How do you factor $x^{3} + 10 x^{2} + 33 x + 34$ $x^3 + 10x^2 + 33x + 34$ ?

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How do you factor $x^{3} + 10 x^{2} + 33 x + 34$ $x^3 + 10x^2 + 33x + 34$ ?

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Answer 1 · 2016-07-08T12:51:49

$x^{3} + 10 x^{2} + 33 x + 34 =^{=} (x^{2} + 8 x + 17) (x + 2)$ $x^3+10x^2+33x+34=^=(x^2+8x+17)(x+2)$

Explanation:

As putting $x = - 2$ $x=-2$ , we get $x^{3} + 10 x^{2} + 33 x + 34 = {(- 2)}^{3} + 10 \times 2^{2} + 33 \times (- 2) + 34 = - 8 + 40 - 66 + 34 = 0$ $x^3+10x^2+33x+34=(-2)^3+10xx2^2+33xx(-2)+34=-8+40-66+34=0$ . Hence, $- 2$ $-2$ is a zero of the function and $(x + 2)$ $(x+2)$ is a factor of $x^{3} + 10 x^{2} + 33 x + 34$ $x^3+10x^2+33x+34$ .

Now dividing the function by $(x + 2)$ $(x+2)$ , as $x^{3} + 10 x^{2} + 33 x + 34 =^{=} x^{2} (x + 2) + 8 x (x + 2) + 17 (x + 2) = (x^{2} + 8 x + 17) (x + 2)$ $x^3+10x^2+33x+34=^=x^2(x+2)+8x(x+2)+17(x+2)=(x^2+8x+17)(x+2)$ , we get $x^{2} + 8 x + 17$ $x^2+8x+17$ .

Now discriminant of $x^{2} + 8 x + 17$ $x^2+8x+17$ is $8^{2} - 4 \cdot 1 \cdot 17 = 64 - 68 = - 4$ $8^2-4*1*17=64-68=-4$ is negative, as such it cannot be factorized in real factors.

And hence factors of the function are

$x^{3} + 10 x^{2} + 33 x + 34 =^{=} (x^{2} + 8 x + 17) (x + 2)$ $x^3+10x^2+33x+34=^=(x^2+8x+17)(x+2)$

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How do you factor $x^{3} + 10 x^{2} + 33 x + 34$ $x^3 + 10x^2 + 33x + 34$ ?

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