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plz help

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Given that x=2 is a root of p(x)=x^4-3x^3-18x^2+90x-100, find the other roots (real and nonreal).

Oct 18, 2020

#1
+10820
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Given that x=2 is a root of p(x)=x^4-3x^3-18x^2+90x-100, find the other roots (real and nonreal).

Hello Guest!

$$x_1=2$$

$$(x^4-3x^3-18x^2+90x-100)$$ : $$(x-2)$$ $$x^3 - x^2 - 20 x + 50$$

$$\underline{x^4-2x^3}$$

$$-x^3-18x^2$$

$$\underline{-x^3\ +\ 2x^2}$$

$$-20x^2+90x$$

$$\underline{-20x^2+40x}$$

$$50x-100$$

$$\underline{50x-100}$$

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$$x_2=-5$$

$$(x^3 -\ x^2 - 20 x + 50)$$ : $$(x+5)$$ =  $$x^2-6x+10$$

$$\underline{x^3+5x^2}$$

$$-6x^2-20x$$

$$\underline{-6x^2-30x}$$

$$10x+50$$

$$\underline{ 10x+50}$$

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$$x^2-6x+10=0$$

$$x=3\pm \sqrt{9-10}$$

$$x_3=3+i\\ x_4=3-i$$

!

Oct 18, 2020