# CBSE Previous Year Question PaperClass 10 Maths | 2018 Section C Q14

###### CBSE Solved Question Paper 3 Mark | Polynomials

Zeroes of a polynomial. A standard text book question. However, it is a medium difficulty question because of the computation involved in arriving at the answer. Two zeroes of a polynomial are given and you have to find all the zeroes of the polynomial.

Question 14: Find all zeroes of the polynomial (2x4 – 9x3 + 5x2 + 3x – 1) if two of its zeroes are (2 + √3) and (2 - √3).

## NCERT Solution to Class 10 Maths

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Given polynomial is 2x4 – 9x3 + 5x2 + 3x – 1

Two of its zeroes: (2 + √3) and (2 - √3)
Because (2 + √3) and (2 - √3) are two zeros of the poynomial, (x – (2 + √3)) (x – (2 - √3)) is a factor of the given polynomial.
i.e., x2 - 2x + √3x - 2x + 4 - 2√3 - √3x + 2√3 - 3 is a factor of the polynomial.
Simplifying the above expression, we get x2 - 4x + 1 is a factor of the polynomial.

Because x2 - 4x + 1 is a factor of the polynomial, x2 – 4x + 1 will divide the polynomial without leaving a reaminder.
Let us do a long division of the polynomial by x2 – 4x + 1 and compute the quotient.
2x2 - x - 1
x2 - 4x + 1  |  2x4 - 9x3 + 5x2 + 3x - 1          First term of quotient $$frac{2x^4}{x^2}$ = 2x2 2x4 - 8x3 + 2x2 - x3 + 3x2 + 3x Second term of quotient $\frac{-x^3}{x^2}$ = -x -x3 + 4x2 - x -x2 + 4x - 1 Third term of quotient $\frac{-x^2}{x^2}$ = -1 -x2 + 4x - 1 0 2x4 - 9x3 + 5x2 + 3x - 1 =$x2 - 4x + 1) (2x2 - x - 1)
2x2 - x - 1 can be factorized as 2x2 - 2x + x - 1
2x (x - 1) + 1(x - 1) = (2x + 1) (x - 1)
So, its zero are x = $-$frac{1}{2}$ and 1 ∴ The zeroes of the polynomial are$2 + √3), (2 - √3), -$\frac{1}{2}$ and 1

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