How do you solve x^3-5x^2+5x-4=0?

Answer 1

firstly check the equation for solution by putting x from +-1 to +-n
on putting x=4 the equation satisfies itself .
Now your one of the answer becomes x=4. Devide whole of the equation with (x-4)
the answer will be =x^2-x+1
For solving this apply Dharacharya's Method for finding roots of a quadratic equation i.e. roots=[-b+-(b^2-4ac)^1/2]/2a
youwill get {-1+-3^1/2 i}/2
so roots of eq. = 4, [-1+3^1/2i]/2, [-1-3^1/2i]/2

Answer 2

x^3 - 5x^2 + 5x - 4 = 0

Let p(x) = x^3 - 5x^2 + 5x - 4

What you want to test are factors of -4, such that p(x) = 0.
You want to test -1, 1, -2, 2, into the function p(x), and see if any of them give you 0.

Once you get a solution (we'll call it r), then it would follow that
(x - r) is a factor, you would do long division to obtain a quadratic, and then you would equate the quadratic to 0 to find the other two solutions.

Answer 3

With a polinomial, the possible answers are going to be factors of the coefficient of the leading term or factors of the coefficient of the last term.
The leading term is 1, so the factors are 1 and -1.
The last term is -4, so the factors are 1,-1,2,-2,4,and-4.

Then, once you have found at least one solution (x=r), you know that one of the factors of the polynomial is (x-r). You can divide the polynomial by (x-r) using long division and get down to a quadratic equation that you can factor, or at least apply the quadratic equation on.

Answer 4

p(x)=x^3-5x^2+5x-4
p(4)=4^3-5(4)^2+5(4)-4=0

x^3-5x^2+5x-4
=x^3-4x^2-x^2+4x+x-4
=x^2(x-4)-x(x-4)+1(x-4)
=(x-4)(x^2-x+1)