If x^3-ax^2+bx+3 leaves a remainder -19 when divided by (x+2) and a remainder 17 when divided by (x-2) prove that a+b=6
Answers
Answer:
Step-by-step explanation:
put x = -2 in the equation x³ - ax² + bx + 3
(-2)³ - a (-2)² + b(-2) + 3 = -19
-8 - 4a² - 2b + 3 = -19
-4a² - 2b = -19 + 5
-4a² - 2b = -14 -----(i)
put x = 2 in the equation x³ - ax² + bx + 3
(2)³ - a (2)² + 2b + 3 = 17
8 - 4a² + 2b + 3 = 17
-4a² + 2b = 17 - 11
-4a² + 2b = 6 ----(ii)
On adding equation (i) and (ii)
-4a² - 2b = -14
-4a² + 2b = 6
-------------------
-8a² = -8
a² = 1
a = ±1
-4(1)² + 2b = 6
-4 + 2b = 6
2b = 10
b = 5
hence a + b = 1 + 5 = 6
Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
We Have :-
p ( x ) = x³ - ax² + bx + 3
When divided by ( x + 2 ) gives - 19
When divided by ( x - 2 ) gives 17
To Prove :-
a + b = 6
Solution :-
When , x + 2 = 0
x = - 2
p ( - 2 ) = ( - 2 )³ - a ( - 2 )² + b ( - 2 ) + 3 = - 19
- 8 - 4a - 2b + 3 = - 19
- 4a - 2b - 5 = - 19
- 4a - 2b = - 14
2a + b = 7 ------------- ( i )
When , x - 2 = 0
x = 2
p ( 2 ) = ( 2 )³ - a ( 2 )² + b ( 2 ) + 3 = 17
8 - 4a + 2b + 3 = 17
- 4a + 2b + 11 = 17
2b - 4a = 6
b - 2a = 3 ----------- ( ii )
Adding ( i ) and ( ii )
2a + b = 7
b - 2a = 3
2b = 10
b = 5
Now using this in equation ( i )
2a + b = 7
2a + 5 = 7
2a = 2
a = 1
a + b = 6
Hence proved
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