Question

if three zeroes of a cubic Polynomial are 2,4 and 5 find the cubic
polynomial.​​

Answer 1

EXPLANATION.

Three zeroes of a cubic polynomial are 2, 4, 5.

As we know that,

Let, us assume that.

⇒ α = 2, β = 4, γ = 5.

General equation of cubic polynomial,

⇒ x³ - (α + β + γ)x² + (αβ + βγ + γα)x - αβγ.

Sum of zeroes of a cubic polynomial.

⇒ α + β + γ = -b/a.

⇒ 2 + 4 + 5 = 11.

Products of zeroes of a cubic polynomial two at a time.

⇒ αβ + βγ + γα = c/a.

⇒ (2)(4) + (4)(5) + (5)(2).

⇒ 8 + 20 + 10 = 38.

Products of zeroes of a cubic polynomial.

⇒ αβγ = -d/a.

⇒ (2)(4)(5) = 40.

Put this value in equation, we get.

⇒ x³ - (11)x² + (38)x - 40.

⇒ x³ - 11x² + 38x - 40 = 0.

Answer 2

$\begin{gathered}\begin{gathered}\bf Given \:\: zeroes \: of \: cubic \: polynomial : \begin{cases} &\sf{ \alpha = 2} \\ &\sf{ \beta = 4}\\ &\sf{ \gamma = 5} \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{cubic \: polynomial} \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline{\bold{Solution :- }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline{\bold{❥︎Step :- 1 }}$

☆ Sum of zeroes taken one at a time,

$\sf \:  ⟼ \: S_1 \: = \alpha + \beta + \gamma$

$\sf \:  ⟼ \: S_1 \: = 2 + 4 + 5$

$\sf \:  ⟼ \: S_1 = 11$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline{\bold{❥︎Step :- 2 }}$

☆ Sum of the zeroes taken two at a time.

$\sf \:  ⟼ \: S_2 \: = \: \alpha \beta + \beta \gamma + \gamma \alpha$

$\sf \:  ⟼ \: S_2 = 2 \times 4 + 4 \times 5 + 5 \times 2$

$\sf \:  ⟼ \: S_2 = 8 + 20 + 10 = 38$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline{\bold{❥︎Step :- 3 }}$

☆ Product of zeroes

$\sf \:  ⟼ \: S_3 = \alpha \beta \gamma$

$\sf \:  ⟼ \: S_3 = 2 \times 4 \times 5 = 40$

$\large\underline{\bold{❥︎Step :- 4 }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\bf So, \: we \: get \: \begin{cases} &\sf{S_1 \: = 11} \\ &\sf{S_2 \: = 38}\\ &\sf{S_3 \: = 40} \end{cases}\end{gathered}\end{gathered}$

$\large\underline{\bold{❥︎Step :- 5 }}$

☆ Now, cubic polynomial is given by

$\sf \:  ⟼ \: f(x) \: = k( {x}^{3} - S_1 {x}^{2} + S_2x - S_3)$

$\sf \:  ⟼\: where \: k \: is \: non \: zero \: real \: number$

☆ On substituting the values, we get

$\bf \:  ⟼ \: f(x) = k({x}^{3} - 11 {x}^{2} + 38x - 40)$

─━─━─━─━─━─━─━─━─━─━─━─━─