(B) Given equation is $x^3 - 3x + 2 = 0$. By testing $x = 1$,we get $1^3 - 3(1) + 2 = 1 - 3 + 2 = 0$. So,$(x - 1)$ is a factor. Dividing $x^3 - 3x + 2

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(B) Given equation is $x^3 - 3x + 2 = 0$. By testing $x = 1$,we get $1^3 - 3(1) + 2 = 1 - 3 + 2 = 0$. So,$(x - 1)$ is a factor. Dividing $x^3 - 3x + 2$ by $(x - 1)$,we get $x^2(x - 1) + x(x - 1) - 2(x - 1) = 0$. $(x - 1)(x^2 + x - 2) = 0$. Factoring the quadratic part: $(x - 1)(x + 2)(x - 1) = 0$. Thus,$(x - 1)^2(x + 2) = 0$. The roots are $1, 1, -2$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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If two roots of the equation $x^3 - 3x + 2 = 0$ are the same,then the roots are:

A
$2, 2, 3$
B
$1, 1, -2$
C
$-2, 3, 3$
D
$-2, -2, 1$

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