Question 2 Evaluate the polynomial x^3+2x^2-8x-2 when x=-2 square 1 pts

To evaluate the polynomial \( x^{3} + 2x^{2} - 8x - 2 \) at \( x = -2 \), substitute \(-2\) for \(x\) in the polynomial and simplify:<br /><br />\[<br />(-2)^{3} + 2(-2)^{2} - 8(-2) - 2<br />\]<br /><br />Calculate each term step by step:<br /><br />1. \((-2)^{3} = -8\)<br />2. \(2(-2)^{2} = 2 \times 4 = 8\)<br />3. \(-8(-2) = 16\)<br />4. The constant term is \(-2\).<br /><br />Now, add these results together:<br /><br />\[<br />-8 + 8 + 16 - 2<br />\]<br /><br />Simplify the expression:<br /><br />\[<br />0 + 16 - 2 = 14<br />\]<br /><br />Therefore, the value of the polynomial when \( x = -2 \) is \(\boxed{14}\).