Use the factor theorem to determine whether g(x) is a factor of p(x) in the following case.
$p (x) = 2 x^{3} + x^{2} - 2 x - 1; g (x) = 2 x + 1$

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Solution

Here $p (x) = 2 x^{3} + x^{2} - 2 x - 1; g (x) = 2 x + 1$
Now
$p (- \frac{1}{2}) = 2 {(- \frac{1}{2})}^{3} + {(- \frac{1}{2})}^{2} - 2 (- \frac{1}{2}) - 1$
$= - \frac{1}{4} + \frac{1}{4} + 1 - 1 = 0$
$∵ p (- \frac{1}{2}) = 0$
i.e. $2 x + 1$ is a factor of $2 x^{3} + x^{2} - 2 x - 1.$

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Use the factor theorem to determine whether g(x) is a factor of p(x) in the following case.p(x)=2x3+x2−2x−1;g(x)=2x+1

Here p(x)=2x3+x2−2x−1;g(x)=2x+1Nowp(−12)=2(−12)3+(−12)2−2(−12)−1=−14+14+1−1=0∵p(−12)=0i.e. 2x+1 is a factor of 2x3+x2−2x−1.

Use the factor theorem to determine whether g(x) is a factor of p(x) in the following case.
$p (x) = 2 x^{3} + x^{2} - 2 x - 1; g (x) = 2 x + 1$

Here $p (x) = 2 x^{3} + x^{2} - 2 x - 1; g (x) = 2 x + 1$
Now
$p (- \frac{1}{2}) = 2 {(- \frac{1}{2})}^{3} + {(- \frac{1}{2})}^{2} - 2 (- \frac{1}{2}) - 1$
$= - \frac{1}{4} + \frac{1}{4} + 1 - 1 = 0$
$∵ p (- \frac{1}{2}) = 0$
i.e. $2 x + 1$ is a factor of $2 x^{3} + x^{2} - 2 x - 1.$