+(-1)\times x+(-1)\times(-6)\)

Step2: Simplify the expression

\(x\times x+x\times(-6)+(-1)\times x+(-1)\times(-6)=x^{2}-6x-x + 6=x^{2}-7x + 6\)

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 \(6\). So the correct option is: 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 \(6\).