If p ( x ) = x^3 - ax^2 + bx + 3 leaves a remainder -19 when divided by ( x + 2 ) and
remainder 17 when divided by ( x - 2) ,then prove that a + b = 6 .
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Answers
Step-by-step explanation:
p(x) = x³ - ax² + bx + 3
- p(x) leaves a remainder -19 when divided by (x + 2)
⇒ x + 2 = 0
⇒ x = -2
p(-2) = -19
Put x = -2,
(-2)³ - a(-2)² + b(-2) + 3 = -19
-8 - a(4) - 2b + 3 = -19
-8 - 4a - 2b + 3 = -19
-4a - 2b = -19 + 5
-4a - 2b = -14
-2(2a + b) = -2(7)
2a + b = 7 ➟ [1]
- p(x) leaves a remainder 17 when divided by (x - 2)
⇒ x - 2 = 0
⇒ x = 2
p(2) = 17
Put x = 2,
(2)³ - a(2)² + b(2) + 3 = 17
8 - a(4) + 2b + 3 = 17
- 4a + 2b + 11 = 17
-4a + 2b = 17 - 11
-4a + 2b = 6
2(-2a + b) = 2(3)
-2a + b = 3 ➟ [2]
Adding both the equations,
2a + b - 2a + b = 7 + 3
2b = 10
b = 10/2
b = 5
Substitute in equation [1],
2a + b = 7
2a + 5 = 7
2a = 2
a = 2/2
a = 1
➜ a + b
= 1 + 5
= 6
Hence, a + b = 6
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