Question:
Write the coefficient of x2 in each of the following
(i) $\frac{\pi}{6} x+x^{2}-1$
(ii) $3 x-5$
(iii) $(x-1)(3 x-4)$
(iv) $(2 x-5)\left(2 x^{2}-3 x+1\right)$
Solution:
(i) The coefficient of $x^{2}$ in $\frac{\pi}{6} x+x^{2}-1$ is 1 .
(ii) The coefficient of $x^{2}$ in $3 x-5$ is $0 .$
(iii) Let $p(x)=(x-1)(3 x-4)$
$=3 x^{2}-7 x+4$
$=3 x^{2}-4 x-3 x+4$
Hence, the coefficient of $x^{2}$ in $p(x)$ is 3 .
(iv) Let $p(x)=(2 x-5)\left(2 x^{2}-3 x+1\right)$
$=2 x\left(2 x^{2}-3 x+1\right)-5\left(2 x^{2}-3 x+1\right)$
$=4 x^{3}-6 x^{2}+2 x-10 x^{2}+15 x-5$
$=4 x^{3}-16 x^{2}+17 x-5$
Hence, the coefficient of $x^{2}$ in $p(x)$ is $-16$.