Question

Two zeroes of cubic polynomial ax3+3x2-bx-6 are -1 and -2. find the third zero and the value of a and b

Answer 1

HELLO DEAR,

ax³ + 3x² - bx - 6

x = -1 , -2

put the values of x in Equation

we get,

a(-1)³ + 3(-1)² - b(-1) - 6 = 0

=> -a + 3 + b - 6 = 0

=> b - a = 3--------(1)

now x = - 2

a(- 2)³ + 3(-2)² - b(-2) - 6 = 0

=> -8a + 12 + 2b - 6 = 0

=> 2b - 8a + 6 = 0

=> b - 4a = -3--------(2)

from--(1) and -----(2)

b - a = 3
b - 4a = -3
(-)___(+)__(+)
———————
3a = 6

=> a = 2 put in --(1)

b - 3 = 2

=> b = 5 now put this value in Equation

ax³ + 3x² - bx - 6 = 0

=> 2x³ + 3x² - 5x - 6 = 0

two zeroes are given (-1 , -2)

(x + 1)(x + 2) = x² + 2x + x + 2 = 0

=> x² + 3x + 2 =0

x² + 3x + 2)2x³ + 3x² - 5x - 6(2x - 3
2x³ + 6x² + 4x
- - -
—————————
-3x² - 9x - 6
-3x² - 9x - 6
+ + +
—————————
0 0 0

hence another zeroes is

2x - 3 = 0

=> x = 3/2

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

heya...!!!



given that:-

it's two of zeroes are -1 , -2

(x + 1)(x + 2)

➾ x² + 2x + x + 2 = 0

➾ x³ + 3x + 2= 0



NOW

put the zeroes in the Equation we get lenear equations,

➾ where, x = - 1


➾ a(-1)³ + 3(-1)² - b(-1) - 6 = 0

➾ -a + 3 + b - 6 = p

➾ b - a = 3 ---------(1)


where, x = -2

➾ a(-2)³ + 3(-2)² - b(-2) - 6 = 0

➾ -8a + 12 + 2b - 6 = 0

➾ 2b - 8a + 6 = 0

➾ b - 4a = -3-------(2)

from--(1) and --(2)

we get,

b - a = +3
b - 4a = -3
(-)___(+)__(+)
————————
3a = 6

➾ a = 2 put in--(1)

we get,

➾ b - 2 = 3

➾ b = 5 , a = 2

now,

ax³ + 3x² - bx - 6 = 0

➾ 2x³ + 3x² - 5x - 6 = 0

➾ 2x³ + (6x² - 3x²) - (9x - 4x) - 6 = 0

➾ 2x³ + 6x² + 4x - 3x² - 9x - 6 = 0

➾ (x + 1)(x + 2)(2x - 3)
———————————
(x + 1)(x + 2)

➾ (2x - 3)

➾ x = 3/2 its another zeroes