if p(x)= 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
Question :-- if p(x)= 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 ?
Concept used :--- Let p(x) be any polynomial of degree greater than or equal to one and let ‘a’ be any real number. If p(x) is divided by the linear polynomial (x – a), then the remainder is p(a).
Solution :--
Case (1) :-- p(x)= x^3-ax^2+bx+3 leaves a remainder -19 when divided by (x+2) ...
So, when we put x = (-2) , it will gives Remainder as (-19) .
→ p(-2) = x^3-ax^2+bx+3 = (-19)
→ 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 ------------------- Equation (1) .
_____________________________
Case (2) :---- p(x)= x^3-ax^2+bx+3 leaves a remainder 17 when divided by (x-2)..
So, when we put x = 2 , we get remainder as 17.
→ p(2) = x^3-ax^2+bx+3 = 17
→ (2)³ - a(2)² + 2b + 3 = 17
→ 8 - 4a + 2b + 3 = 17
→ 4a - 2b = 11-17
→ 4a -2b = -6
→ 2a - b = (-3) ----------------- Equation (2) .
_________________________
Adding Equation (1) and (2) now we get,
→ 4a = 7 + (-3)
→ 4a = 4
→ a = 1 .
Putting value of a in Equation (1) , we get,
→ 2*1 + b = 7
→ b = 7-2 = 5 ..
___________________________
So,
→ a + b = 1 + 5 = 6 . .
✪✪ Hence Proved ✪✪
So, value of a+b is 6..
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