Find a polynomial function with real coefficients of degree 4 and zeros x=1-i and x=√(7). let the leading coefficient be equal to 1. written in factored form over rational numbers.

Step-by-step explanation:

Given that one of the zeroes is 1 - i.

Complex roots occur in pairs.

This means, if a + bi is a zero of a polynomial, then a - bi is also its zero.

Therefore, 1 + i is also a zero of the polynomial.

Factors are x - (1 - i) and x - (1 + i).

Multiply the factors.

[x - (1 - i)][x - (1 + i)] = (x - 1 + i)(x - 1 - i)

Given that one of the zeroes is .

Irrational roots occur in pairs.

This means, if an irrational number m is a zero of a polynomial, then - m is also its zero.

Therefore, is also a zero of the polynomial.

Factors are and .

Multiply the factors.

Hence, the required polynomial is .

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~