Question

obtain all the zeros of 3x³-x²-3x+1.
given that one of its zeros is 1​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: remaining \: zeroes \: be \: a \: and \: b$

So,

We have,

$\rm :\longmapsto\:1,a,b \: are \: zeroes \: of \: 3{x}^{3} - {x}^{2} - 3x + 1$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\bf\implies \:1 + a + b = - \dfrac{( - 1)}{3} = \dfrac{1}{3}$

$\bf\implies \:a + b = - \dfrac{2}{3} - - - (1)$

Also,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{3}}}}$

$\rm :\longmapsto\:1 \times a \times b = - \dfrac{1}{3}$

$\rm :\longmapsto\:ab = - \dfrac{1}{3}$

$\rm :\longmapsto\:3ab = - 1$

$\rm :\longmapsto\:3a\bigg( - \dfrac{2}{3} - a \bigg) = - 1 \: \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\: - 2a - 3 {a}^{2} = - 1$

$\rm :\longmapsto\: {3a}^{2} + 2a - 1 = 0$

$\rm :\longmapsto\: {3a}^{2} + 3a - a - 1 = 0$

$\rm :\longmapsto\:3a(a + 1) - 1(a + 1) = 0$

$\rm :\longmapsto\:(3a - 1)(a + 1) = 0$

$\bf\implies \:a = - 1 \: \: \: or \: \: \: a = \dfrac{1}{3}$

Put the values of a in equation (1) to get values of b.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf a & \bf b = - \dfrac{2}{3} - a \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 1 & \sf \dfrac{1}{3} \\ \\ \sf \dfrac{1}{3} & \sf - 1 \end{array}} \\ \end{gathered}$

Hence,

$\bf :\longmapsto\:Zeroes \: are \: 1, - 1, \dfrac{1}{3}$

Additional Information :-

$\rm :\longmapsto\:If \: \alpha \: and \: \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c \: then$

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

and

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: remaining \: zeroes \: be \: a \: and \: b$

So,

We have,

$\rm :\longmapsto\:1,a,b \: are \: zeroes \: of \: 3{x}^{3} - {x}^{2} - 3x + 1$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\bf\implies \:1 + a + b = - \dfrac{( - 1)}{3} = \dfrac{1}{3}$

$\bf\implies \:a + b = - \dfrac{2}{3} - - - (1)$

Also,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{3}}}}$

$\rm :\longmapsto\:1 \times a \times b = - \dfrac{1}{3}$

$\rm :\longmapsto\:ab = - \dfrac{1}{3}$

$\rm :\longmapsto\:3ab = - 1$

$\rm :\longmapsto\:3a\bigg( - \dfrac{2}{3} - a \bigg) = - 1 \: \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\: - 2a - 3 {a}^{2} = - 1$

$\rm :\longmapsto\: {3a}^{2} + 2a - 1 = 0$

$\rm :\longmapsto\: {3a}^{2} + 3a - a - 1 = 0$

$\rm :\longmapsto\:3a(a + 1) - 1(a + 1) = 0$

$\rm :\longmapsto\:(3a - 1)(a + 1) = 0$

$\bf\implies \:a = - 1 \: \: \: or \: \: \: a = \dfrac{1}{3}$

Put the values of a in equation (1) to get values of b.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf a & \bf b = - \dfrac{2}{3} - a \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 1 & \sf \dfrac{1}{3} \\ \\ \sf \dfrac{1}{3} & \sf - 1 \end{array}} \\ \end{gathered}$

Hence,

$\bf :\longmapsto\:Zeroes \: are \: 1, - 1, \dfrac{1}{3}$

Additional Information :-

$\rm :\longmapsto\:If \: \alpha \: and \: \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c \: then$

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

and

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$