Question

Let $f(x) = 2x^2+5x+1$. If we write $f(x)$ as
$f(x) = a(x+1)(x-2) +b(x-2)(x-1)+c(x-1)(x+1)$
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)

Given, $f(x)=2 x^{2}+5 x+1\,\,\,...(i)$
Also, $f(x)=a(x+1)(x-2)+b(x-2)(x-1) +c(x-1)(x+1) $
$=a\left(x^{2}-x-2\right)+b\left(x^{2}-3 x+2\right) +c\left(x^{2}-1\right)$
$\Rightarrow f(x)=(a+b+c) x^{2}+(-a-3 b) x+(-2 a+2 b-c)\,\,\,...(ii)$
On equating the coefficients of $x^{2}, x$ and
constant term in Eqs. (i) and (ii), we get
$a+b+c =2,-a-3 b=5 $
and-$2 a+2 b-c =1$
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 $c .$