obtain all zeros of polynomial f(x) = x ³+4x ²+x-6 if one of it zero is 1.
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obtain all zeros of polynomial f(x) = x ³+4x ²+x-6 if one of it zero is 1.
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[tex]\large\underline{\sf{Solution-}}[/tex]
Given polynomial is f(x) = x³ + 4x² + x - 6
Let assume that
[tex]\sf \: \alpha , \: \beta, \: \gamma \: be \: zeroes \: of \: f(x) \: such \: that \: \alpha = 1 \\ \\ [/tex]
We know,
[tex]\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]
[tex]\sf \: \implies \: \alpha + \beta + \gamma = - \dfrac{4}{1} = - 4 \\ \\ [/tex]
[tex]\sf \: \: 1+ \beta + \gamma = - 4 \\ \\ [/tex]
[tex]\sf \: \: \beta + \gamma = - 5 \\ \\ [/tex]
[tex]\sf \: \implies \: \gamma = - 5 - \beta - - - (1) \\ \\ [/tex]
Also,
[tex]\boxed{{\sf Product\ of\ the\ zeroes= \: - \: \frac{Constant}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]
[tex]\sf \: \implies \: \alpha \beta \gamma = - \dfrac{( - 6)}{1} = 6 \\ \\ [/tex]
[tex]\sf \: \: 1 \times \beta \gamma = 6 \\ \\ [/tex]
[tex]\sf \: \: \beta \gamma = 6 \\ \\ [/tex]
[tex]\sf \: \: ( - 5 - \gamma ) \gamma = 6 \\ \\ [/tex]
[tex]\sf \: \: - 5 \gamma - { \gamma }^{2} = 6 \\ \\ [/tex]
[tex]\sf \: \: { \gamma }^{2} + 5 \gamma + 6 = 0 \\ \\ [/tex]
[tex]\sf \: \: { \gamma }^{2} + 3\gamma + 2 \gamma + 6 = 0 \\ \\ [/tex]
[tex]\sf \: \: \gamma( \gamma + 3) + 2( \gamma + 3) = 0 \\ \\ [/tex]
[tex]\sf \: \: ( \gamma + 3) \: ( \gamma + 2) = 0 \\ \\ [/tex]
[tex]\bf\implies \: \gamma \: = \: - \: 3 \: \: \: \: \: or \: \: \: \: \: \gamma \: = \: - \: 2 \\ \\ [/tex]
Hence,
[tex]\bf\implies \: \: 1 , \: - 2, \: - 3 \: be \: zeroes \: of \: f(x) = {x}^{3} + {4x}^{2} + x - 6 \\ \\ [/tex]
Answer:
Let, f(x) = x³ + 4x² + x - 6 be the given polynomial
and, α = 1.
We know that,
[tex]⟹ \: \gamma = - 5 - \beta .[/tex]
(-5 - γ)γ = 6.
-5γ - γ² = 6.
γ² + 5γ + 6 = 0.
γ² + 3γ + 2γ + 6 = 0.
γ(γ + 3) + 2(γ + 3) = 0.
(γ + 2)(γ + 3) = 0.
⟹ γ = -2 or γ = -3.
Hence, 1, -2, and -3 are the zeros of f(x) = x³ + 4x² + x - 6.