Examine whether x + 2 is a factor of $x^{3} + 3 x^{2} + 5 x + 6$ and 2x + 4.

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Solution

The zero of x + 2 is -2. Let p(x) = $x^{3} + 3 x^{2} + 5 x + 6$ and s(x) = 2x + 4
Then, $p (- 2) = (- 2)^{3} + 3 (- 2)^{2} + 5 (- 2) + 6$
= -8 + 12 - 10 + 6 = 0
So, by the Factor Theorem, x + 2 is a factor of $x^{3} + 3 x^{2} + 5 x + 6$ .
Again, s(-2) = 2(-2)+4 = 0
So, x + 2 is a factor of 2x + 4

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