form a polynomial f(x) with real coefficients having the given degree and zeros. degree 5; zeros -7; -i;-9+i
enter the polynomial.
f(x)=a(?)
Complex numbers always appear as conjugate pairs, so if you have -i, then you also have +i
and if you have -9+i, there will also be -9 - i
so we know we have factors of (x+7) , (x^2 + 1) and two more
I will use the sum and product rule to find the other
sum of -9+i and -9 - i = -18
product of the above is 81 - i^2 = 81 + 1 = 82
resulting in the quadratic factor
x^2 + 18x + 82
so f(x) = (x+7)(x^2 + 1)(x^2 + 18x + 82)
notice, if expanded this will give you a 5th degree polynomial. If you have to expand it, do it very carefully and patiently.
To form a polynomial with real coefficients, we need to make sure that complex conjugate pairs of zeros occur together. Given that the zeros are -7, -i, and -9+i, we also know that the conjugates of -i and -9+i are present.
The conjugate of -i is i, and the conjugate of -9+i is -9-i.
Therefore, the zeros of the polynomial are: -7, -i, i, -9+i, -9-i.
To form the polynomial, we can use these zeros and their conjugates as factors. The general form of a fifth-degree polynomial is:
f(x) = a(x - r1)(x - r2)(x - r3)(x - r4)(x - r5),
where r1, r2, r3, r4, r5 are the zeros. Plugging in the values of the zeros, we get:
f(x) = a(x + 7)(x + i)(x - i)(x - 9 + i)(x - 9 - i).
Expanding this expression, we have:
f(x) = a(x + 7)(x^2 + 1)(x - 9)^2 + 1.
This is the polynomial with real coefficients of degree 5 and the given zeros.
To form a polynomial with real coefficients, we know that complex zeros must come in conjugate pairs. Given the zeros -7, -i, and -9+i, we have -9+i and its conjugate -9-i as a pair.
To find the polynomial, we can start by writing the factors using the zeros:
(x - (-7))(x - (-i))(x - (-9+i))(x - (-9-i))
Simplifying the factors:
(x + 7)(x + i)(x - 9 + i)(x - 9 - i)
We can further simplify this by expanding the expressions:
(x + 7)(x + i)(x - 9 + i)(x - 9 - i)
= (x^2 + 7x + ix + 7i)(x^2 - 9x + ix - 9i)
Now, let's multiply the binomials:
= [(x^2 + 7x) + (ix + 7i)][(x^2 - 9x) + (ix - 9i)]
= [(x^2 + 7x) + i(x + 7)][(x^2 - 9x) + i(x - 9)]
Expanding further:
= (x^2 + 7x + ix + 7i)(x^2 - 9x + ix - 9i)
= (x^2 + (7 + i)x + 7i)(x^2 + (-9 + i)x - 9i)
Multiplying the remaining binomials:
= x^2(x^2 + (-9 + i)x - 9i) + (7 + i)x(x^2 + (-9 + i)x - 9i) + 7i(x^2 + (-9 + i)x - 9i)
Let's simplify these expressions separately:
x^2(x^2 + (-9 + i)x - 9i) = x^4 - 9x^3 + ix^3 - 9ix^2
(7 + i)x(x^2 + (-9 + i)x - 9i) = 7x^3 + 7ix^2 + ix^3 - (9 + 9i)x^2 - 9(-9 + i)x - 9i^2x
= (7x^3 - (9 + 9i)x^2 - 81x + 9ix^2 + 9(-9 + i)x - 9i^2x)
= 7x^3 - (9 + 9i - 9i^2)x^2 - (81 - 81i)x
7i(x^2 + (-9 + i)x - 9i) = 7ix^2 + i^2x^2 - (9 - 9i)x - 9i(-9 + i)x
= (7i + i^2)x^2 - (9 - 9i)x - (81i - 9i^2)x
= (-7 + i)x^2 + (9 - 9i - (81i - 9i^2))x
= (-7 + i)x^2 + (9 - 90i)x
Finally, add all the simplified expressions:
f(x) = x^4 - 9x^3 + ix^3 - 9ix^2 + 7x^3 - (9 + 9i - 9i^2)x^2 - (81 - 81i)x + (-7 + i)x^2 + (9 - 90i)x
= x^4 - (9 - 7)x^3 + (-9i + i)x^2 + (-81 + 7 - 9 + 9i - 9i - 90i)x
= x^4 - 2x^3 - 8ix^2 - 83x - 99i
Therefore, the polynomial f(x) with real coefficients, a degree of 5, and given zeros -7, -i, and -9+i is:
f(x) = x^4 - 2x^3 - 8ix^2 - 83x - 99i