how to solve x^4-3x^3+2=0?

Question

how to solve x^4-3x^3+2=0?

do I factor this or what?

Answer 1

You could try to factor this, but it's a little difficult.
The best way to go would be to first test if it has any integer solutions.
The equation is 4th degree, so there are 4 solutions for x.
It's final factored form will be something like :
(x - a)(x - b)(x - c)(x - d) = 0, where we have to find a, b, c and d.

Notice that the only integers that can divide into +2 are -1, +1, -2 or +2.
If x = -1, and the expression is zero, then the factor will be (x + 1).
If x = +1, the factor will be (x - 1).
If x = -2, the factor will be (x + 2).
If x = +2, the factor will be (x - 2)

Trying each of these values for x, we find that only x = +1 works.
Therefore, (x - 1) is a factor.

Now divide (x - 1) into (x^4 - 3x^3 + 2) using
synthetic division, and you get : x^3 - 2x^2 - 2x - 2

Now you have -2 as a constant, so you can do the same thing.
All you have to do is find out if x can equal -1, +1, -2 or +2,
in the equation x^3 - 2x^2 - 2x - 2 = 0,
because there might be another simple integer solution.

But alas, there is no other.

So far then, we have : x^4 - 3x^3 + 2 = 0,
which is (x - 1)(x^3 - 2x^2 - 2x - 2) = 0 giving one solution as x = 1.

If you want to go further, there is a way to find the zeros of
a cubic equation represented by x^3 + ax^2 + bx + c = 0.
The values of a, b and c for the equation are : a = -2, b = -2, c = -2.

Calculate Q = (3b - a^2) / 9 = -10 / 9.
Calculate R = (9ab - 27c - 2a^3) / 54 = 53 / 27.
Calculate D = sqrt(Q^3 + R^2) = sqrt(201) / 9.
This is > 0, so there is 1 real zero and 2 complex conjugate zeros.
Calculate S = (R + D)^(1/3) = [53 + 3*sqrt(201)]^(1/3) / 3 = 1.523803
Calculate T = (R - D)^(1/3) = [53 - 3*sqrt(201)]^(1/3) / 3 = 0.729170
Calculate real value of x = S + T - a / 3
= 1.523803 + 0.729170 - (-2) / 3
= 2.9196 to 4 decimal places.

The 2 complex conjugate zeros, which I won't calculate, are given by:
(-1/2)(S + T) - a/3 ± (1/2)*sqrt(3)*(S - T)*i, where i = sqrt(-1).

So the real solutions are x = 1 or x = 2.9196.

Answer 2

Factor theorem comes to mind:
You can guess one solution x=1 immediately, so then you have one factor (x-1) Long division gives the other factor as
(x^3-2x^2-2x-2). See if you can guess another factor of this and repeat the process until you have all factors. I ran into a brick wall at this point, so I graphed it, which gave only 2 zeros at x=1 (which I guessed) and at x=2.2196. Thats it.

Answer 3

A gadget of linear equations is two or better linear equations that are being solved concurrently. customarily, a answer of a gadget in 2 variables is an ordered pair that makes both equations authentic. In different words, it really is the position both graphs intersect, what they have in hardship-loose. So if an ordered pair is a answer to at least one equation, yet not the different, then it isn't a answer to the gadget. A consistent gadget is a gadget that has a minimum of one answer. An inconsistent gadget is a gadget that has no answer There are 3 techniques to sparkling up structures of linear equations in 2 variables: graphing substitution technique eliminating technique

Answer 4

x^4-3x^3+2=0
=>x^4-1-3x^3+3=0
=>(x^2+1)(x^2-1)-3(x^3-1)=0
=>(x^2+1)(x+1)(x-1)-3(x-1)(x^2+x+1)=0
(x-1) {(x^2+1)(x+1)-3(x^2+x+1)=0
=>(x-1)(x^3+x^2+x+1-3x^2-3x-3)=0
=(x-1)(x^3-2x^2-2x-2)=0
Now find the value of x

Answer 5

u go x^4-3x^3+2=0
which is 4x^7+2=0
4x^7=-7
then figure it out

Answer 6

NO YOU CANT FACTORIZE THIS EXPRESSION

Answer 7

sry i cant :(

ill still try though