Solve x⁴ - 6x³ - 15x² + 152x - 240 = 0

How to work it out

1. 1

   A quartic has up to 4 real roots. The rational root theorem says that any integer root of a polynomial with integer coefficients and leading coefficient 1 divides the constant term -240, so the candidates are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±16, ±20, ±24, ±30, ±40, ±48, ±60, ±80, ±120, ±240.

   $$x^{4} - 6x^{3} - 15x^{2} + 152x - 240 = 0$$
2. 2

   Test the candidates until one gives 0. x = 3 is a root. Divide by (x − 3) with synthetic division; the bottom row is the quotient and a 0 remainder.

   $$\begin{array}{c|ccccc} 3 & 1 & -6 & -15 & 152 & -240 \\ & 0 & 3 & -9 & -72 & 240 \\ \hline & 1 & -3 & -24 & 80 & 0 \end{array}$$
3. 3

   x = 4 is a root. Divide by (x − 4) with synthetic division; the bottom row is the quotient and a 0 remainder.

   $$\begin{array}{c|cccc} 4 & 1 & -3 & -24 & 80 \\ & 0 & 4 & 4 & -80 \\ \hline & 1 & 1 & -20 & 0 \end{array}$$
4. 4

   What is left is the quadratic x² + x - 20 = 0. Solve it (factoring or the quadratic formula):

   $$x^{2} + x - 20 = 0 \Rightarrow x = 4,\ -5$$
5. 5

   So the polynomial factors as

   $$x^{4} - 6x^{3} - 15x^{2} + 152x - 240 = (x + 5)(x - 3)(x - 4)^{2}$$

Candidates tested (computed)

Vieta's formulas check the answer (computed)

Questions

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