Solve x^4-6x^3-35x^2+312x-560=0

$$x ^ { 4 } - 6 x ^ { 3 } - 35 x ^ { 2 } + 312 x - 560 = 0$$

Solve for x

$$x=-7$$

$$x=5$$

$$x=4$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -560 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±560,±280,±140,±112,±80,±70,±56,±40,±35,±28,±20,±16,±14,±10,±8,±7,±5,±4,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-6x^{3}-35x^{2}+312x-560 by x-4 to get x^{3}-2x^{2}-43x+140. Solve the equation where the result equals to 0.

$$x^{3}-2x^{2}-43x+140=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 140 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±140,±70,±35,±28,±20,±14,±10,±7,±5,±4,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-2x^{2}-43x+140 by x-4 to get x^{2}+2x-35. Solve the equation where the result equals to 0.

$$x^{2}+2x-35=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 2 for b, and -35 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-35\right)}}{2}$$

Do the calculations.

$$x=\frac{-2±12}{2}$$

Solve the equation x^{2}+2x-35=0 when ± is plus and when ± is minus.

$$x=-7 x=5$$

List all found solutions.

$$x=4 x=-7 x=5$$