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Solve x³ - x² - 25x + 25 = 0

The solutions of x³ - x² - 25x + 25 = 0 are x = -5, 1, 5.

Graph of y = x³ - x² - 25x + 25-6-4-2246-100-80-60-40-2020406080x = -5x = 1x = 5y
Graph of y = x³ - x² - 25x + 25. The curve crosses the x-axis at x = -5, 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 25, so the candidates are ±1, ±5, ±25.

    x3−x2−25x+25=0x^{3} - x^{2} - 25x + 25 = 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.

    11−1−2525010−2510−250\begin{array}{c|cccc} 1 & 1 & -1 & -25 & 25 \\ & 0 & 1 & 0 & -25 \\ \hline & 1 & 0 & -25 & 0 \end{array}
  3. 3

    What is left is the quadratic x² - 25 = 0. Solve it (factoring or the quadratic formula):

    x2−25=0⇒x=5, −5x^{2} - 25 = 0 \Rightarrow x = 5,\ -5
  4. 4

    So the polynomial factors as

    x3−x2−25x+25=(x+5)(x−1)(x−5)x^{3} - x^{2} - 25x + 25 = (x + 5)(x - 1)(x - 5)
LaTeX source of the formulas on this page
\[ x^{3} - x^{2} - 25x + 25 = (x + 5)(x - 1)(x - 5) = 0 \]
\[ x^{3} - x^{2} - 25x + 25 = 0 \]
\[ \begin{array}{c|cccc} 1 & 1 & -1 & -25 & 25 \\ & 0 & 1 & 0 & -25 \\ \hline & 1 & 0 & -25 & 0 \end{array} \]
\[ x^{2} - 25 = 0 \Rightarrow x = 5,\ -5 \]
\[ x^{3} - x^{2} - 25x + 25 = (x + 5)(x - 1)(x - 5) \]

Candidates tested (computed)

Candidate xp(x)Root?
10root

Vieta's formulas check the answer (computed)

RelationFrom the rootsFrom the coefficients
Sum of the roots11
Product of the roots-25-25

Questions

What are the roots of x³ - x² - 25x + 25 = 0?

The solutions of x³ - x² - 25x + 25 = 0 are x = -5, 1, 5.

How do you factor x³ - x² - 25x + 25?

(x + 5)(x − 1)(x − 5).

What is the sum of the roots?

1 (it equals −(-1) by Vieta's formulas).

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    Solve x³ - x² - 25x + 25 = 0: x = -5, 1, 5 | WordGPT