Solve the 次方 equation

KennySir

Hello~

I want to ask something about solve the equation.

When we see the equation,
f(x)=ax^2+bx+c
it can be used the formula to solve it that is,
x=[-b±√(b^2-4ac)]/2a
then can find x or no answer(that no intersection in y=0)!

if we don't use (x+e)(x+f) to solve,
how can I solve the cubic equation,
f(x)=ax^3+bx^2+cx+d
or quartic equation,
or more.

or that,
my goal is going to prove that,
there is no answer in the f(x).
Any other formula or method can help me!

Thank you!

Naozumi

Your question is well-answered in Wikipedia, just by searching cubic equation and quartic equation. These formuale are quite complicated.

For general polynomial equation of degree 5 above, it can be proved that there is no simple formula (only involve finite operations of surds and arithmetic operations on the coefficients of equations). For details, you may need to study tertiary level of Algebra (concepts og groups)

陸小軒

Maybe I can help you...
for example:
f(x)=ax^3+bx^2+cx+d
a=2 b=1 c=-8 d=-4
f(x)=2x^3+x^2-8x-4
[Because the factors of the coefficient of x^3 are+1,+2 and the factors of the constant term-4 are +1,+2,+4(1x4,2x2,4x1)
So,the possbile linear factors of f(x) are x+1,x+2,x+4,2x+1]
By the factor theorem,
f(2)=2(2)^3+2^2-8(2)-4=0
So x-2 is a factor[f(2),this 2 is come form,x-2=0→x=2]

then use long division,2x^3+x^2-8x-4 divided by x-2
so
f(x)=(x-2)(2x^2+5x+2) (←x=[-b±√(b^2-4ac)]/2a)
      =(x-2)(2x+1)(x+2)

Hope that can help understand

davidmaths11

Method to solve Cubic Equation
Cubic Formula

五次及更高次的代數方程沒有一般的代數解法,可以用Galois Group proof