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Question: How do you find all zeros of \({{x}^{3}}+{{x}^{2}}+9x+9\) given zero \(3i\)?...

How do you find all zeros of x3+x2+9x+9{{x}^{3}}+{{x}^{2}}+9x+9 given zero 3i3i?

Explanation

Solution

We have to find the roots of the given polynomial when we have given the one root as 3i3i. We will simplify the given polynomial by taking common terms out. Then by equating each common factor with zero we will find the remaining zeros of the given polynomial.

Complete step-by-step answer:
We have been given a polynomial x3+x2+9x+9{{x}^{3}}+{{x}^{2}}+9x+9.
We have to find the zeros of the given polynomial.
Now, we know that the number of zeros or roots of an equation or polynomial depends on the degree of polynomial. A polynomial with degree three has three zeros, given one zero as 3i3i.
Now, let us try to simplify the given polynomial. Then we will get
⇒(x3+x2)+(9x+9)=0\Rightarrow \left( {{x}^{3}}+{{x}^{2}} \right)+\left( 9x+9 \right)=0
Now, taking common terms out we will get
⇒x2(x+1)+9(x+1)=0\Rightarrow {{x}^{2}}\left( x+1 \right)+9\left( x+1 \right)=0
Again taking common factors out we will get
⇒(x2+9)(x+1)=0\Rightarrow \left( {{x}^{2}}+9 \right)\left( x+1 \right)=0
Now, equating each factor with zero and simplifying the obtained equations we will get
⇒(x2+9)=0\Rightarrow \left( {{x}^{2}}+9 \right)=0 and (x+1)=0\left( x+1 \right)=0
⇒x2=−9\Rightarrow {{x}^{2}}=-9 and x=−1x=-1
⇒x=−9 ⇒x=±3i \begin{aligned} & \Rightarrow x=\sqrt{-9} \\\ & \Rightarrow x=\pm 3i \\\ \end{aligned}
Hence we get the zeros of the given polynomial as 3i,−3i,−13i,-3i,-1.

Note: In this particular question we solve the polynomial by grouping and taking common terms out. This approach is not useful in solving all polynomials. It is difficult to find the zeros of a polynomial by using this method. To find the zeros of a polynomial there is no specific method so we have to solve it by using a hit and trial method.