Question

14. Find all zeroes of the polynomial (2x4 - 9x +
5x2 + 3x - 1) if two of its zeroes are (2 + 73 ) and
(2-√3).
[3]. please answer​

Answer 1

Solution :-

two of the zeroes are (2 + √3) and (2 - √3) .

so, (x - 2 - √3)(x - 2 + √3) is a factor of given polynomial .

or,

→ (x - 2 - √3)(x - 2 + √3) = x² - 4x + 1

then, (x - 1) will divide the polynomial . Dividing by long division method,

x² - 4x + 1 ) 2x⁴ - 9x³ + 5x² + 3x - 1 ( 2x² - x - 1

2x⁴ - 8x³ + 2x²

- x³ + 3x² + 3x

- x³ + 4x² - x

- x² + 4x - 1

- x² + 4x - 1

0

we get,

→ Quotient = 2x² - x - 1 .

then,

→ 2x² - x - 1 = 0

→ 2x² - 2x + x - 1 = 0

→ 2x(x - 1) + 1(x - 1) = 0

→ (x - 1)(2x + 1) = 0

→ x = 1 and (-1/2) .

Hence, all zeroes of given polynomial are {1, (-1/2), (2 + √3) , (2 - √3)} .

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