please help me with 6 question
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Heya !!!
( X + 1 ) and ( X - 2 ) are tow factor of the given polynomial.
X + 1 = 0 ----------X - 2 = 0
X = -1 ------------- X = 2
P(X) = X³ + AX² + 2X + B
P(-1) = (-1)³ + A × (-1)² + 2 × -1 + B
=> -1 + A × 1 - 2 + B = 0
=> -1 + A - 2 + B = 0
=> A + B = 3 --------(1)
And,
P(2) = (2)³ + A × (2)² + 2 × 2 + B = 0
=> 8 + A × 4 + 4 + B = 0
=> 4A + B = -12 ------(2)
From equation (1) we get,
A + B = 3
A = ( 3 - B ) -------(3)
Putting the value of A in equation (2)
4A + B = -12
4 × ( 3 - B ) + B = -12
12 - 4B + B = -12
-3B = -12 - 12
B = 24/3 = 8
Putting the value of B in equation (3)
A = 3 - B = 3 - 8 = -5
★ HOPE IT WILL HELP YOU ★
( X + 1 ) and ( X - 2 ) are tow factor of the given polynomial.
X + 1 = 0 ----------X - 2 = 0
X = -1 ------------- X = 2
P(X) = X³ + AX² + 2X + B
P(-1) = (-1)³ + A × (-1)² + 2 × -1 + B
=> -1 + A × 1 - 2 + B = 0
=> -1 + A - 2 + B = 0
=> A + B = 3 --------(1)
And,
P(2) = (2)³ + A × (2)² + 2 × 2 + B = 0
=> 8 + A × 4 + 4 + B = 0
=> 4A + B = -12 ------(2)
From equation (1) we get,
A + B = 3
A = ( 3 - B ) -------(3)
Putting the value of A in equation (2)
4A + B = -12
4 × ( 3 - B ) + B = -12
12 - 4B + B = -12
-3B = -12 - 12
B = 24/3 = 8
Putting the value of B in equation (3)
A = 3 - B = 3 - 8 = -5
★ HOPE IT WILL HELP YOU ★
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