Question

A polynomial \( p(x) \) in x leaves remainder 5 and 6 when the polynomial \( p(x) \) is divided by \( x+1 \) and \( x+3 \) respectively. If it leaves remainder \( ax+b \) when divided by \( x^{2}+4x+3 \), then \( 4a+6b = \)

• A. 88
• B. 66
• C. 77
• D. 25

Hint:

If a polynomial leaves remainder \(R(x)\) on division by a quadratic divisor, substitute the roots of the divisor into \(R(x)\). This quickly gives equations for the coefficients of the remainder.

Answer 1

The Correct Option is D

Step 1: Express the polynomial in quotient-remainder form.
Since the divisor is \[ x^2+4x+3=(x+1)(x+3), \] the remainder must be a polynomial of degree less than 2. Hence, \[ p(x)=(x+1)(x+3)Q(x)+(ax+b), \] where \(Q(x)\) is the quotient and \(ax+b\) is the remainder.

Step 2: Use the given remainders to form equations.
When \(p(x)\) is divided by \(x+1\), the remainder is 5. By the Remainder Theorem, \[ p(-1)=5. \] Substituting \(x=-1\), \[ (-1+1)(-1+3)Q(-1)+(-a+b)=5. \] Since \((-1+1)=0\), \[ -a+b=5. \] Similarly, when \(p(x)\) is divided by \(x+3\), the remainder is 6. Therefore, \[ p(-3)=6. \] Substituting \(x=-3\), \[ (-3+1)(-3+3)Q(-3)+(-3a+b)=6. \] Since \((-3+3)=0\), \[ -3a+b=6. \] Thus we obtain the system \[ -a+b=5 \qquad ...(1) \] \[ -3a+b=6 \qquad ...(2) \]

Step 3: Find the values of \(a\) and \(b\).
Subtracting (2) from (1), \[ (-a+b)-(-3a+b)=5-6 \] \[ 2a=-1 \] \[ a=-\frac12. \] Substituting in (1), \[ -\left(-\frac12\right)+b=5 \] \[ \frac12+b=5 \] \[ b=\frac92. \]

Step 4: Evaluate \(4a+6b\).
\[ 4a+6b = 4\left(-\frac12\right) + 6\left(\frac92\right) \] \[ =-2+27 \] \[ =25. \] Therefore, \[ \boxed{4a+6b=25} \] {25}