find the zeroes of the polynomial when one of its zeroes is given p(x) x cube – 8×2 +19 x -12 , having one of its zeroes as 4​

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1. [tex]\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}[/tex]

   [tex]\begin{gathered}\begin{gathered}\bf \: To\: find – \begin{cases} &\sf{remaining \: zeroes \: of \: p(x)} \end{cases}\end{gathered}\end{gathered}[/tex]

   [tex]\large\underline{\sf{Solution-}}[/tex]

   We know that

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

   Sum of zeroes taken one at a time is

   [tex] \boxed{ \sf \: \alpha + \beta + \gamma = – \dfrac{b}{a} }[/tex]

   and

   Product of zeroes is

   [tex] \boxed{ \sf \: \alpha \beta \gamma = – \dfrac{d}{a} }[/tex]

   Now,

   Given that

   [tex]\rm :\longmapsto\:p(x) = {x}^{3} – {8x}^{2} + 19x – 12[/tex]

   [tex] \sf \: On \: comparing \: with\: a {x}^{3} + b {x}^{2} + cx + d \: we \: have[/tex]

   [tex] \: \: \: \: \: \: \: \: \bull \sf \: a \: = \: 1[/tex]

   [tex] \: \: \: \: \: \: \: \: \bull \sf \: b = – 8[/tex]

   [tex] \: \: \: \: \: \: \: \: \bull \sf \: c = 19[/tex]

   [tex] \: \: \: \: \: \: \: \: \bull \sf \: d = – 12[/tex]

   Also,

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

   [tex]\rm :\longmapsto\:Let \: \alpha = 4[/tex]

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

   [tex]\rm :\longmapsto\:Let \: remaining \: zeroes \: be \: \beta \: and \: \gamma [/tex]

   Now,

   We know that

   [tex]\rm :\longmapsto\:\sf \: \alpha + \beta + \gamma = – \dfrac{b}{a}[/tex]

   On substituting the values, we get

   [tex]\rm :\longmapsto\:\sf \: 4 + \beta + \gamma = – \dfrac{( – 8)}{1}[/tex]

   [tex]\bf\implies \: \beta + \gamma = 4[/tex]

   [tex]\bf\implies \: \gamma = 4 – \beta – – – (1)[/tex]

   Also,

   [tex]\rm :\longmapsto\:\sf \: \alpha \beta \gamma = – \dfrac{d}{a}[/tex]

   On substituting the values, we get

   [tex]\rm :\longmapsto\:\sf \: 4 \times \beta \gamma = – \dfrac{( – 12)}{1}[/tex]

   [tex]\rm :\longmapsto\: \beta \gamma = 3[/tex]

   [tex]\rm :\longmapsto\: \beta (4 – \beta ) = 3[/tex]

   [tex]\rm :\longmapsto\:4\beta – {\beta}^{2} = 3[/tex]

   [tex]\rm :\longmapsto\: {\beta}^{2} – 4\beta + 3 = 0[/tex]

   [tex]\rm :\longmapsto\: {\beta}^{2} – 3\beta – \beta + 3 = 0[/tex]

   [tex]\rm :\longmapsto\:\beta(\beta – 3) – 1(\beta – 3) = 0[/tex]

   [tex]\rm :\longmapsto\:(\beta – 3)(\beta – 1) = 0[/tex]

   [tex]\bf\implies \:\beta = 3 \: \: \: or \: \: \: \beta = 1[/tex]

   So,

   [tex]\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}[/tex]

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

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