Polynomial x^3 - a x^2 + b x - 6 leaves remainder -8 when divided by ( x - 1 ) and ( x - 2 ) is a factor of it. Find the values of ‘a’ and ‘b’
Answers
Answer:
given: p(x) = x³ - ax² + bx - 6
x-1 leaves -8 as remainder and x-2 is a factor of the given polynomial
Step-by-step explanation:
as x-1 leaves -8 as remainder, p(1) = -8
=> 1³ - a×1² + b × 1 - 6 = - 8
=> 1 - a + b - 6 = - 8
=> b - a = - 8 + 6 - 1
=> b - a = -3................. 1
x-2 is a factor of the polynomial,, p(2) = 0
[ a factor leaves no remainder ]
=> 2³ - a × 2² + b × 2 - 6 = 0
=> 8 - 4a + 2b - 6 = 0
=> 2b - 4a = 6 - 8
=> 2b - 4a = -2
=> b - 2a = -1..............2 (divide through by 2)
on solving equations 1 and 2
b - a = -3
b - 2a = -1
a = - 2 and b = - 5
Answer:
- -2, -5 are the values of a and b respectively.
Step-by-step explanation:
f(x) = x³ - ax² + bx - 6
f(1) = (1)³ - a(1)² + b(1) - 6 = -8
f(2) = (2)³ - a(2)² + b(2) - 6 = -8
f(1) :
› 1 - a + b - 6 = -8
› -a + b - 5 = -8
∴ -a + b = -3 - (1)
f(2) :
› 8 - 4a + 2b - 6 = -8
› -4a + 2b = -2
› -2a + b = -1 - (1)
Solving (1) and (2) :
-a + b = -3
-2a + b = -1
∴ a = -2
Substitute (a) in (1) or (2) :
› -(-2) + b = -3
› 2 + b = -3