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Andrej [43]
2 years ago
9

F(1) =3 and f (n+1)= f(n)^2+3 then find the value of f(4)

Mathematics
1 answer:
Tcecarenko [31]2 years ago
6 0

Answer:

f(4) = 1612

Step-by-step explanation:

using the recursive formula and f(1) = 3 , then

f(2) = f(1)² + 3 = 3² + 3 = 9 + 3 = 12

f(3) = f(2)² + 3 = 144 + 3 = 147

f(4) = f(3)² + 3 = 147² + 3 = 1609 + 3 = 1612

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3 years ago
A polynomial with integer coefficients is of the form \[x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 18.\]You are told that the integer $r$
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Since the polynomial given [x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 18.\] is divisible by (x - r)^2, this means that (x - r)^2 is a factor of the polynomial just like 4 is divisible by 2, we say 2 is a factor.

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Let a be 1 since the coefficients are integers

We then have Answer:

Step-by-step explanation:

Since the polynomial given [x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 18.\] is divisible by (x - r)^2, this means that (x - r)^2 is a factor of the polynomial just like 4 is divisible by 2, we say 2 is a factor.

Since (x - r)^2 is a factor, we equate the factor to zero in order to get the 'x' variable.

(x - r)^2 =0

Taking square root of both sides gives x-r =0 i.e x=r

Substituting x=r into the polynomial to get 'r' we have r^4 + r^3 + r^2 + r + 18=0.

Factorizing this, possible value of r will be -2 and 2

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3 years ago
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