Question

If \(P(x)=x^3+ax^2+bx+c\) is a polynomial such that \(P(1)=0\), \(P(2)=0\), and \(P(3)=6\), then \(6a+4b+3c=\)

• A. \(0\)
• B. \(-8\)
• C. \(-10\)
• D. \(10\)

Hint:

Whenever several roots are given, immediately factorize the polynomial using those roots. This drastically reduces the amount of algebra required.

Answer 1

The Correct Option is B

Step 1: Use the conditions \(P(1)=0\) and \(P(2)=0\). Since 1 and 2 are roots, \[ P(x)=(x-1)(x-2)(x-k) \] for some constant \(k\).

Step 2: Use \(P(3)=6\). \[ P(3) = (3-1)(3-2)(3-k) \] \[ =2(1)(3-k) \] \[ =6 \] Thus, \[ 3-k=3 \] \[ k=0 \] Hence \[ P(x)=x(x-1)(x-2) \]

Step 3: Expand the polynomial. \[ P(x)=x(x^2-3x+2) \] \[ =x^3-3x^2+2x \] Comparing with \[ x^3+ax^2+bx+c \] gives \[ a=-3,\qquad b=2,\qquad c=0 \]

Step 4: Compute the required expression. \[ 6a+4b+3c \] \[ =6(-3)+4(2)+3(0) \] \[ =-18+8 \] \[ =-10 \] \[ \boxed{-10} \]