Question

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

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

Answer 1

Answer: (C)

Exactly one choice for each of $a,b,c$

Given, $f\left(x\right)=2x^2+5x+1$ ......(i)
Also, $f\left(x\right)=a(x+1)(x-2)+b(x-2)(x-1)+c(x-1)(x+1)$
$=a(x^2-x-2)+b(x^2-3x+2)+c(x^2-1)$
$\Rightarrow f\left(x\right)=(a+b+c)x^2+(-a-3b)x+(-2a+2b-c)$ .......(ii)
On equating the coefficients of $x^2,x$ and constant term in equations (i) and (ii), we get
$a+b+c=2,-1-3b=5$
and $-2a+2b-c=1$
On solving above equations, we get
$a=-\frac{35}{4},b=\frac{5}{4}$ and $c=\frac{38}{4}$
Hence, exactly one choice for each of $b$ and $c$.