If x^5 - 3x^4 –ax^3 + 3ax^2 + 2ax + 4 is exactly divisible by (x – 2) then find the value of a.
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Given f(x) = x^5 - 3x^4 - ax^3 + 3ax^2 + 2ax + 4.
Given g(x) = x - 2.
By the remainder theorem,
x - 2 = 0
x = 2.
Plug x = 2 in f(x), we get
f(2) = (2)^5 - 3(2)^4 - a(2)^3 + 3a(2)^2 + 2a(2) + 4 = 0
= 32 - 48 - 8a + 12a + 4a + 4 = 0
= -16 + 4 + 8a = 0
= -12 + 8a = 0
8a = 12
a = 12/8
a = 3/2.
Therefore the value of a = 3/2.
Hope this helps!
Given g(x) = x - 2.
By the remainder theorem,
x - 2 = 0
x = 2.
Plug x = 2 in f(x), we get
f(2) = (2)^5 - 3(2)^4 - a(2)^3 + 3a(2)^2 + 2a(2) + 4 = 0
= 32 - 48 - 8a + 12a + 4a + 4 = 0
= -16 + 4 + 8a = 0
= -12 + 8a = 0
8a = 12
a = 12/8
a = 3/2.
Therefore the value of a = 3/2.
Hope this helps!
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