Question

Let$f\left(x\right)=2x^2+5x+1$. If we write$f\left(x\right)$ as $f\left(x\right)=\left[a\left(x+1\right)\left(x-2\right)+b\left(x-2\right)\left(x-1\right)+c\left(x-1\right)\left(x+1\right)\right]$ for real numbers$a,b,c$, then

• A. There are infinite numbers of choices for a, b and c
• B. Only one choice for a but infinite numbers of choices for b and c
• C. Exactly one choice for each of a, b and c
• D. More than one but finite number of choices for a, b and c

Answer 1

The correct option is C

Exactly one choice for each of a, b and c

Explanation for the correct option:

$f\left(x\right)=2x^2+5x+1$ and $f\left(x\right)=\left[a\left(x+1\right)\left(x-2\right)+b\left(x-2\right)\left(x-1\right)+c\left(x-1\right)\left(x+1\right)\right]$

Equating and comparing the coefficients

$$\begin{gathered} 2x^2+5x+1=a\left(x+1\right)\left(x-2\right)+b\left(x-2\right)\left(x-1\right)+c\left(x-1\right)\left(x+1\right) \\ \Rightarrow 2x^2+5x+1=a\left(x^2-x-2\right)+b\left(x^2-3x+2\right)+c\left(x^2-1\right) \\ \Rightarrow 2x^2+5x+1=x^2\left(a+b+c\right)+x\left(-a-3b-c\right)-2a+2b-c \end{gathered}$$

This gives

$$\begin{gathered} a+b+c=2 \\ a+3b=-5 \\ -2a+2b-c=1 \end{gathered}$$

Solving the equations, we get

$a=-4,b=\frac{-1}{3},c=\frac{9}{3}$

Hence, option (C) is the correct answer