How to find the roots of $-x^3+3x^2-7x+5 = 0$?

Question

I would like to understand how to go about solving something like this, not just get the solution but some kind of methodology (that hopefully makes as much intuitive sense as possible); I honestly don't even know where to start.

$-x^3+3x^2-7x+5 = 0$

Answer 1

Using the Rational Root Theorem, the only possible rational roots are $x = \pm 1$ and $x = \pm 5$.

Now, which of these (if any) are in fact roots? (Hint: one of them is a root).

After you find one root $x = r$, you can factor the polynomial into $(x-r)$ times a quadratic, which you have a formula to solve.

Answer 2

Hint:

Since the sum of the coefficients is $0$, one of the roots is $1$. The other roots can be found by dividing $-x^3+3x^2-7x+5$ by $x-1.$

Answer 3

$$-x^3+3x^2-7x+5 = 0$$ By inspection $x=1$ is a root then divide by $x-1$ to find other two roots

Answer 4

By inspection, $x = 1$ is a root. Thus, for $x \neq 1$, $$ 0 = -x^3 + 3x^2 - 7x + 5 = -x^3 + 3x^2 - 3x + 1 - (4x - 4) = -(x - 1)^3 -4(x - 1) $$ That is, $$ (x-1)[-4 - (x - 1)^2] = 0 \quad \Rightarrow \quad (x - 1)^2 = -4 \quad x = 1\pm 2i $$

Answer 5

If you submit x=1 in the polynomial, you get 0. So that makes x=1 as one of the roots. We need to find 2 other, because it is a polynomial of degree 3. So we divide the polynomial by x-1, by polynomial long divison method.The polynomial then breaks into below form..

$${(x-1)(-x^2-4x+5)}=0$$

Or,$$(1-x)(x+5)(x-1)=0$$

Answer 6

HINT : By Remainder theorem, we observe that $x-1$ is factor of the polynomial you have given. So It will be very easy to proceed, divide polynomial by x-1 by Horner's formula, you will get a quadratic polynomial, and then factorise quadratic polynomial if It is reducible over Z$[x]$.