Math, asked by shivanshudwivedi39, 8 months ago

Find a cubic palynomial whose zeroes are -2 -3 and -1 ​

Answers

Answered by Anonymous
7

\large\bf\underline \blue {To \:  \mathscr{f}ind:-}

  • we need to find the cubic polynomial.

 \huge\bf\underline \purple{ \mathcal{S}olution:-}

 \bf\underline{\red{Given:-}}

  • zero of required cubic polynomial are -2 , -3 and -1

  { \blue{ \mathscr{  \underline{Let : -  }}}}

Let α , β and Ɣ be the zeroes of the required cubic polynomial.

  • Let α = -2
  • β = -3
  • Ɣ = -1

⇝(α + β + Ɣ) = -2 + (-3) + (-1)

⇝(α + β + Ɣ) = -2 - 3 - 1

⇝(α + β + Ɣ) = -6 ⠀....2)

⇝(αβ + βƔ + Ɣα) = -2 × (-3) + (-3) × (-1) + (-1) ×(-2)

⇝(αβ + βƔ + Ɣα) = 6 + 3 + 2

⇝(αβ + βƔ + Ɣα) = 11 ⠀....1)

⇝ (αβƔ) = -2 × (-3) × (-1)

⇝ (αβƔ) = -6 ⠀....3)

  • Formula for quadratic polynomial is:-

▶x³ -(α + β + Ɣ)x² + (αβ + βƔ + Ɣα)x -(αβƔ)

Putting values of (α + β + Ɣ), (αβ + βƔ + Ɣα) and (αβƔ) in the formula .

↛x³ -(-6)x² + (11)x - (-6)

↛x³ + 6x² + 11x + 6

Hence

The required cubic polynomial is:-

  • x³ + 6x² + 11x + 6

━━━━━━━━━━━━━━━━━━━━━━━━━

Answered by Anonymous
5

  \bf\huge\underline\red{answer} :  -

 \bf\underline \blue{standard \: form \: for \: cubic \: polynomial} :

 \implies \bf \orange{ {ax}^{3}  +  {bx}^{2}  + cx + d}

OR

 \implies \bf  \ {x}^{3}  - ( \alpha  +  \beta  +  \gamma ) {x}^{2}  + ( \alpha  \beta  +  \beta  \gamma  +  \gamma  \alpha )x - ( \alpha  \beta  \gamma )........(1)

 \bf \large \ \: its \: zeroes \: are \:  \alpha ,  \:  \beta  \: and \:  \gamma

 \bf \:  \:  {\alpha  =  - 2}

 \bf \: \beta  =  - 3

 \bf \:  \gamma  =  - 1

Now,

 \bf  \green{ \alpha  +  \beta  +  \gamma  =  - 2 - 3 - 1 =  - 6}

 \bf \green{ \alpha  \beta  \gamma  = ( - 2)( - 3)( - 1) =  - 6}

 \bf \green{ \alpha  \beta  +  \beta  \gamma  +  \gamma  \alpha  = ( - 2)( - 3) + ( - 3)( - 1) + ( - 1)( - 2) = 6 + 3 + 2 = 11}

substitute these values in eq.(1)

 \implies \bf  \ {x}^{3}  - ( \alpha  +  \beta  +  \gamma ) {x}^{2}  + ( \alpha  \beta  +  \beta  \gamma  +  \gamma  \alpha )x - ( \alpha  \beta  \gamma )

 \implies \bf \:  {x}^{3}  - ( - 6) {x}^{2}  + (11)x - ( - 6)

 \implies \bf \:  {x}^{3}  + 6 {x}^{2}  + 11x + 6

 \bf \huge \red{cubic \: polynomial \: is \:  {x}^{3}  + 6 {x}^{2}  + 11x + 6}

Similar questions