Question:
The zeros of the polynomial $p(x)=2 x^{2}+5 x-3$ are
(a) $\frac{1}{2}, 3$
(b) $\frac{1}{2},-3$
(c) $\frac{-1}{2}, 3$
(d) $1, \frac{-1}{2}$
Solution:
The given polynomial is $p(x)=2 x^{2}+5 x-3$.
Putting $x=\frac{1}{2}$ in $p(x)$, we get
$p\left(\frac{1}{2}\right)=2 \times\left(\frac{1}{2}\right)^{2}+5 \times \frac{1}{2}-3=\frac{1}{2}+\frac{5}{2}-3=3-3=0$
Therefore, $x=\frac{1}{2}$ is a zero of the polynomial $p(x)$.
Putting x = –3 in p(x), we get
$p(-3)=2 \times(-3)^{2}+5 \times(-3)-3=18-15-3=0$
Therefore, x = –3 is a zero of the polynomial p(x).
Thus, $\frac{1}{2}$ and $-3$ are the zeroes of the given polynomial $p(x)$.
Hence, the correct answer is option (b).