By actual division , shows that $x^{2} - 3$ is a factor of $2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12.$

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Solution

As we can see the remainder comes out to be zero. Hence we can say that $x^{2} - 3$ is a factor of $2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12.$

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Similar questions

2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2

x^{2} - 3 x + 2

2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2

x^{2} - 3 x + 2

g (x)

p (x)

p (x) = 2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12 g (x) = x^{2} - 3

2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2

x^{2} - 3 x + 2

2 x^{4} + 2 x^{2} - x + 2

x^{2} - 3 x + 2

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