Question

Obtain all the zeroes of 6x^4+17x^3-29x^2-2x+8 if two of its zeroes are -1/2 and 2/3


plz someone solve the question if any one is intelligent​

Answer 1

All the zeroes of the polynomial are -1/2, 2/3, 1 and -4

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Answer 2

Answer:

We know that two of its zeroes are -1/2 and 2/3 So x=-1/2 =>2x+1 is a factor of the above polynomial and thus 3x+2 is also a factor of the polynomial

Lets divide this polynomial by (2x+1)

(2x+1)√ 6x^4+17x^3-29x^2-2x+8 |(3x^3+7x^2-18x+8)

- 6x^4-3x^3

______________

14x^3-29x^2-2x+8

- 14x^3-7x^2

________________

-36x^2-2x+8

+36x^2+18x

_____________

16x+8

-16x-8

_________

0

The quotient thus obtained is

3x^3+7x^2-18x+8 . Now divide this by (3x-2)

(3x-2)√3x^3+7x^2-18x+8 |(x^2+3x-4)

-3x^3+2x^2

_____________

9x^2-18x+8

- 9x^2 +6x

____________

-12x+8

12x-8

_______

0

Thus the quotient obtained is

${x}^{2} + 3x - 4 = 0 = > {x}^{2} + 4x - x - 4$

$= 0 = > x(x + 4) - 1(x + 4) = 0 = >$

$(x - 1)(x + 4) = 0 = > x = 1\: or \: - 4$

Hence other two zeroes of this polynomial are 1and -4 respectively. Hope it helps you.