Question

how to solve ax^4+bx^3+cx^2+dx+e

Answer 1

Quartic equations that take one of these four forms can be solved using only the quadratic equation, with no need to apply the more complicated formulas above.

x4 + bx3 + cx2 = 0
Hint: Rewrite it as x2(x2 + bx + c) = 0.

x4 + cx2 + e = 0
Hint: Use the quadratic formula to solve for x2.

x4 + 2mx3 + (m2 + 2n)x2 + 2mnx + e = 0
Hint: Rewrite it as x4 + 2mx3 + (m2 + 2n)x2 + 2mnx + n2 = n2 - e.
Factor the left side into (x2 + mx + n)2, then take the square root of both sides.


Example=Solve the quartic equation x4 - 4x3 + 5x2 - 4x + 4 = 0.

x4 - 4x3 = -5x2 + 4x - 4
x4 - 4x3 + (4 + 2p)x2 - 4px + p2 = (4 + 2p)x2 - 4px + p2 -5x2 + 4x - 4

Now solve the cubic -8p3 + 20p2 = 0. The solutions are p = 0, 0, 5/2. You can use any real value of p to plug into the quartic. For this example, we will use 0 since it is easier to work with.

x4 - 4x3 + 4x2 = -x2 + 4x - 4
square root(x4 - 4x3 + 4x2) = square root(-x2 + 4x - 4)
x2 - 2x = ±i(x - 2)

This yields two quadratic equations with complex coefficients:

x2 + (-2 + i)x - 2i = 0

Answer 2

ax⁴ + bx³ +cx² +dx +e is 4th degree polynomial e.g byquadratic polynomial.

we can't find exactly what is the roots of given polynomial . but at least one roots given then we find out by relation between roots and coefficient .

here ax⁴ + bx³ +cx² +dx+e is given polynomial let A , B, C and D are the roots of this .then,
A + B + C + D = -b/a
(A + B)( C + D) + AB + CD = c/a
AB(C+D) +CD(A +B) = -d/a
ABCD = e/a

Let ∅ is roots of given polynomial given then question ask find another roots of this polynomial.
then, you should be use relation.
you will have four equation and three variable then solution possible. i know its complicated . but I think this is one option for finding roots when directly factorisation not possible .