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Question: How do you find a polynomial function that has zeros \(1+\sqrt{3}\), \(1-\sqrt{3}\)?...

How do you find a polynomial function that has zeros 1+31+\sqrt{3}, 1−31-\sqrt{3}?

Explanation

Solution

In this question we have the zeros of a polynomial which are the values which on substitution give us the value of the polynomial as zero. Since the degree of the polynomial is not mentioned we will consider the simplest polynomial with the distinct zeros which are 1+31+\sqrt{3} and 1−31-\sqrt{3} respectively.

Complete step by step solution:
We have the zeros of the equation given to us as 1+31+\sqrt{3} and 1−31-\sqrt{3}.
On considering the variable in the polynomial as xx, we can write the solution of the polynomial as:
x=1+3x=1+\sqrt{3} and x=1−3x=1-\sqrt{3}
On transferring the terms from the right-hand side to the left-hand side on both the equations, we get:
x−(1+3)=0x-\left( 1+\sqrt{3} \right)=0 and x−(1−3)=0x-\left( 1-\sqrt{3} \right)=0
Now the polynomial equation can be derived from the solutions of it by multiplying them therefore, the polynomial equation is:
⇒(x−(1+3))(x−(1−3))\Rightarrow \left( x-\left( 1+\sqrt{3} \right) \right)\left( x-\left( 1-\sqrt{3} \right) \right)
On opening the brackets, we get:
⇒(x−1−3)(x−1+3)\Rightarrow \left( x-1-\sqrt{3} \right)\left( x-1+\sqrt{3} \right)
Now on multiplying the terms, we get:
⇒x2−x+x3−x+1−3−x3+3−3=0\Rightarrow {{x}^{2}}-x+x\sqrt{3}-x+1-\sqrt{3}-x\sqrt{3}+\sqrt{3}-3=0
On cancelling the similar terms, we get:
⇒x2−x−x+1−3=0\Rightarrow {{x}^{2}}-x-x+1-3=0
On simplifying the terms, we get:
⇒x2−2x−2=0\Rightarrow {{x}^{2}}-2x-2=0
Which is the required polynomial equation. On writing it in terms of the function, we get:
f(x)=x2−2x−2f(x)={{x}^{2}}-2x-2, which is the required polynomial equation.

Note:
In this question we have a polynomial equation of degree 22. These types of polynomials are also called quadratic equations.
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, we get the value as 00.
The roots of a quadratic equation can be found using the formula (x,y)=−b±b2−4ac2ac(x,y)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2ac}
Where (x,y)(x,y) are the roots of the equation and a,b,ca,b,c are the coefficients of the terms in the quadratic equation.