Can polynomials have odd roots (like f(x)= cube root of x)?? Fast!! I’ll give branliest!
2 answers:
I think you're asking if it's possible to have a cube root, fifth root, 7th root, etc of a number as a solution to f(x). The answer is yes it's possible.
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Example:
f(x) = x^3 - 29
This function has one real-number root of (cube root of 29) and the other two roots are complex or imaginary roots.
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Answer:
To prove that 3·4ⁿ + 51 is divisible by 3 and 9, we have;
3·4ⁿ is divisible by 3 and 51 is divisible by 3
Where we have;
= 3·4ⁿ + 51
= 3·4ⁿ⁺¹ + 51
-
= 3·4ⁿ⁺¹ + 51 - (3·4ⁿ + 51) = 3·4ⁿ⁺¹ - 3·4ⁿ
-
= 3( 4ⁿ⁺¹ - 4ⁿ) = 3×4ⁿ×(4 - 1) = 9×4ⁿ
∴ -
is divisible by 9
Given that we have for S₀ = 3×4⁰ + 51 = 63 = 9×7
∴ S₀ is divisible by 9
Since -
is divisible by 9, we have;
-
=
-
is divisible by 9
Therefore is divisible by 9 and
is divisible by 9 for all positive integers n
Step-by-step explanation: