Find a polynomial $$ f(x) $$ of degree 4 with real coefficients and the following zeros. 3 (multiplicity 2 ), $$ -i $$ $$ f(x)=\square $$

Question

UpStudy AI · Accepted Answer

The given zeros are $$3$$ with multiplicity $$2$$ and $$-i$$. Since the coefficients of the polynomial must be real, the complex conjugate $$i$$ must also be a zero.
The zeros of the polynomial are:
- $$3$$ (with multiplicity $$2$$), which contributes the factor $$(x - 3)^2$$
- $$-i$$ and $$i$$, which contribute the factors $$(x + i)$$ and $$(x - i)$$

Thus, the polynomial can be expressed as:
$$
f(x) = (x - 3)^2 (x + i)(x - i)
$$
First, we will simplify the complex factors:
$$
(x + i)(x - i) = x^2 + 1
$$
Now, we can write the polynomial as:
$$
f(x) = (x - 3)^2 (x^2 + 1)
$$
Next, we will expand $$(x - 3)^2$$ and then multiply it by $$(x^2 + 1)$$:
$$
(x - 3)^2 = x^2 - 6x + 9
$$
Now, we will multiply this by $$(x^2 + 1)$$:
$$
f(x) = (x^2 - 6x + 9)(x^2 + 1)
$$
We will now expand the expression:
$$
f(x) = x^2(x^2 + 1) - 6x(x^2 + 1) + 9(x^2 + 1)
$$
Calculating each term:
1. $$x^2 \cdot x^2 = x^4$$
2. $$x^2 \cdot 1 = x^2$$
3. $$-6x \cdot x^2 = -6x^3$$
4. $$-6x \cdot 1 = -6x$$
5. $$9 \cdot x^2 = 9x^2$$
6. $$9 \cdot 1 = 9$$

Now, combine these:
$$
f(x) = x^4 - 6x^3 + (1 + 9)x^2 - 6x + 9
$$
This simplifies to:
$$
f(x) = x^4 - 6x^3 + 10x^2 - 6x + 9
$$
The polynomial $$ f(x) $$ of degree 4 with the given zeros is: $$ f(x) = x^4 - 6x^3 + 10x^2 - 6x + 9 $$

Find a polynomial of degree 4 with real coefficients and the following zeros.
3 (multiplicity 2 ),

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Algebra Questions

Find a polynomial of degree 4 with real coefficients and the following zeros. 3 (multiplicity 2 ),