Question

Which of the following lanthanoids show +4 oxidation state to acquire noble gas configuration?

Answer 1

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➀ Given p( x ) = 3x

Let p ( x ) = 0

So, 3x = 0

x = 0/3 = 0 , p ( x ) = 3x is zero

∴ No. of zeroes is one

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➁ Given $\:p(x) = {x}^{2} + 5x + 6$is a quadratic polynomial.

It has almost two zeroes.

To find zeroes, let p( x ) = 0

➜${x}^{2} + 5x + 6 = 0$

➜${x}^{2} + 3x + 2x + 6 = 0$

➜$x(x + 3) + 2(x + 3) = 0$

➜$(x + 3)(x + 2) = 0$

➜$x + 30 = 0 \: \: or \: \: x + 2 = 0$

➜$x = - 3 \: \: or \: \: x = - 2$

∴ The zeroes of the polynomial are -3 and -2

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➂ 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 and -3

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➃ Given $p(x) = {x}^{4} - 16$ is a biquadratic polynomial. It has almost two zeroes.

Let p ( x ) = 0

➜ ${x}^{4} - 16 = 0$

➜${( {x}^{2}) }^{2} - {4}^{2} = 0$

➜$( {x}^{2} - 4)( {x}^{2} + 4) = 0$

➜$(x + 2)(x - 2)( {x}^{2} + 4) = 0$

➜$(x + 2) = 0 \: \: or \: (x - 2) = 0 \: \: or( {x}^{2} + 4) = 0$

➜ x = - 2 (or) x = 2 (or) ${x}^{2}$= - 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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