CBSE Previous Year Question Paper
Class 10 Maths | 2018 Section C Q14

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).

Video Explanation

Explanatory Answer

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