Use the Factor Theorem to determine whether g(x) is a factor of p(x)
$p (x) = x^{3} - 4 x^{2} + x + 6, g (x) = x - 3$

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Solution

g(x) = x - 3, substitute x = 3 in p(x).
$p (x) = x^{3} - 4 x^{2} + x + 6 \Rightarrow p (3) = (3)^{3} - 4 (3)^{2} + 3 + 6 = 27 - 36 + 3 + 6 = 0$
Therefore, g(x) is a factor of $x^{3} - 4 x^{2} + x + 6$ .

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Similar questions

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

g (x)

f (x)

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

Use the factor theorem to determine and state whether $g (x)$ is a factor of $p (x)$ in each of the following cases:

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

g (x)

f (x)

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

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