A Polynomial Equation | Solving Special Cubics 😊

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[@Qermaq]

Fun I've been practicing ones like this. I noted that if x is real it's got to be negative, and hopefully it's a factor of 4. Once I found that (x + 2) is a factor of the cubic, it was easy there on.

[@nizamkanchon3311]

Plz upload videos on linear algebra, vector , and matrix

[@DonRedmond-jk6hj]

you could also use the rat'l root thm and synthetic division. It takes 4 lines.

[@lawrencejelsma8118]

I wonder why high school math teachers never taught his class rational root theorem of factors of the last coefficient "plug and try" to know x= -2 is a factor. Then - 2 Synthetic divides into 1 1 0 4 as x^2 - x + 2 and then 0 remainder. So by high school Junior or Sophomore coursework students quickly obtain (x + 2)(x^2 - x + 2) that can be easily checked by multiplying those two factors with the last factor a tricky quadratic factor.

[@lagomoof]

3rd method (incomplete): Observe that x cannot be zero. Divide through by 4x^3. Substitute y = 1/x and now the cubic in y is in the correct form for Cardano's method.

[@nasrullahhusnan2289]

Due to small constant, by scrutinizing the equation it is clear that x=-2

[@phill3986]

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[@nasrullahhusnan2289]

x³+x²+4=(x²+ax+b)(x+c)
=x³+(c+a)x²+(ac+b)x+bc
c+a=1 (1)
ac+b=0 (2)
bc=4 (3)
Multiply (2) by c: ac²+bc=0
Noting (3) ac²+4=0
ac=-4 (4)
Multiply (1) by c²: c³+ac²=c²
Noting (4) c³-4=c²
c²(c-1)=4
=2²(2-1)
c=2, b=2, a=-1
Thus 0=x³+x²+4
=(x²--x+2)(x+2)
x=-2 or x²-x+2=0
x=½[1±sqrt(-7)]
=½[1±isqrt(7)]

[@ftthgvhg2313]

Nice