Find the valur of a and b so that the polynomial (x^3 - 10x^2 +ax +b ) is exactly divisible by (x-1) as well as (x-2)
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Heya !!!
( X -1) and ( X -2) are two factor of the given polynomial.
( X -1) = 0 --------------- ( X -2) = 0
X = 1 ------------------ X = 2
P(X) = X³ - 10X² + AX + B
P(1) = (1)³ - 10 × (1)² + A × 1 + B
=> 1 - 10 + A + B = 0
=> A + B = 9 ----------(1)
Also,
X = 2
P(2) = (2)³ -10 × (2)² + A × 2 + B
=> 8 - 40 + 2A + B = 0
=> 2A + B = 32 ---------(2)
From equation (1) we get ,
A + B = 9
B = 9 - A --------(3)
Putting the value of A in equation (2)
2A + B = 32
2A + 9 - A = 32
A = 32-9
A = 23
Putting the value of A in equation (3)
B = 9 - A
B = 9 - 23 = -14
★ HOPE IT WILL HELP YOU ★
( X -1) and ( X -2) are two factor of the given polynomial.
( X -1) = 0 --------------- ( X -2) = 0
X = 1 ------------------ X = 2
P(X) = X³ - 10X² + AX + B
P(1) = (1)³ - 10 × (1)² + A × 1 + B
=> 1 - 10 + A + B = 0
=> A + B = 9 ----------(1)
Also,
X = 2
P(2) = (2)³ -10 × (2)² + A × 2 + B
=> 8 - 40 + 2A + B = 0
=> 2A + B = 32 ---------(2)
From equation (1) we get ,
A + B = 9
B = 9 - A --------(3)
Putting the value of A in equation (2)
2A + B = 32
2A + 9 - A = 32
A = 32-9
A = 23
Putting the value of A in equation (3)
B = 9 - A
B = 9 - 23 = -14
★ HOPE IT WILL HELP YOU ★
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