Form a polynomial f(x) with real coefficients, with leading coefficient 1, having the given degree and zeros.
Degree: 4; zeros: -4+3i and 1; multiplicity of 2
if -4 + 3 i
then -4 - 3 i is also a zero
so
(x + 4 - 3 i) (x + 4 + 3 i) ( x-1) (x-1) = y
alternative way:
f(x) = (x-1)^2 (a quadratic)
complex roots always come in conjugate pairs, so -4 - 3i is another
sum of the complex roots = -8
product of the complex roots = (-4+3i)(-4 - 3i)
= 16 - 9i^2 = 25
so the quadratic part is x^2 + 8x + 25
f(x) = (x^2 + 8x + 25)(x-1)^2
of course you could also expand
f(x) = (x - (-4+3i))(x - (-4-3i))(x-1)^2 to get the same thing
this is how Damon did it
To form a polynomial \(f(x)\) with real coefficients, with the leading coefficient 1, and having the given degree and zeros, we will start by considering each zero and its multiplicity.
Given zeros:
1. \(x = -4+3i\) (complex conjugate)
2. \(x = 1\) (multiplicity of 2)
For the complex zero \(x = -4+3i\), its conjugate is also a zero. So, we have two complex zeros:
1. \(x_1 = -4+3i\)
2. \(x_2 = -4-3i\) (complex conjugate of \(x_1\))
For the zero \(x = 1\), we know it has a multiplicity of 2. This means it is a repeated zero. So, we have:
3. \(x_3 = 1\) (repeated twice)
To obtain the polynomial \(f(x)\), we will use the fact that the product of linear factors gives the polynomial. Thus, we can write \(f(x)\) in factored form as:
\[f(x) = a(x - x_1)(x - x_2)(x - x_3)(x - x_3)\]
Using the given zeros, the factored form becomes:
\[f(x) = (x - (-4+3i))(x - (-4-3i))(x - 1)^2\]
Expanding this expression, we get:
\[f(x) = (x + 4 - 3i)(x + 4 + 3i)(x - 1)(x - 1)\]
Simplifying further by multiplying the binomials, we have:
\[f(x) = (x^2 + 8x + 25)(x - 1)^2\]
Expanding the squared term using the FOIL method, we obtain:
\[f(x) = (x^2 + 8x + 25)(x^2 - 2x + 1)\]
Finally, we multiply the quadratic factors:
\[f(x) = x^4 + 6x^3 + 9x^2 + 6x + 25\]
Hence, the polynomial \(f(x)\) with real coefficients, leading coefficient 1, and the given zeros is:
\[f(x) = x^4 + 6x^3 + 9x^2 + 6x + 25\]
To form a polynomial with given degree and zeros, we can use the fact that if a complex number is a zero of a polynomial with real coefficients, then its conjugate is also a zero of the polynomial.
Degree: 4
Zeros: -4 + 3i (multiplicity of 2) and 1
Since -4 + 3i is a zero, its conjugate, -4 - 3i, is also a zero. Therefore, the zeros of the polynomial are -4 + 3i, -4 - 3i, and 1 (with a multiplicity of 2).
To find the polynomial, we can first determine its linear terms using the zeros:
(x - (-4 + 3i))(x - (-4 - 3i))(x - 1)(x - 1)
Simplifying this:
(x + 4 - 3i)(x + 4 + 3i)(x - 1)(x - 1)
Now, we can multiply these factors together to find the polynomial:
(x^2 + 8x + 25)(x - 1)^2
Expanding this further:
(x^2 + 8x + 25)(x^2 - 2x + 1)
Multiplying out gives:
x^4 + 6x^3 + 15x^2 - 2x^3 - 16x^2 - 40x + x^2 + 8x + 20
Combining like terms:
x^4 + 4x^3 - 3x^2 - 32x + 20
So, the polynomial with the given degree and zeros is f(x) = x^4 + 4x^3 - 3x^2 - 32x + 20.