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Given ax3 + 4x2 + 3x - 4 and x3 - 4x + a leave the same remainder when divided by x - 3.
Let p(x) = ax3 + 4x2 + 3x - 4 and g(x) = x3 - 4x + a
By remainder theorem, if f(x) is divided by (x − a) then the remainder is f(a)
Here when p(x) and g(x) are divided by (x − 3) the remainders are p(3) and g(3) respectively.
Also given that p(3) = g(3) → (1)
Put x = 3 in both p(x) and g(x)
Hence equation (1) becomes,
a(3)3 + 4(3)2 + 3(3) - 4 = (3)3 - 4(3) + a
⇒ 27a + 36 + 9 − 4 = 27 − 12 + a
⇒ 27a + 41 = 15 + a
⇒ 26a = 15 − 41 = − 26
∴ a = −1
Let p(x) = ax3 + 4x2 + 3x - 4 and g(x) = x3 - 4x + a
By remainder theorem, if f(x) is divided by (x − a) then the remainder is f(a)
Here when p(x) and g(x) are divided by (x − 3) the remainders are p(3) and g(3) respectively.
Also given that p(3) = g(3) → (1)
Put x = 3 in both p(x) and g(x)
Hence equation (1) becomes,
a(3)3 + 4(3)2 + 3(3) - 4 = (3)3 - 4(3) + a
⇒ 27a + 36 + 9 − 4 = 27 − 12 + a
⇒ 27a + 41 = 15 + a
⇒ 26a = 15 − 41 = − 26
∴ a = −1
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