Write a polynomial function of least degree with integral coefficients that has the given zeros
1 answer:
Answer:
+ 3x² - 4
Step-by-step explanation:
Note that complex zeros occur in conjugate pairs
If 2i is a zero then - 2i is a zero
The zeros are x = 1, x = - 1, x = 2i, x = - 2i, thus the factors are
(x - 1), (x + 1), (x - 2i) and (x + 2i)
The polynomial is expressed as the product of the factors, thus
f(x) = (x - 1)(x + 1)(x - 2i)(x + 2i) ← expanding in pairs
= (x² - 1)(x² - 4i²) → i² = - 1
= (x² - 1)(x² + 4) ← distribute
= + 4x² - x² - 4
= + 3x² - 4
You might be interested in
Answer:
All options are correct.
Step-by-step explanation:
In option (A), the given equation is
Add 37 on both sides.
Subtract x from both sides.
LHS=RHS for all values of x. It means equation x-37=x-37 has infinite many solutions.
Similarly.
In option (B), the given equation is
LHS=RHS for all values of x. It means equation 73x-37=73x-37 has infinite many solutions.
In option (C), the given equation is
LHS=RHS for all values of x. It means equation 37x-37=37x-37 has infinite many solutions.
In option (D), the given equation is
LHS=RHS for all values of x. It means equation 74x-37=74x-37 has infinite many solutions.
Therefore, all options are correct.