Question

If \( x^2 + x - 6 \) is a factor of \( 2x^3 + x^2 + ax + b \), then \( 6a + 13b = \) ?

• A. \( 305 \)
• B. \( 133 \)
• C. \( \mathbf{0} \)
• D. \( -1 \)

Hint:

Use remainder theorem by plugging in roots of the factor into the polynomial to create equations.

Answer 1

The Correct Option is C

Factor \( x^2 + x - 6 = (x - 2)(x + 3) \) Given it divides \( 2x^3 + x^2 + ax + b \), let’s perform polynomial division or use remainder theorem: Substitute roots into the cubic: Let \( f(x) = 2x^3 + x^2 + ax + b \) \[ f(2) = 16 + 4 + 2a + b = 0 \Rightarrow 2a + b = -20 \quad \text{(1)} \] \[ f(-3) = -54 + 9 -3a + b = 0 \Rightarrow -3a + b = 45 \quad \text{(2)} \] Solving equations (1) and (2): From (1): \( b = -20 - 2a \) Substitute into (2): \[ -3a + (-20 - 2a) = 45 \Rightarrow -5a = 65 \Rightarrow a = -13,\quad b = 6 \] Then \( 6a + 13b = 6(-13) + 13(6) = -78 + 78 = \boxed{0} \)