Find the fourth degree polynomial with zeros calculator

Fourth-degree polynomials, equations of the form

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The Quartic Formula
Special Cases of Quartics
Number Line

Ax4 + Bx3 + Cx2 + Dx + E = 0

where A is not equal to zero, are called quartic equations. If you divide both sides of the equation by A you can simplify the equation to

x4 + bx3 + cx2 + dx + e = 0.

A quartic equation with real number coefficients can have either four real roots, two real roots and two complex roots, or four complex roots. Complex roots occur in conjugate pairs. To solve a general quartic equation, you need to solve associated cubic and quadratic equations in a multi-step process. Certain special quartics can be solved with simpler methods.

You can apply the quartic formula following the instructions below, or use the quartic equation solver calculator on the left.

The Quartic Formula

Given a general fourth degree equation x4 + bx3 + cx2 + dx + e = 0, you can rearrange terms to form the equation

x4 + bx3 = -cx2 - dx - e.

Now add the expression (b2/4 + 2p)x2 + bpx + p2 to both sides:

x4 + bx3 + (b2/4 + 2p)x2 + bpx + p2 = (b2/4 + 2p - c)x2 + (bp - d)x + p2 - e.

The left hand side is now a perfect square: (x2 + (b/2)x + p)2. You want to find a real number p such that the right hand side is also a square. In order for the right hand side to be a square quadratic, the discriminant must be zero. That is,

(bp - d)2 - 4(b2/4 + 2p - c)(p2 - e) = 0.

This simplifies to a cubic equation in p:

-8p3 + 4cp2 + (8e - 2bd)p + d2 - 4ce + b2e = 0.

Since every cubic equation has at least one real root, you can find a suitable value of p to resolve the quartic. After you plug in the value of p, you take the square root of both sides to create two quadratic equations. This gives you a total of four solutions

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

This yields two quadratic equations with complex coefficients:

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

Using the quadratic equation, the roots of the first equation are 2 and -i, and the roots of the second are 2 and i. These four roots are the roots of the original quartic.

Special Cases of Quartics

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.

By the fundamental Theorem of Algebra, any polynomial of degree #4# can be written in the form:

# P(x) = A(x-alpha)(x-beta)(x-gamma) (x-delta) #

Where, #alpha,beta,gamma,delta# are the roots (or zeros) of the equation #P(x)=0#

We are given that #-sqrt(11)# and #2i# are solutions (presumably, although not explicitly stated, of #P(x)=0#, thus, wlog, we can write:

# alpha = -sqrt(11) #
# beta = 2i #

We also know that complex roots must appear as complex conjugate pairs, so #-2i# must be another root, so in addition:

# gamma = -2i #

And so we have:

# P(x) = A(x+sqrt(11))(x-2i)(x+2i) (x-delta) #

We can multiply the complex products to get real coefficients quadratic:

# (x-2i)(x+2i) = x^2-(2i)^2 #
# " " = x^2-4i^2 #
# " " = x^2+4 #

And using this we get:

# P(x) = A(x+sqrt(11))(x^2+4) (x-delta) #

We require that #P(x)# has integer coefficients, rather than irrational coefficients, and this would require that the product #(x+sqrt(11))# and #(x-delta)# has integer coefficients, Now: