I Solved A Polynomial Equation | Math Olympiads 

For the 2nd method, I recommend substitution for y = x + 3. It is not required expansion of 4th-power term.

I approached this as follows. I rewrote the original expression as (x+1)^2-1+(x+2)^2=1-(x+3)^4. Decomposing the two differences of squares and rearranging I get: (x+2)*(x^2+5x+4)=-(x^2+6x+8)*(x^2+6x+10). Further decomposition brings to (x+2)*(x+4)*(x+1)=-(x+2)*(x+4)*(x^2+6x+10). From here we see that x=-2 and x=-4 are two solutions. After simplification we remain with x+1=-(x^2+6x+10). Rearranging and putting together the like terms gives x^2+7x+11=0. From here we get the last two real solutions

X=-7+(5^(1/2))/2,-7-(5^(1/2))/2

1st method's quartic equation can be solved by the method of undetermined coefficients. x⁴+13x³+61x²+122x+88=(x²+Ax+B)(x²+Cx+D)=0 x⁴+Ax³+Cx³+ACx²+Bx²+Dx²+ADx+BCx+BD=x⁴+13x³+61x²+122x+88 x⁴+(A+C)x³+(AC+B+D)x²+(AD+BC)x+BD=x⁴+13x³+61x²+122x+88 A+C=13; AC+B+D=61; AD+BC=122; BD=88. Let's assume that the coefficients are integers. BD=88>0, so B and D must be both positive OR negative at the same time. 88=1*88=2*44=4*22=8*11 (±). 1) BD=1*88: A+C=13; AC+1+88=61; A+88C=122. A=13-C=122-88C. C-13=88C-122; 87C=109; C=109/87. C isn't an integer, so BD≠1*88. 2) BD=2*44: A+C=13; AC+2+44=61; 2A+44C=122. A=13-C=61-22C. C-13=22C-61; 21C=48; C=48/21=16/7. C isn't an integer, so BD≠2*44. 3) BD=4*22: A+C=13; AC+4+22=61; 4A+22C=122. A=13-C=30,5-5,5C. 2C-26=11C-61; 9C=35; C=35/9. C isn't an integer, so BD≠4*22. 4) BD=8*11: A+C=13; AC+8+11=61; 8A+11C=122. 8A=104-8C=122-11C. 8C-104=11C-122; 3C=18; C=6 ✅; A=13-C=13-6=7. A=7; B=11; C=6; D=8, so x⁴+13x³+61x²+122x+88=(x²+7x+11)(x²+6x+8)=0. A) x²+7x+11=0. D=49-4*11=5>0; x1,2=(-7±√5)/2. B) x²+6x+8=0. D=36-4*8=4>0; x3=(-6+2)/2=-4/2=-2; x4=(-6-2)/2=-8/2=-4. So, the equation's roots is (-7±√5)/2; -4; -2.

I got -4 and -2 just fine, but my other solutions ended up being complex.

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It is obvious that -2 is the root. It turns into cubic. Maybe we can use cubic formula to solve it then. Problem solved. 🙂🙂🙂🙂🙂🙂