Verify weather $2,\ 3\ and \frac{1}{2}$ are the zeroes of the polynomial $P(x)=2x^{3}-11x^{2}+17x-6$.
Given: The polynomial $P(x)=2x^{3}-11x{2}+17x-6$.
To do: To verify weather $2,\ 3\ and\ \frac{1}{2}$ are the zeroes of the given polynomial.
Solution:
$p(x)=2x^{3}-11x^{2}+17x-6$
Putting $x=2$,
$p(2)=2(2)^{3}-11(2)^{2}+17(2)-6$
$=2(8)-11(4)+34-6$
$=16-44+28$
$=44-44=0$
Putting $x=3$,
$p(3)=2( 3)^{3}-11( 3)^{2}+17( 3)-6$
$=2(27)-11(9)+51-6$
$=54-99+51-6$
$=99-99=0$
Putting $x=\frac{1}{2}$,
$p( \frac{1}{2})=2( \frac{1}{2})^{3}-11( \frac{1}{2})^{2}+17( \frac{1}{2})-6$
$=2( \frac{1}{8})-11( \frac{1}{8})+ \frac{17}{2}-6$
$= \frac{2}{8}- \frac{11}{8}+ \frac{17}{2}-6$
$= \frac{1}{4}- \frac{11}{4}+ \frac{17}{2}-6$
$=\frac{( 1-11)}{4}+\frac{17}{2}-6$
$=-\frac{10}{4}+\frac{17}{2}-6$
$=\frac{(-10+34-24)}{4}$
$=\frac{(34-34)}{4}$
$=\frac{0}{4}$
$=0$
Therefore, $2,\ 3$ and $\frac{1}{2}$ are the zeroes of the given polynomial.
Hence verified
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