Chapter 2: Polynomials

Q&A -Ask Doubts and Get Answers

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.

Answer:

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.

VIDEO EXPLANATION


Related Questions for Study

What our students and parents say about us!

Choose EduSakshamยฎ
Embrace Better Learning

edusaksham-mobile-app