Solve x^3+40x^2+304x-1920=0

$${ x }^{ 3 } +40 { x }^{ 2 } +304x-1920=0$$

Solve for x

$$x=-20$$

$$x=4$$

$$x=-24$$

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 -1920 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1920,±960,±640,±480,±384,±320,±240,±192,±160,±128,±120,±96,±80,±64,±60,±48,±40,±32,±30,±24,±20,±16,±15,±12,±10,±8,±6,±5,±4,±3,±2,±1$$

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}+40x^{2}+304x-1920 by x-4 to get x^{2}+44x+480. Solve the equation where the result equals to 0.

$$x^{2}+44x+480=0$$

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, 44 for b, and 480 for c in the quadratic formula.

$$x=\frac{-44±\sqrt{44^{2}-4\times 1\times 480}}{2}$$

Do the calculations.

$$x=\frac{-44±4}{2}$$

Solve the equation x^{2}+44x+480=0 when ± is plus and when ± is minus.

$$x=-24 x=-20$$

List all found solutions.

$$x=4 x=-24 x=-20$$