Question

If two zeros $x^{3}+x^{2}-5x-5$ are $\sqrt{5} and -\sqrt{5}$, then its third zero is
(a) 1
(b) -1
(c) 2
(d) -2

Answer 1

SOLUTION :

The correct option is (b) : - 1 .

Let α = √5 , β = - √5 and γ are the three Zeroes of the cubic  polynomial.

Given :  The cubic  polynomial f(x) = x³ + x² - 5x - 5  

On comparing with  ax³ + bx² + cx + d  

a = 1, b= 1, c = - 5 , d = - 5

Sum of zeroes of cubic  polynomial= −coefficient of x² / coefficient of x³

α + β + γ = −b/a

√5  +(-√5) +  γ = −1/1

√5  - √5 +  γ = − 1

γ = − 1

Hence, the third zero (γ) is − 1 .

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

$\underline{\large\bf{\mathfrak{Hello!}}}$

$If \: two \: zeros \: of \: polynomial \: \\ {x}^{3} + {x}^{2} - 5x - 5 \: is \: \sqrt{5} \: and \: - \sqrt{5} \\ \\ So, \: factor \: will \: be \\ => (x - \sqrt{5} )(x + \sqrt{5} ) \\ => {x}^{2} - 5 \\ Let \: the \: third \: zero \: be \: 'x' \\ \\ Now, \: dividing \: {x}^{2} - 5 \: with \: {x}^{3} + {x}^{2} - 5x - 5 \\ \\ \frac{{x}^{3} + {x}^{2} - 5x - 5}{ {x}^{2} - 5 } = 0 \\ \\\frac{{x}^{3} - 5x + {x}^{2} - 5}{ {x}^{2} - 5 } = 0 \\ \\\frac{x({x}^{2} - 5) +1({x}^{2} - 5)}{ {x}^{2} - 5 } = 0 \\ \\ \frac{ ({x}^{2} - 5)(x + 1)}{ {x}^{2} - 5 } = 0 \\ \\ x + 1 = 0 \\ \\ x = -1 \\ \\ So, \: third \: zero \: is \: -1$

$\boxed{ Correct \: answer \: is \: (b) \: -1 }$

$\bf{\mathfrak{Hope \: this \: helps...:)}}$