Form a polynomial f(x) with real coefficients having the given degree and zeros
21) Degree: 4, zeroes: 2i and -3i
22) Degree: 3, Zeroes i and -10
for every a+bi there is a a-bi solution
(x-2i)(x+2i)(x-3i)(x+3i) = 0
(x^2 +4)(x^2+9) = 0
x^4 + 13 x^2 + 36 = 0
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(x-i)(x+i)(x+10) = 0 etc
To form a polynomial with the given degree and zeros, we use the fact that complex zeros occur in conjugate pairs for polynomials with real coefficients.
For Question 21:
Since the given zeros are 2i and -3i, the conjugate pairs are (2i, -2i) and (-3i, 3i).
To obtain the polynomial, we multiply out these pairs:
(x - 2i)(x + 2i) and (x + 3i)(x - 3i)
Simplifying:
(x^2 - (2i)^2) and (x^2 - (3i)^2)
(x^2 + 4) and (x^2 + 9)
Now multiply these two expressions to obtain the polynomial:
(x^2 + 4)(x^2 + 9)
Expanding further:
x^4 + 9x^2 + 4x^2 + 36
Finally, combining like terms:
f(x) = x^4 + 13x^2 + 36
For Question 22:
Since the given zeros are i and -10, we don't have any complex conjugate pairs since i is already a complex number.
To obtain the polynomial, we use the zero product property:
For the zero i, the factor is (x - i)
For the zero -10, the factor is (x + 10)
Multiply these factors to obtain the polynomial:
(x - i)(x + 10)
Expanding:
x^2 + 10x - ix - 10i
Combining like terms:
x^2 + 10x - ix - 10i
We know that i is the square root of -1, so we can rewrite the polynomial as:
x^2 + 10x - ix - 10i
Finally, we can rearrange the terms with the imaginary unit (i) separate from the real terms:
x^2 + (10 - i)x - 10i
Therefore, the polynomial is:
f(x) = x^2 + (10 - i)x - 10i