THIS IS WORTH 50 POINTS!!I’ve been stuck on this for awhile now, I don’t know if anyone will understand this, but I’m praying that

THIS IS WORTH 50 POINTS!! I’ve been stuck on this for awhile now, I don’t know if anyone will understand this, but I’m praying that someone does, I’ll send more pictures of the problem!


[tex]THIS IS WORTH 50 POINTS!! I’ve been stuck on this for awhile now, I don’t know if anyone will und[/tex]

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  1. Notice that when [tex]x=-1[/tex], you have

    [tex](-1)^3+2(-1)+3=-1-2+3=0[/tex]

    so [tex]x=-1[/tex] is a solution, and the cubic has [tex]x+1[/tex] as a factor. This means you can divide to get another quadratic factor and thus two more solutions.

    [tex]\dfrac{x^3+2x+3}{x+1}=x^2-x+3[/tex]

    So,

    [tex]x^3+2x+3=(x+1)(x^2-x+3)=0[/tex]

    You can use the quadratic formula here:

    [tex]x=\dfrac{1\pm\sqrt{1-12}}2=\dfrac12\pm i\dfrac{\sqrt{11}}2[/tex]

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