form a polynomial with real coefficients have given degree and zeros.
degree 5, zeros 9, -i; 8+i
please show work
complex roots are in pairs, so we have
(x-9)(x+i)(x-i)(x-(8+i))(x-(8-i))
(x-9)(x^2+1)((x-8)^2 + 1)
. . .
To form a polynomial with real coefficients of degree 5 and given zeros 9, -i, and 8+i, it must also have the conjugates of the complex zeros as zeros.
First, we know that one of the zeros is 9. So, one factor of the polynomial is (x - 9).
Next, we know that one of the zeros is -i. So, one factor of the polynomial is (x + i).
Since the complex zeros appear in pairs, the conjugate of -i is +i. So, another factor of the polynomial is (x - i).
Similarly, the conjugate of 8+i is 8-i. So, another factor of the polynomial is (x - 8 - i).
Now we multiply these factors together to get the polynomial:
(x - 9) * (x + i) * (x - i) * (x - 8 - i) * (x - 8 + i).
To simplify this expression, we can multiply these factors step by step.
= (x - 9) * (x^2 - i^2) * ((x - 8)^2 - i^2)
= (x - 9) * (x^2 - (-1)) * ((x - 8)^2 - (-1))
= (x - 9) * (x^2 + 1) * ((x - 8)^2 + 1)
= (x^3 - 9x^2 + x - 9) * (x^2 + 1) * (x^2 - 16x + 65).
Now, we can multiply these expressions further:
= (x^3 - 9x^2 + x - 9) * (x^4 - 16x^3 + 65x^2 + x^2 - 16x + 65)
= (x^3 - 9x^2 + x - 9) * (x^4 - 16x^3 + 66x^2 - 16x + 65).
Finally, we can combine like terms and simplify further if needed.
To form a polynomial with a given degree and zeros, we need to use the concept of the complex conjugate. The complex conjugate of any complex number a + bi is a - bi.
Given the zeros 9, -i, and 8 + i, we can start by using the complex conjugate to find the remaining zeros.
The zeros are:
1. 9
2. -i
3. 8 + i
The conjugates of these zeros are:
1. 9 (conjugate of a real number remains the same)
2. i (conjugate of -i is i)
3. 8 - i (conjugate of 8 + i is 8 - i)
Now we have all the zeros of the polynomial:
1. 9
2. i
3. 8 - i
To find the polynomial, we multiply these zeros together. Since the polynomial has a degree of 5, we have two more zeros to find.
Multiplying the complex zeros together:
(9)(i)(8 - i) = 72i - 9i^2 = -9i^2 + 72i
The remaining zeros will be the complex conjugates of each other, so we have:
4. -9i^2 + 72i
5. -9i^2 - 72i
Simplifying:
-9i^2 + 72i = -9(-1) + 72i = 9 + 72i
-9i^2 - 72i = -9(-1) - 72i = 9 - 72i
Now we have all five zeros:
1. 9
2. i
3. 8 - i
4. 9 + 72i
5. 9 - 72i
To form the polynomial, we use the fact that if a number is a zero, then (x - zero) is a factor.
Therefore, the polynomial with these zeros is:
(x - 9)(x - i)(x - (8 - i))(x - (9 + 72i))(x - (9 - 72i))
Expanding this polynomial will give the final answer.