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Question
By using factor theorem in the following example, determine whether q(x) is a factor p(x) or not.
p(x) = x3 − x2 − x − 1, q(x) = x − 1
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Solution
p(x) = x3 − x2 − x − 1
Divisor = q(x) = x − 1
∴ p(1) = (1)3 - (1)2 − 1 − 1
= 1 − 1 − 1 − 1
= − 2 ≠ 0
Since p(1) ≠ 0, so by factor theorem q(x) = x − 1 is not a factor of polynomial p(x) = x3 − x2 − x − 1.
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