Question

How do you form a polynomial $f\left( x \right)$ with real coefficients having a given degree and zeros? Degree $4$ ; zeros $- 5 + 2i$ ; $3$ multiplicity $2$ . How do you form a polynomial $f\left( x \right)$ with real coefficients having a given degree and zeros? Degree $5$ ; zeros: $- 4; - i; - 3 + i$

Answer 1

In this problem we are asked to form the polynomial with real coefficients having given degree and zeros. Also we have given two zeros with the multiplicity of some integer. And we have asked to form another polynomial with degree $5$ and with the use of given zeros. So we need to find two polynomials $f\left( x \right)$ .

Complete Step by Step Solution:
$1)$ We have to find the polynomial $f\left( x \right)$ with real coefficients having degree $4$ and zeros $- 5 + 2i$ ; $3$ multiplicity $2$ .
$\Rightarrow x = - 5 + 2i,x = - 5 - 2i,x = 3,x = 3$ , since our degree is $4$ so there are four zeros.
Now, $f\left( x \right) = \left( {x + 5 - 2i} \right)\left( {x + 5 + 2i} \right){\left( {x - 3} \right)^2}$
Now let’s write our $f\left( x \right)$ as the multiple of two terms $A,B$ , where $A = \left( {x + 5 - 2i} \right)\left( {x + 5 + 2i} \right)$ and $B = {\left( {x - 3} \right)^2}$
$\therefore f\left( x \right) = A.B - - - - - \left( 1 \right)$
Let us expand $A$ and $B$ using distributive property and ${\left( {a - b} \right)^2}$ formula, we get
$A = {x^2} + 5x + 2ix + 5x + 25 + 10i - 2ix - 10i - 4{i^2}$ , but we know that ${i^2} = - 1$
Substitute ${i^2} = - 1$ in the expansion of $A$ , we get
$A = {x^2} + 10x + 29$
Expanding $B$ implies,
$B = {x^2} - 6x + 9$
Substitute the value of $A,B$ in the equation (1), we get
$f\left( x \right) = \left( {{x^2} + 10x + 29} \right)\left( {{x^2} - 6x + 9} \right)$
Using distributive property, we get
$f\left( x \right) = {x^4} - 6{x^3} + 9{x^2} + 10{x^3} - 60{x^2} + 90x + 29{x^2} - 174x + 261$ , simplifying the terms we get,
$f\left( x \right) = {x^4} + 4{x^3} - 22{x^2} - 84x + 261$ , this is the polynomial with degree four.
$2)$ We have to find the polynomial $f\left( x \right)$ with real coefficients having degree $5$ and zeros $- 4; - i; - 3 + i$ .
$\Rightarrow x = - 4,x = \pm i,x = - 3 \pm i$ , since we are having degree $5$ .
Now, $f\left( x \right) = \left( {x + 4} \right)\left( {x - i} \right)\left( {x + i} \right)\left( {x + 3 - i} \right)\left( {x + 3 + i} \right)$
Multiplying second term and third term together and multiplying fourth term and fifth term together, we get
$f\left( x \right) = \left( {x + 4} \right)\left( {{x^2} + 1} \right)\left( {{x^2} + 6x + 10} \right)$
Now, using distributive property, we have

$f\left( x \right) = {x^5} + 10{x^4} + 3{x^3} + 50{x^2} + 34x + 40$ , this is the polynomial with degree $5$ .

Note: If there are two brackets, then apply distributive property. While applying distribution property we need to be careful about the multiplication. Here we have given complex numbers in some zeros, so when multiplying two complex numbers we have to be clear about the imaginary part. Product of two imaginary parts adds a negative sign to the result.