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Solve x⁴ + 19x³ + 135x² + 425x + 500 = 0

The solutions of x⁴ + 19x³ + 135x² + 425x + 500 = 0 are x = -5, -5, -5, -4 (repeated roots counted once each: -5, -4).

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Graph of y = x⁴ + 19x³ + 135x² + 425x + 500-6.5-6-5.5-5-4.5-4-3.5-3-2.55101520x = -5x = -4y
Graph of y = x⁴ + 19x³ + 135x² + 425x + 500. The curve crosses the x-axis at x = -5, x = -4, the roots found above.

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 500, so the candidates are ±1, ±2, ±4, ±5, ±10, ±20, ±25, ±50, ±100, ±125, ±250, ±500.

    x4+19x3+135x2+425x+500=0x^{4} + 19x^{3} + 135x^{2} + 425x + 500 = 0
  2. 2

    Test the candidates until one gives 0. x = -4 is a root. Divide by (x + 4) with synthetic division; the bottom row is the quotient and a 0 remainder.

    −41191354255000−4−60−300−500115751250\begin{array}{c|ccccc} -4 & 1 & 19 & 135 & 425 & 500 \\ & 0 & -4 & -60 & -300 & -500 \\ \hline & 1 & 15 & 75 & 125 & 0 \end{array}
  3. 3

    x = -5 is a root. Divide by (x + 5) with synthetic division; the bottom row is the quotient and a 0 remainder.

    −5115751250−5−50−125110250\begin{array}{c|cccc} -5 & 1 & 15 & 75 & 125 \\ & 0 & -5 & -50 & -125 \\ \hline & 1 & 10 & 25 & 0 \end{array}
  4. 4

    What is left is the quadratic x² + 10x + 25 = 0. Solve it (factoring or the quadratic formula):

    x2+10x+25=0⇒x=−5x^{2} + 10x + 25 = 0 \Rightarrow x = -5
  5. 5

    So the polynomial factors as

    x4+19x3+135x2+425x+500=(x+5)3(x+4)x^{4} + 19x^{3} + 135x^{2} + 425x + 500 = (x + 5)^{3}(x + 4)
LaTeX source of the formulas on this page
\[ x^{4} + 19x^{3} + 135x^{2} + 425x + 500 = (x + 5)^{3}(x + 4) = 0 \]
\[ x^{4} + 19x^{3} + 135x^{2} + 425x + 500 = 0 \]
\[ \begin{array}{c|ccccc} -4 & 1 & 19 & 135 & 425 & 500 \\ & 0 & -4 & -60 & -300 & -500 \\ \hline & 1 & 15 & 75 & 125 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -5 & 1 & 15 & 75 & 125 \\ & 0 & -5 & -50 & -125 \\ \hline & 1 & 10 & 25 & 0 \end{array} \]
\[ x^{2} + 10x + 25 = 0 \Rightarrow x = -5 \]
\[ x^{4} + 19x^{3} + 135x^{2} + 425x + 500 = (x + 5)^{3}(x + 4) \]

Candidates tested (computed)

Candidate xp(x)Root?
11080no
-1192no
22058no
-254no
45832no
-40root

Vieta's formulas check the answer (computed)

RelationFrom the rootsFrom the coefficients
Sum of the roots-19-19
Product of the roots500500

Questions

What are the roots of x⁴ + 19x³ + 135x² + 425x + 500 = 0?

The solutions of x⁴ + 19x³ + 135x² + 425x + 500 = 0 are x = -5, -5, -5, -4 (repeated roots counted once each: -5, -4).

How do you factor x⁴ + 19x³ + 135x² + 425x + 500?

(x + 5)(x + 4).

What is the sum of the roots?

-19 (it equals −19 by Vieta's formulas).

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    Solve x⁴ + 19x³ + 135x² + 425x + 500 = 0: x = -5, -4 | WordGPT