Question

How do you write a polynomial in standard form given the zeros $x=0,0,2,3$ ?

Answer 1

Hint: Try to get the factors by treating zeros as the roots of the polynomial (assuming the polynomial is equal to 0). Multiply the factors to get the required polynomial. Rearrange the polynomial as the sum of decreasing power of ‘x’, to get the polynomial in standard form.

Complete step by step answer:
If we have zeros of the polynomial means they are the roots when the polynomial is equal to zero.
Considering $x=0,0,2,3$ as four roots of the polynomial, the factors are $\left( x-0 \right)$, $\left( x-0 \right)$, $\left( x-2 \right)$ and $\left( x-3 \right)$.
By multiplying all the factors we can get the polynomial as
$\left( x-0 \right)\left( x-0 \right)\left( x-2 \right)\left( x-3 \right)$
Taking two at a time
$\begin{align}
  & =\left( \left( x-0 \right)\left( x-0 \right) \right)\left( \left( x-2 \right)\left( x-3 \right) \right) \\
 & \Rightarrow \left( x\cdot x \right)\left( {{x}^{2}}-3x-2x+6 \right) \\
 & \Rightarrow {{x}^{2}}\left( {{x}^{2}}-5x+6 \right) \\
 & \Rightarrow {{x}^{4}}-5{{x}^{3}}+6{{x}^{2}} \\
\end{align}$
Standard form of a polynomial: We should write the polynomial as the sum of the decreasing power of the variable, ’x’.
Here, since it’s already written in decreasing power of ‘x’ hence the standard form of the polynomial is
$={{x}^{4}}-5{{x}^{3}}+6{{x}^{2}}$
This is the required polynomial.

Note:
Zeros should be considered as the roots for getting the required factors by assuming the polynomial is equal to ‘0’. After the multiplication of factors rearrange should be done if necessary. Final polynomials should be in the sum of decreasing power of ‘x’. For example the polynomial ${{x}^{4}}-5{{x}^{3}}+6{{x}^{2}}$ can’t be written as $6{{x}^{2}}-5{{x}^{3}}+{{x}^{4}}$ as it is not in the standard form. Since we are getting a fourth degree polynomial hence the written order should be first fourth degree term then third degree term then second degree term and so on.