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 Math Help > Basic Algebra > Factoring > Solving cubic, quartic, higher degree > Ferrari's method of solving a quartic equation

### Quartic equation

This is the method for solving any quartic equation.
Say we have:
ax4+bx3+cx2+dx+e=0 .
Divide by a , to rewrite this as:
x4+fx3+gx2+hx+j=0 .
Put y=x+f/4 , so x=yf/4 , so

 x4 = y4−f3y+3f2y2/8−f3y/16+f4/256 x3 = y3−3y2f/4+3yf2/16−f3/64 x2 = y2−2yf/4+f2/16

so x4+fx3+gx2+hx+j=y4fy3+3f2/y2/8f3y/16+f4/256+fy33y2f2/4+3yf3/16f4/64+gy2gyf/2+gf2/16+hyhf/4+j
=y4+y3(f+f) y2(3f2/8)32/4+4) +y(f3/16+3f3/16gf/2+h) +
 � �
f4/256f4/64+gf2/16hf/4+j
 � �
So we can rewrite the original equation as:
y4+py2+qy+r=0
with p=3f2/8+g ,
q=f3/8gf/2+h ,
r=3f4/256+gf2/16hf/4+j .
Now, we can write:
y4+py2=qyr
y4+2py2+p2=py2qyr+p2
(y2+p)2=py2qyr+p2 .
Now, for any z ,
(y2+p+z)2=((y2+p)+z)2
=(y2+p)2+2(y2+p)z+z2
=py2qyr+p2+2z(y2+p)+z2
=(p+2z)y2qy+(p2r+2pz+z2) (*)
The right hand side of (*) is a quadratic in y ; and we can choose z so that it is a perfect square, i.e. so that the discriminant is zero, ie:
(q)24(p+2z)(p2r+2pz+z2)=0 .
We can rewrite this as (q24p3+4pr)+(16p2+8r)z20pz28z3=0 .
This is a cubic equation in z . So we can solve it for z using the cubic equation formula (Cardano's method).
When we have solved this to find a value for z , we can substitute in this value of z to (*). This makes the right hand side of (*) a perfect square, so we can take the square root of both sides of (*). (*) is then a quadratic equation in y , which we can solve using the quadratic solution formula. The values of y will easily give us values of x , i.e. solutions of the original equation.
There can be 0, 1, 2, 3, or 4 real solutions.

### Internet references

Maths.org: Ferrari's method of solving quartic polynomials

Wikipedia: Quartic equation -- look at the quick and memorable solution from first principles, near the bottom.