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Solve x³ - 5x² - x + 5 = 0

The solutions of x³ - 5x² - x + 5 = 0 are x = -1, 1, 5.

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Graph of y = x³ - 5x² - x + 5-2-1123456-202040x = -1x = 1x = 5y
Graph of y = x³ - 5x² - x + 5. The curve crosses the x-axis at x = -1, x = 1, x = 5, the roots found above.

How to work it out

  1. 1

    A cubic has up to 3 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 5, so the candidates are ±1, ±5.

    x3−5x2−x+5=0x^{3} - 5x^{2} - x + 5 = 0
  2. 2

    Test the candidates until one gives 0. x = 1 is a root. Divide by (x − 1) with synthetic division; the bottom row is the quotient and a 0 remainder.

    11−5−1501−4−51−4−50\begin{array}{c|cccc} 1 & 1 & -5 & -1 & 5 \\ & 0 & 1 & -4 & -5 \\ \hline & 1 & -4 & -5 & 0 \end{array}
  3. 3

    What is left is the quadratic x² - 4x - 5 = 0. Solve it (factoring or the quadratic formula):

    x2−4x−5=0⇒x=5, −1x^{2} - 4x - 5 = 0 \Rightarrow x = 5,\ -1
  4. 4

    So the polynomial factors as

    x3−5x2−x+5=(x+1)(x−1)(x−5)x^{3} - 5x^{2} - x + 5 = (x + 1)(x - 1)(x - 5)
LaTeX source of the formulas on this page
\[ x^{3} - 5x^{2} - x + 5 = (x + 1)(x - 1)(x - 5) = 0 \]
\[ x^{3} - 5x^{2} - x + 5 = 0 \]
\[ \begin{array}{c|cccc} 1 & 1 & -5 & -1 & 5 \\ & 0 & 1 & -4 & -5 \\ \hline & 1 & -4 & -5 & 0 \end{array} \]
\[ x^{2} - 4x - 5 = 0 \Rightarrow x = 5,\ -1 \]
\[ x^{3} - 5x^{2} - x + 5 = (x + 1)(x - 1)(x - 5) \]

Candidates tested (computed)

Candidate xp(x)Root?
10root

Vieta's formulas check the answer (computed)

RelationFrom the rootsFrom the coefficients
Sum of the roots55
Product of the roots-5-5

Questions

What are the roots of x³ - 5x² - x + 5 = 0?

The solutions of x³ - 5x² - x + 5 = 0 are x = -1, 1, 5.

How do you factor x³ - 5x² - x + 5?

(x + 1)(x − 1)(x − 5).

What is the sum of the roots?

5 (it equals −(-5) by Vieta's formulas).

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    Solve x³ - 5x² - x + 5 = 0: x = -1, 1, 5 | WordGPT