Solve the equation (x^2+x+1)*(x^2+x+2)=12

[Math Tutor]

[tex](x^2+x+1)*(x^2+x+2)=12[/tex]

[Guest]

Spoiler: show

Let [tex]y=x^2+x[/tex], the equation becomes an easily solvable quadratic [tex](y+1)(y+2)=12[/tex].
[tex]y= 2[/tex], or [tex]-5[/tex], so [tex]x^2+x = 2[/tex], or [tex]-5[/tex]. Solving these quadratics give
[tex]x=-2, 1,(-1+i\sqrt{19})/2,(-1-i\sqrt{19})/2[/tex]

R. Baber.

[Guest]

Alternatively

Spoiler: show

Use the difference of two squares formula to rewrite the LHS as [tex]((x+1/2)^2+5/4)^2-(1/2)^2[/tex], it is now easy to rearrange and solve for [tex]x[/tex].

or

Spoiler: show

Expand and rearrange to get [tex]x^4+2 x^3+4 x^2+3 x-10=0[/tex], by considering the factors of [tex]10[/tex], you can guess that [tex]x=1[/tex] and [tex]x=-2[/tex] are solutions, which means [tex]x-1[/tex] and [tex]x+2[/tex] are factors of [tex]x^4+2 x^3+4 x^2+3 x-10[/tex], and so [tex]x^4+2 x^3+4 x^2+3 x-10 = (x-1)(x+2)(x^2+x+5)=0[/tex] allowing you to calculate the complex roots.

R. Baber.

[Guest]

Great!