Question

If the polynomial \[ ax^4+bx^3+2x^2+4 \] is divisible by \[ x^2-x-2, \] then \(a+b=\)

• A. \(1\)
• B. \(-1\)
• C. \(3\)
• D. \(-6\)

Hint:

If a polynomial is divisible by another polynomial, substitute the roots of the divisor into the polynomial.

Answer 1

The Correct Option is A

Concept: If a polynomial is divisible by: \[ x^2-x-2 \] then its roots must satisfy the polynomial. Factorize: \[ x^2-x-2=(x-2)(x+1) \] Roots are: \[ x=2,\;-1 \]

Step 1: Substitute \(x=2\).
Given polynomial: \[ P(x)=ax^4+bx^3+2x^2+4 \] Since divisible: \[ P(2)=0 \] \[ 16a+8b+2(4)+4=0 \] \[ 16a+8b+12=0 \] Divide by \(4\): \[ 4a+2b+3=0 \qquad ...(1) \]

Step 2: Substitute \(x=-1\).
\[ P(-1)=0 \] \[ a-b+2(1)+4=0 \] \[ a-b+6=0 \] \[ a-b=-6 \qquad ...(2) \]

Step 3: Solve equations.
From (2): \[ a=b-6 \] Substitute into (1): \[ 4(b-6)+2b+3=0 \] \[ 4b-24+2b+3=0 \] \[ 6b-21=0 \] \[ b=\frac72 \] Then: \[ a=\frac72-6 \] \[ =-\frac52 \] So: \[ a+b=-\frac52+\frac72 \] \[ =\frac22 \] \[ =1 \] Thus, the required answer is: \[ \boxed{1} \]