Find a polynomial f(x) of degree 3 with real coefficients and the following zeros.
-1,3+i
f(x)=
complex roots always come in pairs which are conjugates of each other
so there are three roots, ...
-1, 3+i, and 3-i
f(x) = (x+1)(x - 3 - i)(x - 3 + i)
= (x+1)(x^2 -6x + 10)
If you have to expand it, do so.
Thanks
To find a polynomial with the given zeros, we can use the fact that complex roots always come in conjugate pairs.
Since -1 is a zero, we know that (x+1) is a factor of the polynomial.
Since 3+i is a zero, we know that (x - (3+i)) is a factor of the polynomial.
To find the factor for the conjugate root, we need to find the conjugate of 3+i, which is 3-i. Therefore, (x - (3-i)) is also a factor of the polynomial.
Multiplying these factors together, we get:
(x+1)(x - (3+i))(x - (3-i))
Expanding this expression, we get:
(x+1)(x - 3 - i)(x - 3 + i)
Let's simplify this expression:
(x+1)(x^2 - 3x - ix - 3x + 9 + 3i - ix + 3i + i^2)
Simplifying further, we get:
(x+1)(x^2 - 6x + 10)
Finally, expanding the expression, we get:
x^3 - 6x^2 + 10x + x^2 - 6x + 10
Combining like terms, we get:
x^3 - 5x^2 + 4x + 10
Therefore, the polynomial f(x) is:
f(x) = x^3 - 5x^2 + 4x + 10
To find a polynomial with the given zeros, we'll use the fact that if a polynomial has a complex zero, then its conjugate must also be a zero.
The given zeros are -1, 3+i, and their conjugate 3-i.
To find the polynomial, we multiply the factors corresponding to each zero together:
(x - (-1)) * (x - (3+i)) * (x - (3-i))
(x + 1) * (x - 3 - i) * (x - 3 + i)
Simplifying this expression gives us the polynomial:
f(x) = (x + 1)(x - 3 - i)(x - 3 + i)
To expand it further, we can use the FOIL method:
f(x) = (x^2 - 3x + ix - 3x + 9 - 3i - ix + 3i - i^2)(x + 1)
Simplifying this expression further:
f(x) = (x^2 - 6x + 9 - i^2)(x + 1)
Since i^2 = -1, we can simplify it even more:
f(x) = (x^2 - 6x + 9 + 1)(x + 1)
f(x) = (x^2 - 6x + 10)(x + 1)
Thus, the polynomial f(x) is:
f(x) = x^3 - 6x^2 + 10x + x^2 - 6x + 10
Combining like terms, we get:
f(x) = x^3 - 5x^2 + 4x + 10