Multiple Choice 10 points k^2-11k+28 (k-4)(k-7) (k+4)(k+7) (k+4)(k-7) (k+9)(k-7)

The correct answer is $(k-4)(k-7)$.<br /><br />To solve this problem, we need to factor the quadratic expression $k^{2}-11k+28$.<br /><br />The factored form of a quadratic expression is $(k-a)(k-b)$, where $a$ and $b$ are the roots of the equation.<br /><br />To find the roots, we can use the quadratic formula:<br />$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />In this case, $a = 1$, $b = -11$, and $c = 28$.<br /><br />Plugging these values into the quadratic formula, we get:<br />$k = \frac{11 \pm \sqrt{121 - 112}}{2}$<br />$k = \frac{11 \pm \sqrt{9}}{2}$<br />$k = \frac{11 \pm 3}{2}$<br /><br />So the roots are $k = 4$ and $k = 7$.<br /><br />Therefore, the factored form of the quadratic expression is $(k-4)(k-7)$.