How do you find a polynomial function $f$ with real coefficients of the indicated degree that possesses the given zeros: degree $2$; $4 + 3i$?
Answer
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Hint: in the question we are given that a second-degree polynomial function has one of its roots as $4 + 3i$, and we have to find out the polynomial equation from this information, therefore we will use the property of conjugate of imaginary root to find the polynomial.
Complete step-by-step answer:
We have one of the given roots of the polynomial as $4 + 3i$ and since it is a degree $2$ polynomial, it is a quadratic equation.
Now we know that if one imaginary number is a solution to a polynomial then the conjugate of the number is also a solution to the equation.
Now the conjugate of the solution $4 + 3i$ is $4 - 3i$, which is also the root to the equation therefore, the roots can be written as:
$x = 4 + 3i$ and $x = 4 - 3i$
On rephrasing the equation, we can write it as:
$x - (4 + 3i) = 0$ and $x - (4 - 3i) = 0$
Now the quadratic equation can be derived from the solutions of it by multiplying them therefore, the quadratic equation is:
$ \Rightarrow (x - (4 + 3i))(x - (4 - 3i))$
On opening the brackets, we get:
$ \Rightarrow (x - 4 - 3i)(x - 4 + 3i)$
Now on multiplying the terms, we get:
$ \Rightarrow {x^2} - 4x + 3xi - 4x + 16 - 12i - 3xi + 12i - 9{i^2}$
Now we know that ${i^2} = - 1$ therefore, on simplifying, we get:
$ \Rightarrow {x^2} - 8x + 16 - 9( - 1)$
On simplifying, we get:
$ \Rightarrow {x^2} - 8x + 16 + 9$
On adding the terms, we get:
$ \Rightarrow {x^2} - 8x + 25$, is the required quadratic equation with zeros $4 \pm 3i$ thus the function can be written as:
$ \Rightarrow f(x) = {x^2} - 8x + 25$
Note:
It is to be remembered that zeros of the equation represent the root or the solution of the equation, it is the term which when substituted in the equation, and we get the value as $0$.
The roots of a quadratic equation can be found using the formula $(x,y) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2ac}}$
Where $(x,y)$ are the roots of the equation and $a,b,c$ are the coefficients of the terms in the quadratic equation.
Complete step-by-step answer:
We have one of the given roots of the polynomial as $4 + 3i$ and since it is a degree $2$ polynomial, it is a quadratic equation.
Now we know that if one imaginary number is a solution to a polynomial then the conjugate of the number is also a solution to the equation.
Now the conjugate of the solution $4 + 3i$ is $4 - 3i$, which is also the root to the equation therefore, the roots can be written as:
$x = 4 + 3i$ and $x = 4 - 3i$
On rephrasing the equation, we can write it as:
$x - (4 + 3i) = 0$ and $x - (4 - 3i) = 0$
Now the quadratic equation can be derived from the solutions of it by multiplying them therefore, the quadratic equation is:
$ \Rightarrow (x - (4 + 3i))(x - (4 - 3i))$
On opening the brackets, we get:
$ \Rightarrow (x - 4 - 3i)(x - 4 + 3i)$
Now on multiplying the terms, we get:
$ \Rightarrow {x^2} - 4x + 3xi - 4x + 16 - 12i - 3xi + 12i - 9{i^2}$
Now we know that ${i^2} = - 1$ therefore, on simplifying, we get:
$ \Rightarrow {x^2} - 8x + 16 - 9( - 1)$
On simplifying, we get:
$ \Rightarrow {x^2} - 8x + 16 + 9$
On adding the terms, we get:
$ \Rightarrow {x^2} - 8x + 25$, is the required quadratic equation with zeros $4 \pm 3i$ thus the function can be written as:
$ \Rightarrow f(x) = {x^2} - 8x + 25$
Note:
It is to be remembered that zeros of the equation represent the root or the solution of the equation, it is the term which when substituted in the equation, and we get the value as $0$.
The roots of a quadratic equation can be found using the formula $(x,y) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2ac}}$
Where $(x,y)$ are the roots of the equation and $a,b,c$ are the coefficients of the terms in the quadratic equation.
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