Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = x^{3} - 8, g (x) = - 2$

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Solution

Let:
$p (x) = x^{3} - 8 g (x) = x - 2 x = 2$

By the factor theorem, g(x) is a factor of p(x) if p(2) = 0.

Thus, we have:

$p (x) = x^{3} - 8 = 2^{3} - 8 = 8 - 8 = 0$

Hence, (x - 2) is a factor of the given polynomial.

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Similar questions

Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6, g (x) = 2 x - 3$

p (x) = 2 \sqrt{2} x^{2} + 5 x + \sqrt{2}, g (x) = x + \sqrt{2}

Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = 2 \sqrt{2} x^{2} + 5 x + \sqrt{2}, g (x) = x + \sqrt{2}$

(i) p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1

(i i) p (x) = x^{3} + 3 x^{2} + 3 x + 1, g (x) = x + 2

(i i i) p (x) = x^{3} - 4 x^{2} + x + 6, g (x) = x - 3

g (x)

p (x)

p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1

p (x) = x^{3} + 3 x^{2} + 3 x + 1, g (x) = x + 2

p (x) = x^{3} - 4 x^{2} + x + 6, g (x) = x - 3

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