Answer:
C. -3 must be a root of the polynomial 2·x² + 9·x + 9
Step-by-step explanation:
Given that the remainder when (2·x² + 9·x + 9) is divided by (x + 3) equals zero, then (x + 3) is a factor of 2·x² + 9·x + 9 and -3 is a root of the polynomial
Explaining by using the long division of the polynomial is presented as follows;
<u>2·x + 3</u>
(2·x² + 9·x + 9)/(x + 3)
<u>- (2·x² + 6·x)</u>
3·x + 9
-(3·x + 9)
0
Therefore, (2·x + 3) × (x + 3) = 2·x² + 9·x + 9
The solution of the 2·x² + 9·x + 9 = 0 gives the root of the polynomial
The root is given by, 2·x² + 9·x + 9 = (2·x + 3) × (x + 3) = 0
Therefore, the roots are;
2·x = -3 or x = -3/2 and x = -3
Therefore;
-3 must be a root of the polynomial 2·x² + 9·x + 9.