Question

simon is factoring the polynomial.
$x^2 - 4x - 12$
$(x - 6)(x + \underline{quadquad})$
what value should simon write on the line?
$\bigcirc$ $-6$
$\bigcirc$ $-2$
$\bigcirc$ $2$
$\bigcirc$ $6$

Explanation:

Step1: Recall factoring rules

To factor \(x^2 + bx + c\), we find two numbers that multiply to \(c\) and add to \(b\). Here, the polynomial is \(x^2 - 4x - 12\), and we have one factor \((x - 6)\). Let the missing number be \(a\), so we expand \((x - 6)(x + a)\).

Step2: Expand the product

Expanding \((x - 6)(x + a)\) gives \(x^2 + ax - 6x - 6a = x^2 + (a - 6)x - 6a\). This should match \(x^2 - 4x - 12\).

Step3: Equate coefficients

For the constant term: \(-6a = -12\). Solving for \(a\), divide both sides by \(-6\): \(a=\frac{-12}{-6}=2\)? Wait, no, wait. Wait, the middle term: \(a - 6=-4\). So \(a=-4 + 6 = 2\)? Wait, no, wait the constant term: \(-6a=-12\) so \(a = 2\)? Wait, but let's check again. Wait, the polynomial is \(x^2-4x - 12\). If we factor it, we need two numbers that multiply to \(-12\) and add to \(-4\). The numbers are \(-6\) and \(2\), because \(-6\times2=-12\) and \(-6 + 2=-4\). So the factors are \((x - 6)(x + 2)\). Wait, so the missing number is \(2\)? Wait, but let's check the expansion: \((x - 6)(x + 2)=x^2+2x-6x - 12=x^2-4x - 12\), which matches. So the missing value is \(2\).

Answer:

C. 2 (assuming the options are labeled as A. -6, B. -2, C. 2, D. 6)