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Solve x⁴ + 18x³ + 121x² + 360x + 400 = 0

The solutions of x⁴ + 18x³ + 121x² + 360x + 400 = 0 are x = -5, -5, -4, -4 (repeated roots counted once each: -5, -4).

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Graph of y = x⁴ + 18x³ + 121x² + 360x + 400-6.5-6-5.5-5-4.5-4-3.5-3-2.52468101214x = -5x = -4y
Graph of y = x⁴ + 18x³ + 121x² + 360x + 400. 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 400, so the candidates are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±25, ±40, ±50, ±80, ±100, ±200, ±400.

    x4+18x3+121x2+360x+400=0x^{4} + 18x^{3} + 121x^{2} + 360x + 400 = 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.

    −41181213604000−4−56−260−400114651000\begin{array}{c|ccccc} -4 & 1 & 18 & 121 & 360 & 400 \\ & 0 & -4 & -56 & -260 & -400 \\ \hline & 1 & 14 & 65 & 100 & 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.

    −4114651000−4−40−100110250\begin{array}{c|cccc} -4 & 1 & 14 & 65 & 100 \\ & 0 & -4 & -40 & -100 \\ \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+18x3+121x2+360x+400=(x+5)2(x+4)2x^{4} + 18x^{3} + 121x^{2} + 360x + 400 = (x + 5)^{2}(x + 4)^{2}
LaTeX source of the formulas on this page
\[ x^{4} + 18x^{3} + 121x^{2} + 360x + 400 = (x + 5)^{2}(x + 4)^{2} = 0 \]
\[ x^{4} + 18x^{3} + 121x^{2} + 360x + 400 = 0 \]
\[ \begin{array}{c|ccccc} -4 & 1 & 18 & 121 & 360 & 400 \\ & 0 & -4 & -56 & -260 & -400 \\ \hline & 1 & 14 & 65 & 100 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -4 & 1 & 14 & 65 & 100 \\ & 0 & -4 & -40 & -100 \\ \hline & 1 & 10 & 25 & 0 \end{array} \]
\[ x^{2} + 10x + 25 = 0 \Rightarrow x = -5 \]
\[ x^{4} + 18x^{3} + 121x^{2} + 360x + 400 = (x + 5)^{2}(x + 4)^{2} \]

Candidates tested (computed)

Candidate xp(x)Root?
1900no
-1144no
21764no
-236no
45184no
-40root

Vieta's formulas check the answer (computed)

RelationFrom the rootsFrom the coefficients
Sum of the roots-18-18
Product of the roots400400

Questions

What are the roots of x⁴ + 18x³ + 121x² + 360x + 400 = 0?

The solutions of x⁴ + 18x³ + 121x² + 360x + 400 = 0 are x = -5, -5, -4, -4 (repeated roots counted once each: -5, -4).

How do you factor x⁴ + 18x³ + 121x² + 360x + 400?

(x + 5)(x + 4).

What is the sum of the roots?

-18 (it equals −18 by Vieta's formulas).

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    Solve x⁴ + 18x³ + 121x² + 360x + 400 = 0: x = -5, -4 | WordGPT