Question

How do you write a cubic polynomial function with zeros \[-3,2\] and \[1\] ?

Answer 1

Hint: To find the cubic polynomial when its zeros are given, we have to follow a few steps. First, we have to find the sum of the zeros of the polynomial taken one at a time. After that our second step is to find the sum of the zeros of the polynomial taken two at a time and then find the product of the zeros of the given cubic polynomial. Then substitute the value in the equation \[{{x}^{3}}-{{S}_{1}}{{x}^{2}}+{{S}_{2}}x-P\].
Here, \[{{S}_{1}}\] is sum of the zeros of the polynomial taken one at a time,\[{{S}_{2}}\]is sum of the zeros of the polynomial taken two at a time and \[P\] product of the zeros of the of the given cubic polynomial.

Complete step-by-step answer:
In this question we can see that the zeros given are \[-3,2\] and \[1\]. Now, let us find the sum of the zeros taken one at a time, taken two at a time and the product to find the cubic polynomial.
Sum of the zeros taken one at a time,
Let \[{{x}_{1}}=-3,{{x}_{2}}=2,{{x}_{3}}=1\]
\[{{S}_{1}}={{x}_{1}}+{{x}_{2}}+{{x}_{3}}\]
\[{{S}_{1}}=-3+2+1=0\]
Sum of the zeros taken two at a time,
\[{{S}_{2}}={{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{1}}{{x}_{3}}\]
\[{{S}_{2}}=-3\times 2+2\times 1+(-3\times 1)\]
\[{{S}_{2}}=-6+2+(-3)\]
\[{{S}_{2}}=-7\]
 Product of the zeros
\[P={{x}_{1}}{{x}_{2}}{{x}_{3}}\]
\[P=-3\times 2\times 1=-6\]
Thus, we get \[{{S}_{1}}=0\] \[{{S}_{2}}=-7\] \[P=-6\]
Now, we will substitute the values in the general equation which is \[{{x}^{3}}-{{S}_{1}}{{x}^{2}}+{{S}_{2}}x-P\]
After substitution we get \[{{x}^{3}}-(0){{x}^{2}}+(-7)x-(-6)\]
\[\Rightarrow {{x}^{3}}-7x+6\]
Therefore, the cubic polynomial function with zeros \[-3,2\] and \[1\] is \[{{x}^{3}}-7x+6\] .

Note: Keep in mind the equation used above to find the cubic polynomial using zeros. Also, check the calculations once you have completed them, chances of mistakes are there. One more way to solve the question is since you know the values of the zeros.
Since \[{{x}_{1}}=-3,{{x}_{2}}=2,{{x}_{3}}=1\] are the zeros of same polynomial therefore,\[{{x}_{1}}={{x}_{2}}={{x}_{3}}=x\]
\[\begin{align}
  & {{x}_{1}}=-3,{{x}_{2}}=2,{{x}_{3}}=1 \\
 & {{x}_{1}}+3=0,{{x}_{2}}-2=0,{{x}_{3}}-1=0 \\
 & ({{x}_{1}}+3)({{x}_{2}}-2)({{x}_{3}}-1)=0 \\
 & \Rightarrow (x+3)(x-2)(x-1)=0 \\
 & \\
\end{align}\]
After solving the above equation, we get the same answer
\[{{x}^{3}}-7x+6\]