Question

find the zeroes of the polynomial when one of its zeroes is given p(x) x cube - 8x2 +19 x -12 , having one of its zeroes as 4​

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{A \: polynomial \: p(x) = {x}^{3} - 8 {x}^{2} + 19x - 12 } \\ &\sf{having \: one \: zero \: as \: 4} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{remaining \: zeroes \: of \: p(x)}  \end{cases}\end{gathered}\end{gathered}$

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

We know that

$\sf \: If \: p(x) \: = \: a {x}^{3} + b {x}^{2} + cx + d \: having \: zerors \: \alpha, \beta, \gamma \: then$

Sum of zeroes taken one at a time is

$\boxed{ \sf \: \alpha + \beta + \gamma = - \dfrac{b}{a} }$

and

Product of zeroes is

$\boxed{ \sf \: \alpha \beta \gamma = - \dfrac{d}{a} }$

Now,

Given that

$\rm :\longmapsto\:p(x) = {x}^{3} - {8x}^{2} + 19x - 12$

$\sf \: On \: comparing \: with\: a {x}^{3} + b {x}^{2} + cx + d \: we \: have$

$\: \: \: \: \: \: \: \: \bull \sf \: a \: = \: 1$

$\: \: \: \: \: \: \: \: \bull \sf \: b = - 8$

$\: \: \: \: \: \: \: \: \bull \sf \: c = 19$

$\: \: \: \: \: \: \: \: \bull \sf \: d = - 12$

Also,

Given that one zero of p(x) is 4.

$\rm :\longmapsto\:Let \: \alpha = 4$

Since, p(x) is a cubic polynomial, so it has 3 zeroes.

$\rm :\longmapsto\:Let \: remaining \: zeroes \: be \: \beta \: and \: \gamma$

Now,

We know that

$\rm :\longmapsto\:\sf \: \alpha + \beta + \gamma = - \dfrac{b}{a}$

On substituting the values, we get

$\rm :\longmapsto\:\sf \: 4 + \beta + \gamma = - \dfrac{( - 8)}{1}$

$\bf\implies \: \beta + \gamma = 4$

$\bf\implies \: \gamma = 4 - \beta - - - (1)$

Also,

$\rm :\longmapsto\:\sf \: \alpha \beta \gamma = - \dfrac{d}{a}$

On substituting the values, we get

$\rm :\longmapsto\:\sf \: 4 \times \beta \gamma = - \dfrac{( - 12)}{1}$

$\rm :\longmapsto\: \beta \gamma = 3$

$\rm :\longmapsto\: \beta (4 - \beta ) = 3$

$\rm :\longmapsto\:4\beta - {\beta}^{2} = 3$

$\rm :\longmapsto\: {\beta}^{2} - 4\beta + 3 = 0$

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

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

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

$\bf\implies \:\beta = 3 \: \: \: or \: \: \: \beta = 1$

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf \beta & \bf \gamma \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 3 \\ \\ \sf 3 & \sf 1 \end{array}} \\ \end{gathered}$

• Hence, the zeroes of f(x) are 4, 1 and 3.