The roots of the equation $x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$ are

2, 2, 2, 2

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3, 1, 3, 1

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1, 2, 1, 2

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1, 1, 1, 1

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Solution

The correct option is D
1, 1, 1, 1

$x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$
By hit and trial, x=1 is a root of above equation. Using long division algorithm, we can say
⇒ $(x - 1) (x^{3} - 3 x^{2} + 3 x - 1) = 0$
⇒ $(x - 1) (x - 1)^{3} = 0$
⇒ $x = 1, 1, 1, 1$

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Similar questions

The roots of the equation $x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$ are

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0

\sqrt{- 1}

- 1

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 = 0

{tan}^{- 1} y = 4 {tan}^{- 1} x (| x | < tan \frac{π}{8})

Solve the equation $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0$ , given one root is $\sqrt{(- 1)}$ . Find the value of other three roots.

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