Using factor theorem, show that $x^{2} + 5 x + 6$ is a factor of $x^{4} + 5 x^{3} + 9 x^{2} + 15 x + 18$ .

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Solution

To show that $x^{2} + 5 x + 6$ is a factor of $x^{4} + 5 x^{3} + 9 x^{2} + 15 x + 18$

When we multiply $x^{2} + 5 x + 6$ with some other factors, we should get $x^{4} + 5 x^{3} + 9 x^{2} + 15 x + 18$

So, when we multiply $x^{2} + 5 x + 6$ with $(x^{2} + 3)$ we get,

$(x^{2} + 5 x + 6) \times (x^{2} + 3)$

$= x^{4} + 5 x^{3} + 6 x^{2} + 3 x^{2} + 15 x + 18$

$= x^{4} + 5 x^{3} + 9 x^{2} + 15 x + 18$

Hence proved.

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Similar questions

Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = x^{4} - x^{2} - 12, g (x) = x + 2$

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f (x) = 5 x^{3} + x^{2} - 5 x - 1, g (x) = x + 1

x^{2} - 5 x + 6

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(2 \sqrt{2} x^{2} + 5 x + \sqrt{2})

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