Question

Find the zeroes of the given polynomials.

(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x+2) (x+3)
(iv) p(x) = x⁴ - 16
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Answer 1

༒ Qυєѕтiσи ༒

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Find the zeroes of the given polynomials.

(i) p(x) = 3x

(ii) p(x) = x² + 5x + 6

(iii) p(x) = (x+2) (x+3)

(iv) p(x) = x⁴ - 16

(iv) p(x) = x⁴ - 16

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★彡 Sσℓυтiσи 彡★

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① p(x) = 3x

Let p(x) = 0

So, 3x = 0

⟹ x = 0/3

⟹ x = 0

Zeroes of p(x) = 3x is zero

⛬ No. of zeroes is one.

② p(x) = x² + 5x + 6

Given p(x) = x² + 5x + 6 is a quadratic polynomial.

It has almost two zeroes.

To find zeroes, let p(x) = 0

⟹ x² + 5x + 6 = 0

⟹ x² + 3x + 2x + 6 = 0

⟹ x(x +3) + 2 (x + 3) = 0

⟹ (x + 3) (x + 2) = 0

⟹ x + 3 =0 or x + 2 = 0

⟹ x = -3 or x = -2

⛬ the zeroes of the polynomial are -3 & -2.

③ p(x) = (x+2) (x+3)

Given p(x) = (x + 2) (x + 3)

It is a quadratic polynomial.

It has almost two zeroes.

Let p(x) = 0

⟹ (x + 2) (x + 3) = 0

⟹ (x + 2) = 0 or (x + 3) = 0

⟹ x = -2 or x = -3

⛬ the zeroes of the polynomial are -2 & -3.

④ p(x) = x⁴ - 16

Given p(x) = x⁴ - 16 is a biquadratic polynomial.

It has almost two zeroes.

Let p(x) = 0

⟹ x⁴ - 16 = 0

⟹ (x²)² - 4² =0

⟹ (x² - 4) (x² + 4) = 0

⟹ (x + 2) (x - 2) (x² + 4) = 0

⟹ (x+2)=0 or (x-2)=0 or (x²+4)=0

⟹ x = -2 or x = 2 or x² = -4

x = ± √-4

⛬ The zeroes of the polynomial are 2, -2, ± √-4.

We don't consider √-4. Since it is not real.

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