Question

find all the zeroes of the polynomial p(x) = x4 – 7x³ + 9x² + 13x – 4, if two of its zeroes are 2 + √ 3 and 2 - √ 3.

Answer 1

Answer:

Given, p(x)=x^4 - 7x³ + 9x² + 13x - 4

two zeros of p(x) = 2 + √3 , 2 - √3

= (x-2-√3) (x-2+√3)

= (x-2)² - (√3)²

= x²+ 4 - 4x - 3

= x²- 4x +1 = g(x)

now divide p(x) with g(x):

x²- 4x +1 ) x^4 - 7x³ +9x² +13x - 4 ( x² - 3x - 4

x^4 - 4x³ + x²

(-)__(+)___(-)__________

0 - 3x³ + 8x² +13x - 4

- 3x³ +12x² - 3x

(+)___(-)___(+)______

0 - 4x² +16x - 4

- 4x² +16x - 4

(+)___(-)__(+)___

{ 0. }

now take q(x) , = x² - 3x -4

now spliting of middle terms,

= x² - 4x + x - 4

= x(x - 4) + 1(x-4)

= (x - 4) (x + 1)

x = 4 , -1

therefore, the other zeros of p(x) = 4 , -1