Question: Factor completely. Select “Prime” if the polynomial cannot be factored. \[ 36 + x^2 - 12x \]…

Factor completely. Select “Prime” if the polynomial cannot be factored.

\[ 36 + x^2 - 12x \]

\[ 36 + x^2 - 12x = \boxed{\phantom{0}} \]

Prime

Solution

To factor the polynomial \(36 + x^2 - 12x\), we need to rearrange it in descending order: \[ x^2 - 12x + 36 \] This is a quadratic trinomial. We will try to factor it into two binomials of the form \((x - a)(x - b)\). To do this, we need to find two numbers \(a\) and \(b\) such that: 1. \(a + b = -12\) 2. \(a \cdot b = 36\) Let’s check the pairs of factors of 36 and see if any add up to -12: - \(1 \cdot 36\) (sum is 37) - \(2 \cdot 18\) (sum is 20) - \(3 \cdot 12\) (sum is 15) - \(4 \cdot 9\) (sum is 13) - \(6 \cdot 6\) (sum is 12, but for correct sign should be -6 and -6) The numbers are \(-6\) and \(-6\): \[ \begin{cases} a = -6 & \\ b = -6 & \end{cases} \] Thus, the quadratic can be factored as: \[ (x - 6)(x - 6) = (x - 6)^2 \] Since it is factored into a perfect square and is not prime, the polynomial \(x^2 - 12x + 36\) factors completely as \((x - 6)^2\).