For a polynomial f (x) with real coefficients having the given degree and zeros.
Degree 4; zeros 4+2i ; 4 multiplicity 2
the complex roots occur in conjugate pairs, so 4-2i is also a root. So,
f(x) = (x-4)^2 (x-(4+2i)) (x-(4-2i))
= (x-4)^2 ((x-4)-2i)((x-4)+2i)
= (x-4)^2 ((x-4)^2 - (2i)^2)
= (x-4)^2 (x^2-8x+16 + 4)
= (x-4)^2 (x^2-8x+20)
To find the polynomial with the given zeros and multiplicities, we can use the concept of complex conjugate pairs.
Since the polynomial has degree 4 and the zero 4 + 2i, its complex conjugate is also a zero. Therefore, the complex conjugate of 4 + 2i is 4 - 2i.
We are also given that the zero 4 has multiplicity 2. This means that it appears twice in the polynomial.
Thus, the zeros and their multiplicities are:
- 4 + 2i (multiplicity 1)
- 4 - 2i (multiplicity 1)
- 4 (multiplicity 2)
To find the polynomial, we can write it in factored form as follows:
f(x) = (x - (4 + 2i))(x - (4 - 2i))(x - 4)(x - 4)
Using the concept of conjugate pairs, we can simplify this expression:
f(x) = ((x - 4) - 2i)((x - 4) + 2i)(x - 4)(x - 4)
= ((x - 4)^2 - (2i)^2)(x - 4)(x - 4)
= ((x - 4)^2 + 4)(x - 4)(x - 4)
= (x^2 - 8x + 16 + 4)(x - 4)(x - 4)
= (x^2 - 8x + 20)(x - 4)(x - 4)
Expanding this expression further, we get:
f(x) = (x^2 - 8x + 20)(x^2 - 8x + 20)
Now, if we multiply the two quadratic factors together, we obtain the polynomial in standard form:
f(x) = x^4 - 8x^3 + 20x^2 - 8x^3 + 64x^2 - 160x + 20x^2 - 160x + 400
= x^4 - 16x^3 + 104x^2 - 320x + 400
Therefore, the polynomial f(x) with the given zeros and multiplicities is:
f(x) = x^4 - 16x^3 + 104x^2 - 320x + 400.
To find the polynomial with the given degree and zeros, we'll have to use the multiplication rule for zeros of a polynomial with real coefficients.
First, let's break down the given information:
- Degree: 4
- Zeros: 4+2i (complex conjugate) and 4 (with multiplicity 2)
Since the degree of the polynomial is 4, we know that the polynomial will have 4 zeros in total, counting multiplicities.
The zero 4 has a multiplicity of 2, which means it appears twice as a zero. This implies that the factor (x - 4) will appear twice in the polynomial.
The zero 4+2i is a complex conjugate, which means its conjugate, 4-2i, will also be a zero. Both these complex zeros are connected to the polynomial using the conjugate rule. Therefore, the factors (x - (4+2i)) and (x - (4-2i)) will appear in the polynomial.
Now, we can multiply these factors together to get the polynomial.
Step 1: Multiply the complex factor.
(x - (4+2i))(x - (4-2i))
To simplify this, we can use the difference of squares formula:
(a - b)(a + b) = a^2 - b^2
Applying this formula, we get:
(x - 4 - 2i)(x - 4 + 2i) = (x - 4)^2 - (2i)^2
= (x - 4)^2 - 4i^2
= (x - 4)^2 - 4(-1)
= (x - 4)^2 + 4
Step 2: Multiply the real factor and the result from step 1.
(x - 4)^2 * ((x - 4)^2 + 4)
Expanding this expression, we get:
(x^2 - 8x + 16) * (x^2 - 8x + 16 + 4)
= (x^2 - 8x + 16) * (x^2 - 8x + 20)
Finally, to get the polynomial, we can expand this product:
(x^2 - 8x + 16)(x^2 - 8x + 20)
= x^4 - 8x^3 + 16x^2 - 8x^3 + 64x^2 - 128x + 16x^2 - 128x + 256
= x^4 - 16x^3 + 96x^2 - 256x + 256
Therefore, the polynomial f(x) is:
f(x) = x^4 - 16x^3 + 96x^2 - 256x + 256