Question

How do you find a fourth-degree polynomial equation, with integer coefficients, for which $ - \sqrt {11} $ and $2i$ are solutions?

Answer 1

Hint: As per the above question we are asked to find a fourth degree polynomial equation with two given solutions. We know that by the fundamental theorem of algebra any polynomial of degree $4$can be written as: $P(x) = A(x - \alpha )(x - \beta )(x - \gamma )(x - \delta )$ where $\alpha ,\beta ,\gamma ,\delta $ are the roots or the zeros of the equation, $P(x) = 0$.

Complete step-by-step solution:
Here we have $ - \sqrt {11} $ and $2i$ are the solutions and we have to find a fourth degree polynomial. Let us assume $\alpha = - \sqrt {11} $ and $\beta = 2i$. Also we know that in fourth degree polynomial complex roots must appear as complex conjugate pairs, so we can see that a negative root is there also along with positive one i.e. $\gamma = - 2i$.
Now by substituting all the above values we get: $P(x) = A(x + \sqrt {11} )(x - 2i)(x + 2i)(x - \delta )$.
To get real coefficients of quadratic we have to multiply the complex roots and we get; $(x - 2i)(x + 2i) = x(x + 2i) - 2i(x + 2i)$.
On further simplifying it gives us ${x^2} - 4{i^2} = {x^2} - 4$, since $i$is an imaginary number and it is equal to $\sqrt { - 1} $.
So now we can put this value in the place of two imaginary roots, and using this we get: $P(x) = A(x + \sqrt {11} )({x^2} + 4)(x - \delta )$.
We have to find $P(x)$ with integer coefficients but we have irrational coefficients too, So we have to find the product of $(x + \sqrt {11} )$ and $(x - \delta )$. This can be done as follows; $(x + \sqrt {11} )(x - \delta ) = {x^2} + (\sqrt {11} - \delta )x - \sqrt {11} \delta $
Also if we assume $\delta = - \sqrt {11} $ then we get integer coefficients; $(x + \sqrt {11} )(x - \sqrt {11} ) = {x^2} - {(\sqrt {11} )^2}$
So we get ${x^2} - 11$.
Hence the required answer is $P(x) = A({x^2} - 11)({x^2} + 4)$, where A is an arbitrary number.

Note: We should note that $i$is an imaginary number and it has both positive and negative value. Also arbitrary number is a number which is any number defined to be but it has no specific value chosen yet. Fourth degree polynomials are also called quartic polynomials. They have zero to four roots.