Use the factor theorem to determine whether g(x) is a factor of p(x) in the following case.
$p (x) = 2 x^{4} - 7 x^{3} - 13 x^{2} + 63 x - 45; g (x) = x - 1$

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Solution

Here
$p (x) = 2 x^{4} - 7 x^{3} - 13 x^{2} + 63 x - 45; g (x) = x - 1$
Now
$p (1) = 2 (1)^{4} - 7 (1)^{3} - 13 (1)^{2} + 63 (1) - 45$
$= - 65 + 65 = 0$
$p (1) = 0$
i.e. $x - 1$ is a factor of $2 x^{4} - 7 x^{3} - 13 x^{2} + 63 x - 45$

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Q. Using factor theorem, factorize:

p (x) = 2 x^{4} - 7 x^{3} - 13 x^{2} + 63 x - 45

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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)=2x4−7x3−13x2+63x−45;g(x)=x−1

Herep(x)=2x4−7x3−13x2+63x−45;g(x)=x−1Nowp(1)=2(1)4−7(1)3−13(1)2+63(1)−45=−65+65=0p(1)=0i.e. x−1 is a factor of 2x4−7x3−13x2+63x−45

Use the factor theorem to determine whether g(x) is a factor of p(x) in the following case.
$p (x) = 2 x^{4} - 7 x^{3} - 13 x^{2} + 63 x - 45; g (x) = x - 1$

Here
$p (x) = 2 x^{4} - 7 x^{3} - 13 x^{2} + 63 x - 45; g (x) = x - 1$
Now
$p (1) = 2 (1)^{4} - 7 (1)^{3} - 13 (1)^{2} + 63 (1) - 45$
$= - 65 + 65 = 0$
$p (1) = 0$
i.e. $x - 1$ is a factor of $2 x^{4} - 7 x^{3} - 13 x^{2} + 63 x - 45$