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Answers
Answer:
f(8/5) = 0
Step-by-step explanation:
Given,
f(-1) = 0
f(1) + f(2) = 0
And f(x) is a monic quadratic polynomial.
So,
f(x) = ax^2 + bx + c
We know that in a monic quadratic polynomial, a = 1.
Then,
f(x) = 1x^2 + bx + c = x^2 + bx + c
Now, by applying the value of f(-1) in the expression f(x) , we get,
(-1)^2 + b(-1) + c = 0 [: f(-1) = 0]
=> 1 - b + c = 0
=> b = c + 1 ....... (i)
Now, let us apply f(1) + f(2) = 0 in the expression f(x), we get,
(1)^2 + b(1) + c + (2)^2 + b(2) + c = 0
[: f(1) + f(2) = 0]
=> 1 + b + c + 4 + 2b + c = 0
=> 3b + 2c + 5 = 0 ...... (ii)
Now from equation (i) and equation (ii) , we get,
=> 3(c + 1) + 2c + 5 = 0
=> 3c + 3 + 2c + 5 = 0
=> 5c + 8 = 0
=> 5c = -8
=> c = -8/5
Hence, c = -8/5
Now, by applying the value of c = -8/5 in the equation (i), we get,
=> b = -8/5 + 1
By taking LCM of terms of RHS, we get,
=> b = (-8 + 5) / 5
=> b = -3/5
Hence, now we get the expression f(x) as,
f(x) = x^2 -3/5(x) - 8/5
Now, by applying the value of f(8/5) in the new expression f(x), we get,
f(8/5) = (8/5)^2 - 3/5 × 8/5 - 8/5
f(8/5) = 64/25 - 24/25 - 8/5
Let's take LCM of the terms of RHS, then we get,
f(8/5) = 64/25 - 24/25 - 40/25
=> f(8/5) = (64 - 24 - 40) / 25
=> f(8/5) = (64 - 64) / 25
=> f(8/5) = 0/25
=> f(8/5) = 0
Hence, the value of
f(8/5) = 0
Answer:
f( - 1) = 0
⟹ - 1 is a zero
⟹x 1 is a factor
let say other zero = a
⟹(x - a) is a factor
f(x) = ( x+ 1)(x - a)
f(1) + f(2) = 0
⟹(1 + 1)(1 - a) + (2 + 1)(2 - a) = 0
⟹2 - 2a + 6 - 3a = 0
⟹a = 8/3
8/3 is other zero
hence 8/3 is zero