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:

First, rewrite the quadratic as \(x^2 - 7x + 6\). Factoring this: find two numbers that multiply to 6 and add to -7, which are -1 and -6. So the factors are \((x - 1)(x - 6)\). The area of a rectangle with dimensions \((x - 1)\) and \((x - 6)\) is \((x - 1)(x - 6) = x^2 - 6x - x + 6\), which is the sum of \(x^2\), \(-x\), \(-6x\), and 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.