Question

If √5 and -√5 are the two zeroes of the polynomial x 4 +4x 3 -2x 2 -20x-15. Find its other
zeroes.

Answer 1

Answer:

The remaining zeroes of the polynomial x⁴ + 4x³ – 2x² – 20x – 15 are – 1 and – 3.

Step-by-step explanation:

Given that, √5 and – √5 are the two zeroes of the polynomial x⁴ + 4x³ – 2x² – 20x – 15.

So, (x – √5) and (x + √5) are the factors of the polynomial x⁴ + 4x³ – 2x² – 20x – 15.

Multiply these two factors :

⟹ (x – √5) × (x + √5)

⟹ x² + √5x – √5x – 5

⟹ x² – 5

Now, dividing the polynomial with this quadratic polynomial :

Refer the attachment!

So, even x² + 4x + 3 is the factor of the polynomial.

Now, let’s split the middle term of the quadratic polynomial :

⟹ x² + 4x + 3

⟹ x² + x + 3x + 3

⟹ x (x + 1) + 3 (x + 1)

⟹ (x + 1) (x + 3)

Finding the zeroes :

★ ( x + 1 ) :

⟹ x + 1 = 0

⟹ x = – 1

★ ( x + 3 ) :

⟹ x + 3 = 0

⟹ x = – 3

___________

Therefore, the zeros of the polynomial x⁴ + 4x³ – 2x² – 20x – 15 are – 1 and – 3 other than √5 and – √5.

Answer 2

Given :-

⠀

• Two zeros of p(x) = √5 and-√5

⠀

Solution :-

p(x) = x⁴+4x³-2x²-20x-15

⠀

$\purple{\tt \to g(x) = (x + \sqrt{5} )(x - \sqrt{5)}}$

$\tt \to g(x) = {x}^{2} - \sqrt{5} {}^{2}$

$\tt \to g(x) = {x}^{2} - 5$

On dividing p(x) by g (x)

$\tt = ({x}^{4} + 4 {x}^{3} - 2 {x}^{2} - 20x - 15 )\div {x}^{2} - 5$

$\tt = {x}^{2} + 4x + 3$

$\tt\implies {x}^{2} + 4x + 3$

$\tt \implies {x}^{2} + 3x + x + 3$

$\tt \implies x(x + 3) + 1(x + 3)$

$\tt\implies (x + 1)(x + 3)$

Zeroes :-

$\implies\tt x + 1 = 0$

$\purple{\implies\tt x = - 1}$

$\implies \tt x + 3 = 0$

$\purple{\implies \tt x = - 3}$

⠀

• Therefore, Zeroes of polynomial are √5, -√5, -3 and -1.