### Chapter 2: Polynomials

Q
##### Polynomials CBSE NCERT Solutions

Question:

If two zeroes of the polynomial x4 – 6x- 26x2 + 138x – 35 are $2 \pm \sqrt{3}$ ,  find other zeroes.

Solution: Suppose P(x) = x4 – 6x- 26x2 + 138x – 35.

Since x4 – 6x- 26x2 + 138x – 35 is a polynomial of degree 4, so there will be a total of 4 roots.

$2+\sqrt{3}$ and $2-\sqrt{3}$ are zeroes of polynomial P(x).

$\therefore [ x - (2 + \sqrt{3})] [ x - (2 - \sqrt{3})] = 0$

$\Rightarrow [ (x-2) + \sqrt{3}] [ (x-2) - \sqrt{3}] = 0$

$\Rightarrow [ (x-2)^{2} - (\sqrt{3})^{2}] = 0$

$\Rightarrow$ [(x2 - 4x + 4) – 3] = 0

$\Rightarrow$ x2 - 4x + 1 = 0

(x2 - 4x + 1  ) = 0  is a factor of given polynomial P(x).

Now, when we will divide P(x) by (x2 - 4x + 1) the quotient obtained will also be a factor of P(x) and the remainder will be 0.

Therefore, x4 – 6x- 26x2 + 138x – 35 = (x2 - 4x + 1)(x2 - 2x - 35)

We can find zeroes of the polynomial (x2 - 2x - 35) using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -2 and product is -35 x 1 = -35.

Therefore, we can write,

x2 - 7x + 5x - 35 = 0

$\Rightarrow$ x ( x - 7) + 5( x - 7) = 0

$\Rightarrow$ (x - 7)(x + 5) = 0

So, x = 7 and x = −5.

Hence, all four zeroes of polynomial P(x) = x4 – 6x- 26x2 + 138x – 35 are $2+\sqrt{3}$$2-\sqrt{3}$,  7 and −5.

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