Question

obtain all the zeroes of the polynomial p(x)=x⁴+4x³-7x²-22x+24 if two of its zeroes are 1 and 2​

Answer 1

Answer:

$given \: p(x) = {x}^{4} + {4x}^{3} - {7x}^{2} - 22x + 24$

two zeros of p ( x ) = 1 , 2

= (x - 1) (x - 2)

= x² - 2x -x +2

= x² - 3x +2 = g(x)

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

x² - 3x +2 ) x^4 +4x³ - 7x² - 22x +24 ( x² +7x +12

x^4 - 3x³ +2x²

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

0 + 7x³ - 9x² - 22x +24

7x³ - 21x²+14x

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

0 + 12x² - 36x +24

+ 12x² - 36x +24

(-)____(+)___(-)____

{ 0 }

now take q(x) , = x² +7x +12

spliting of middle terms,

= x² +3x + 4x +12

= x (x + 3) + 4 (x + 3)

= (x + 3) (x + 4)

x = - 3 , - 4

therefore, the other two zeros are -3 ,-4