the zeros of the polynomial ax cube + 3 x square - b x - 6 are -1 and -2 find the values of a and b also find the third zero
Answers
Answer:
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
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Step-by-step explanation:
ax³ + 3x² - bx - 6
x = -1 , -2
a(-1)³ + 3(-1)² - b(-1) - 6 = 0
=> -a + 3 + b - 6 = 0
=> b - a = 3 (1)
a(- 2)³ + 3(-2)² - b(-2) - 6 = 0
=> -8a + 12 + 2b - 6 = 0
=> 2b - 8a + 6 = 0
b - a = 3
b - 4a = -3
3a = 6
=> a = 2
b - 3 =
ax³ + 3x² - bx - 6 = 0
=> 2x³ + 3x² - 5x - 6 = 0
(x + 1)(x + 2) = x² + 2x + x + 2 = 0
=> x² + 3x + 2 =0
x² + 3x + 2)2x³ + 3x² - 5x - 6(2x - 3
2x³ + 6xsq + 4x
-3x² - 9x - 6
-3x² - 9x - 6
2x - 3 = 0
=> x = 3/2