Obtain all zeroes of 6x^4-13x^3-34x^2+47x+30, if two of its zeroes are -2 and 3
Answers
Answer:
Given, p(x) = 6x^4 - 13x³ -34x²+47x+30
two zeros of p(x) = -2 , 3
= (x + 2) (x - 3)
= x² -3x +2x -6
= x² -1x -6 = g(x)
now divide p(x) with g(x),
x² - x - 6 ) 6x^4 -13x³ - 34x²+47x +30 ( 6x² -7x -5
6x^4 - 6x³ - 36x²
(-)___(+)___(+)___________
0 - 7x³ +2x² + 47x +30
- 7x³ +7x² +42x
(+)___(-)__(-)________
0 - 5x² + 5x + 30
- 5x² +5x + 30
(+)___(-)__(-)_____
{ 0 }
now take q(x), = 6x² -7x -5
spliting of the middle terms,
= 6x² +3x -10x -5
= 3x(2x + 1) - 5 (2x + 1)
= (2x + 1) (3x - 5)
x = -1/2 , 5/3
therefore, the other two zeros of p(x) = -1/2 , 5/3