The number of polynomials having zeros -2 and 5 is.
Answer
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Hint: For solving this problem, we obtain the sum and product of roots individually from the given data and then place the values of variables in the generalized form to obtain the final polynomial. Once we obtain the polynomial we can easily say about the numbers of such polynomials.
Complete step-by-step answer:
In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
If we have two zeros of a quadratic equation then the polynomial could be formed by using the simplified result which could be stated as:
${{x}^{2}}-(a+b)x+ab$, where a and b are two zeroes of the equation.
According to the problem statement, the two zeros of a polynomial are -2 and 5.
So, the sum of zeroes of the polynomial
$\begin{align}
& =a+b \\
& =-2+5 \\
& =3 \\
\end{align}$
Now, product of zeroes of the polynomial
$\begin{align}
& =ab \\
& =-2\times 5 \\
& =-10 \\
\end{align}$
Putting the obtained values in the above expression, we get
$\begin{align}
& {{x}^{2}}-(a+b)x+ab={{x}^{2}}-3x+\left( -10 \right) \\
& \therefore {{x}^{2}}-3x-10=0 \\
\end{align}$
Therefore, only one polynomial is formed using -2 and 5 which is ${{x}^{2}}-3x-10=0$.
Note: This problem could be alternatively solved by using the concept that zeros can also be expressed as factors. Now, we have -2 and 5 as zeros. Therefore, the polynomial would be represented as $\left( x+2 \right)\cdot \left( x-5 \right)={{x}^{2}}-3x-10$ which is only one polynomial.
Complete step-by-step answer:
In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
If we have two zeros of a quadratic equation then the polynomial could be formed by using the simplified result which could be stated as:
${{x}^{2}}-(a+b)x+ab$, where a and b are two zeroes of the equation.
According to the problem statement, the two zeros of a polynomial are -2 and 5.
So, the sum of zeroes of the polynomial
$\begin{align}
& =a+b \\
& =-2+5 \\
& =3 \\
\end{align}$
Now, product of zeroes of the polynomial
$\begin{align}
& =ab \\
& =-2\times 5 \\
& =-10 \\
\end{align}$
Putting the obtained values in the above expression, we get
$\begin{align}
& {{x}^{2}}-(a+b)x+ab={{x}^{2}}-3x+\left( -10 \right) \\
& \therefore {{x}^{2}}-3x-10=0 \\
\end{align}$
Therefore, only one polynomial is formed using -2 and 5 which is ${{x}^{2}}-3x-10=0$.
Note: This problem could be alternatively solved by using the concept that zeros can also be expressed as factors. Now, we have -2 and 5 as zeros. Therefore, the polynomial would be represented as $\left( x+2 \right)\cdot \left( x-5 \right)={{x}^{2}}-3x-10$ which is only one polynomial.
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