Question

How do I find the complete factored form of a polynomial with a degree of 3, having a leading coefficient with some zeros i and 1.

Answer 1

Hint: If all the coefficient of a polynomial is real and if some roots of the equation are complex. The all complex roots have their conjugate as a root of the equation. If a+ib is the root of a given polynomial equation that implies a-ib is also a root of that equation .

Complete step by step answer:
In the given question the degree of the equation is 3 that means it is a cubic equation and total 3 roots are possible for the equation. Given 2 roots are i and 1. We know that all complex roots of a polynomial equation have their conjugate as the root of the equation.
To find out the conjugate of a complex number we just multiply -1 with the imaginary part of the number, the real part of the number remains unchanged. For example, the conjugate of 3+4i is 3-4i.
So if 3+4i is the root of a polynomial equation then 3-4i must be a root of that polynomial equation.
In the given question, i is the root of a cubic equation that implies conjugate of i must be a root of the cubic equation.
Conjugate of i=-i.
So the three roots of the equation are 1,i,-i.
We know that if roots of any equation are ${{a}_{1,}}{{a}_{2}},{{a}_{3}},.........,{{a}_{n}}$ then we can write the equation as
$C\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)\left( x-{{a}_{3}} \right)..........\left( x-{{a}_{n}} \right)$ where C is the leading coefficient. Leading coefficient is the highest power of x.
Here in the question the leading coefficient is 2 and roots are 1, i,and - i.

So the we can write the equation as $2\left( x-1 \right)\left( x-i \right)\left( x-(-i) \right)$=$2\left( x-1 \right)\left( x-i \right)\left( x+i \right)$ .
We can solve the equation further but above form is the factored form of the equation.

Note: Always remember that a polynomial equation having real coefficient has an even number of complex roots that means the equation can’t have 1 or 3 or 5 complex roots because complex roots always occur with its conjugate . To write in factored form , if the n roots of a nth degree polynomial are ${{a}_{1,}}{{a}_{2}},{{a}_{3}},.........,{{a}_{n}}$ then the factored form of the equation is $\prod\limits_{i=1}^{n}{\left( x-{{a}_{i}} \right)}$.