Question

how could brent use a rectangle to model the factors of $x^2 - 7x + 6$?
\bigcirc 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.
\bigcirc 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.
\bigcirc 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.
\bigcirc 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: Factor the quadratic

Factor \(x^2 - 7x + 6\). We need two numbers that multiply to \(6\) and add to \(-7\). Those numbers are \(-1\) and \(-6\), so \(x^2 - 7x + 6=(x - 1)(x - 6)\).

Step2: Expand the factors

Expand \((x - 1)(x - 6)\) using the distributive property (FOIL). First: \(x\cdot x = x^2\), Outer: \(x\cdot(-6)=-6x\), Inner: \(-1\cdot x=-x\), Last: \(-1\cdot(-6) = 6\). Then sum these terms: \(x^2-6x - x+6=x^2-7x + 6\), which matches the original quadratic.

Step3: Analyze the options

• Option 1: Dimensions \(x - 3\) and \(x - 4\) expand to \(x^2-7x + 12\), not \(x^2-7x + 6\), and the area description is incorrect.
• Option 2: Dimensions \(x + 7\) and \(x - 1\) expand to \(x^2+6x - 7\), not matching, and area description is wrong.
• Option 3: Dimensions \(x - 1\) and \(x - 6\) expand as shown, and the area sum is \(x^2,-x,-6x,6\), which is correct.
• Option 4: Dimensions \(x - 4\) and \(x + 3\) expand to \(x^2 - x - 12\), not matching, and area description is incorrect.

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 \(6\).