Question

Find a cubic polynomial whose zeros are 1/2 , 1 and -3

Answer 1

‡ Given that,

Find a cubic polynomial whose zeros are 1/2 , 1 and -3

‡ Let,

• α + ß + γ = 1/2
• αß + ßγ + αγ = 1
• αßγ = - 3

Form of Quadratic Polynomial is :

$\tt\:⟹ x³ - (α + ß + γ)x² + (αß + ßγ + αγ)x - αßγ = 0$

• Substitute the zeroes.

$\tt\:⟹ {x}^{3} - ( - \frac{1}{2} ) {x}^{2} + (1)x - ( - 3) = 0$

$\tt\:⟹ {x}^{3} + \frac{1}{2} {x}^{2} + x + 3 = 0$

$\tt\:⟹ \frac{2 {x}^{3} + {x}^{2} + 2x + 6 }{2} = 0$

$\tt\:⟹2 {x}^{3} + {x}^{2} + 2x + 6 = 0 \times 2$

$\tt\:⟹2 {x}^{3} + {x}^{2} + 2x + 6 = 0$

HENCE, IT IS SOLVED..

MARK AS BRAINLIEST