### Chapter 2: Polynomials

Q
##### Polynomials CBSE NCERT Solutions

Question:

Obtain all other zeroes of 3x4 + 6x3 - 2x2 - 10x - 5, if two of its zeroes are $\frac{\sqrt{5}}{3}$  and  $-\frac{\sqrt{5}}{3}$.

Solutions: Suppose P(x) = 3x4 + 6x3 - 2x2 - 10x – 5.

Since 3x4 + 6x3 - 2x2 - 10x – 5 is a polynomial of degree 4, so there will be a total of 4 roots.

$\frac{\sqrt{5}}{3}$   and $-\frac{\sqrt{5}}{3}$  are zeroes of polynomial P(x).

$\therefore (x - \frac{\sqrt{5}}{3})$  $(x+\frac{\sqrt{5}}{3})$ $= 0 \Rightarrow x^{2} - \frac{5}{3} = 0 \Rightarrow (3x^{2} -5 ) = 0$

(3x2 – 5 ) = 0  is a factor of given polynomial P(x).

Now, when we will divide P(x) by (3x− 5) the quotient obtained will also be a factor of P(x) and the remainder will be 0.

Therefore, 3x+ 6x−2x−10x – 5 = (3x– 5)(x2 + 2x + 1)

We can find zeroes of the polynomial (x2 + 2x + 1) using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is 1 and product is 1 x 1 = 2.

Therefore, we can write,

x2 + x + x + 1 = 0

$\Rightarrow$ x ( x + 1) + 1( x + 1) = 0

$\Rightarrow$ (x + 1)(x + 1) = 0

So, x = −1 and x = −1.

Hence, all four zeroes of polynomial P(x) = 3x4 + 6x3 - 2x2 - 10x – 5 are $\frac{\sqrt{5}}{3}$ , $-\frac{\sqrt{5}}{3}$  , −1 and −1.

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