MATHEMATICS: Finding zeros of a quartic function?

Question

Can you teach me the step-by-step process in finding the zeros of a quartic function? Kindly use this:


P(x) = x^4 - 4x^3 + 3x^2 + 4x - 4 where one zero is 2


Thanks.

Answer 1

Step 1 :

One zero is 2, so (x - 2) is a factor.
Divide (x - 2) into x^4 - 4x^3 + 3x^2 + 4x - 4.

The result is : P(x) = (x - 2)(x^3 - 2x^2 - x + 2)

Step 2 :

If there are any more integer factors, they
can only be -2, -1, 1 or 2 because of the
constant 2 in x^3 - 2x^3 - x + 2.

Now try each one in turn, by dividing
x^3 - 2x^2 - x + 2 by (x + 2), (x + 1),
(x - 1) and (x - 2) by synthetic division.

(x + 2) doesn't go evenly.
(x + 1) works, giving (x^2 - 3x + 2)
as the other factor.
You could try the others at this stage,
on x^2 - 3x + 2, but it's not necessary
here, as it factors quite easily into
(x - 2)(x - 1).

Thus, P(x) = (x - 2)^2(x - 1)(x + 1).
Zeros are : x = 2 or x = 1 or x = -1.

Answer 2

Here is the method :
---------------------------
The general form of a quartic equation is :

A*x^4 + B*x^3 + C*x^2 + D*x + E = 0


Start by dividing the equation by the leading coeeficient, to make it of the form :


x^4 + a*x^3 + b*x^2 + c*x + d = 0.


Case #1 : If d = 0, then the roots are 0 and the roots of a cubic equation.
---------

Case #2 : Suppose d is a non-zero
---------

Substitute x = y - a/4, expand, and simplify, to get

y^4 + e*y^2 + f*y + g = 0

where

e = b - 3*a^2/8,
f = c + a^3/8 - a b/2,
g = d - 3*a^4/256 + a^2 *(b/16) - ac/4.

Case #1 : if g = 0, then
----------

You can factor the quartic into y times a cubic. The roots of the original equation are then x = -a/4 and the roots of that cubic with a/4 subtracted from each.


Case #2 : if f = 0, then
---------

the quartic in y is actually a quadratic equation in the variable y^2. You can solve this using your favorite method, and then take the two square roots of each of the solutions for y^2 to find the four values of y which work. Subtract a/4 from each to get the four roots x.


Now, let's assume that d, f and g are all non-zero.
---------------------------------------------------

Let's use Euler's method :
---------------------------

The related auxiliary cubic equation :

z^3 + (e/2)*z^2 + ((e^2 - 4g)/16)*z - f^2/64 = 0

has 3 roots. We can find these roots by solving the cubic equation. None of these is zero because f isn't zero. Let p and q be the square roots of two of those roots, and set

r = -f/(8 p*q).

Then p^2, q^2, and r^2 are the three roots of the above cubic equation. More important is the fact that the four roots of the original quartic are

x = p + q + r - a/4,
x = p - q - r - a/4,
x = -p + q - r - a/4, and
x = -p - q + r - a/4.

Method #2
----------

The quartic in y must factor into two quadratics with real coefficients, since any complex roots must occur in conjugate pairs. So we write,


y^4 + e*(y^2) + f*y + g = (y^2 + h y + j)(y^2 - hy + g/j)



Since f and g aren't zero, neither j nor h is zero either. Without loss of generality, we can assume h > 0 (otherwise swap the two factors). Then, equating coefficients of y^2 and y,


e = g/j + j - h^2,
f = h (g/j - j).


so,


g/j + j = e + h^2,

g/j - j = f/h.

Adding and subtracting these two equations,


2g/j = e + h^2 + f/h,

2j = e + h^2 - f/h.

Multiplying these together,


4g = e^2 + 2e*(h^2) + h^4 - f^2/h^2.

Rearranging,


h^6 + 2 e*(h^4) + (e^2 - 4g)*h^2 - f^2 = 0.


This is a cubic equation in h2, with known coefficients, since we know e, f, and g. We can use the solution for cubic equations shown above to find a positive real value of h^2, whose existence is guaranteed by Descartes' Rule of Signs, and then take its positive square root. This value of h will give a value of j from 2j = e + h^2 - f/h. Once you know h and j, you know the quadratic factors of the quartic in y. Notice that you do not need all the roots h^2 of the cubic, and that any positive real one will do. Further notice that h^2 = 4z from the previous section.

Once you have factored the quartic into two quadratics, finishing the finding of the roots is simple, using the Quadratic Formula. Once the roots y are found, the corresponding x's are gotten from x = y - a/4.

Answer 3

P(x) = x^4 - 4x^3 + 3x^2 + 4x - 4

First, Find the first derivative or P'(x)

4x^3 - 12x^2 + 6x + 4 (divide the whole term by 2)
2x^3 - 6x^2 + 3x + 2

then do the synthetic division process

2 - 6 + 3 + 2 l_2_l
4 -4 -2
--------------------
2 - 2 -1 0

2x^2 - 2x - 1

then use the quadritic formula

the answers should be 2, -0. 37, and1.37

if you will use the quadratic formula you would get an answer with a radical.

Answer 4

Using synthetic division and the fact that x - 2 is a factor:-
"""2|1""""- 4""""3""""4"""- 4
"""""|""""""2"""- 4"""-2"""""4
"""""|1""""-2"""- 1""""2""""0

P(x) = (x - 2) (x³ - 2x² - x + 2)

Synthetic division may be used again to show that x - 1 is a factor of x³ - 2x² - x + 2:-

""""1|1""""-2""""-1""""2
"""""|"""""""1""""-1"""-2
"""""|1""""-1""""-2""""0

P(x) = ( x - 2 ) ( x - 1 ) ( x ² - x - 2 )
P(x) = ( x - 2 ) ( x - 1 ) ( x - 2 ) ( x + 1 )
P(x) = ( x - 1 ) ( x + 1 ) ( x - 2 ) ²
zeros are when x = 1 , x = - 1 , x = 2

Answer 5

Use the long division method for polynomials. You know one zero is 2 so that means one term is (x-2). Divide P(x) by (x-2) to get your answer.

Answer 6

divide your polynomial by x-2 and work with the resulting cubic. find a discussion in your book about "finding rational roots (or zeros)."