Question

Obtain other zeroes of the polynomial
​

Answer 1

$\large\underline{\bf{Answer-}}$

All the zeroes are : 1, √5, -√5 and $\dfrac{-1}{2}$

$\large\underline{\bf{Explanation-}}$

Given :

• Polynomial = 2x⁴ - x³ - 11x² + 5x+ 5
• zeroes = √5 and - √5

To find :

• Other zeroes

Solution :

Som of zeroes = √5 + ( - √5 )

$\implies$ Sum of zeroes = 0

Product of zeroes = √5 ( -√5 )

$\implies$ Product of zeroes = -5

Polynomial = x² - Sx + P

$\implies$ x² -5

Now, we have to divide Polynomial p(x) by x² - 5

__________________________

★ Refer to the attachment for division method ★

_________________________

For other zeroes,

$\implies$ 2x² - x - 1 = 0

Splitting middle term,

$\implies$ 2x² - 2x + x - 1 = 0

$\implies$ 2x ( x - 1 ) + 1 ( x - 1 ) = 0

$\implies$ (2x+1) ( x-1) = 0

$\implies$ x = $\dfrac{-1}{2}$ and x = 1

$\therefore$ All the zeroes are : 1, √5, -√5 and $\dfrac{-1}{2}$

Answer 2

Answer−

All the zeroes are : 1, √5, -√5 and -1/2

Explanation−

Given :

Polynomial = 2x⁴ - x³ - 11x² + 5x+ 5

zeroes = √5 and - √5

To find :

Other zeroes

Solution :

Som of zeroes = √5 + ( - √5 )

⟹ Sum of zeroes = 0

Product of zeroes = √5 ( -√5 )

⟹ Product of zeroes = -5

Polynomial = x² - Sx + P

⟹ x² -5

Now, we have to divide Polynomial p(x) by x² - 5

__________________________

★ Refer to the attachment for division method ★

_________________________

For other zeroes,

⟹ 2x² - x - 1 = 0

splitting middle term

⟹ 2x² - 2x + x - 1 = 0

⟹ 2x ( x - 1 ) + 1 ( x - 1 )

⟹ (2x+1) ( x-1) = 0

⟹ x=-1/2 and x =1

therefore all the zeroes are:1,√5,√-5 and -1/2