Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x³ – 8, g(x) = x – 2

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Solution

Let:
p(x) = x³ – 8
Now,
$g (x) = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$
By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:
$p (2) = (2^{3} - 8) = 0$
Hence, (x $-$ 2) is a factor of the given polynomial.

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Using factor theorem, show that g(x) is a factor of p(x), when
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