Use the factor theorem to determine and state whether $g (x)$ is a factor of $p (x)$
$p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1$

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Solution

$p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1$
$g (x) = 0 \Rightarrow x + 1 = 0 \Rightarrow x = - 1$
$∴$ Zero of $g (x)$ is $- 1$
Now, $p (- 1) = 2 (- 1)^{3} + (- 1)^{2} - 2 (- 1) - 1$
$= - 2 + 1 + 2 - 1 = 0$
$∴$ By factor theorem, $g (x)$ is a factor of $p (x)$ .

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Use the factor theorem to determine and state whether g(x) is a factor of p(x) p(x)=2x3+x2−2x−1,g(x)=x+1

p(x)=2x3+x2−2x−1,g(x)=x+1g(x)=0⇒x+1=0⇒x=−1∴ Zero of g(x) is −1Now, p(−1)=2(−1)3+(−1)2−2(−1)−1=−2+1+2−1=0∴ By factor theorem, g(x) is a factor of p(x).

Use the factor theorem to determine and state whether $g (x)$ is a factor of $p (x)$
$p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1$

$p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1$
$g (x) = 0 \Rightarrow x + 1 = 0 \Rightarrow x = - 1$
$∴$ Zero of $g (x)$ is $- 1$
Now, $p (- 1) = 2 (- 1)^{3} + (- 1)^{2} - 2 (- 1) - 1$
$= - 2 + 1 + 2 - 1 = 0$
$∴$ By factor theorem, $g (x)$ is a factor of $p (x)$ .