Question

Find a quadratic polynomial whose zeroes are $ -3 $ and 4. \[\]

Answer 1

Hint: We recall zeroes of polynomial and quadratic polynomial .We use the fact that the if two zeroes of quadratic polynomials are $ \alpha ,\beta $ then the quadratic polynomial can be written as $ {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta $ \[\]

Complete step by step answer:
We know that a polynomial is an algebraic expression with a single variable. The highest power on the polynomial is called a degree. The values of the variable for which the polynomial returns zero, we call the values zeroes or roots of the polynomial. If $ p\left( x \right)={{a}_{0}}+{{a}_{1}}x+...+{{a}_{n}}{{x}^{n}} $ with degree $ n $ then we shall get zeroes which can find by solving $ p\left( x \right)=0 $ .\[\]
We also know that we call a polynomial quadratic if the degree of the polynomial is 2. The general form of quadratic polynomial is $ a{{x}^{2}}+bx+c $ . Here $ a $ is the coefficient of $ {{x}^{2}} $ which cannot be zero, $ b $ is the coefficient of $ x $ and $ c $ is the constant term. \[\]
We know how to make a quadratic polynomial from its two zeros $ \alpha ,\beta $ . We are given in the question that he zeroes of the polynomial are $ \alpha =-3 $ and $ \beta =4 $ .We first find the sum of the zeroes as
\[\alpha +\beta =-3+4=1\]
We also find the product of the polynomials
\[\alpha \beta =\left( -3 \right)4=-12\]
We take the coefficient of $ {{x}^{2}} $ as $ a=1 $ , we take negative of sum of zeroes as the coefficient of $ x $ that is $ b=-\left( \alpha +\beta \right) $ and we take the product of zeroes as the constant term that is $ c=\alpha \beta $ . So the required quadratic polynomial is;
\[\begin{align}
  & \Rightarrow a{{x}^{2}}+bx+c \\
 & \Rightarrow 1\times {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \\
 & \Rightarrow {{x}^{2}}-1\times x-12 \\
 & \Rightarrow {{x}^{2}}-x-12 \\
\end{align}\]

Note:
We can alternatively solve using factor theorem which states that a polynomial $ p\left( x \right) $ has a factor $ \left( x-a \right) $ if and only if $ p\left( a \right)=0 $ in other words $ a $ is a zero of $ p\left( x \right) $ . Hence $ x-\left( -3 \right)=x+3,x-4 $ will be factors and the only two factors of the quadratic polynomial. So we can find the required quadratic polynomial as $ \left( x+3 \right)\left( x-4 \right)={{x}^{2}}-x-12 $ , When we equate the quadratic polynomial to zero, we get quadratic equation $ a{{x}^{2}}+bx+c=0 $ .