Question

If we have a polynomial as \[p\left( u \right)=4{{u}^{2}}+8u\] then find the zeros of the polynomial \[p\left( u \right)\].

Answer 1

Hint: We solve this problem by making the given polynomial to ‘0’ because the zeros of a polynomial are defined as the values of \[u\] for which the polynomial becomes ‘0’. Then we use a suitable method to solve this problem.

Complete step-by-step solution
We are given that the polynomial as
\[p\left( u \right)=4{{u}^{2}}+8u\]
We know that the zeros of a polynomial are defined as the values of \[u\] for which the polynomial becomes ‘0’.
By using this definition we can write
\[\begin{align}
  & \Rightarrow p\left( u \right)=0 \\
 & \Rightarrow 4{{u}^{2}}+8u=0 \\
\end{align}\]
Here, by taking \[u\] common out from the above equation we get
\[\Rightarrow 4u\left( u+2 \right)=0\]
We know that if multiplication of two numbers is zero then either of them should be ‘0’
By using this result we can write
\[\Rightarrow u=0 or -2\]
Therefore the zeros of the given polynomial are $0, -2$.

Note: This problem can be solved in another method.
For a quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] the formula for roots of equation is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Here we got that
\[\Rightarrow 4{{u}^{2}}+8u=0\]
By using the above formula we get
\[\begin{align}
  & \Rightarrow u=\dfrac{-8\pm \sqrt{64-0}}{2\times 4} \\
 & \Rightarrow u=\dfrac{-8\pm 8}{8} \\
\end{align}\]
Here, positive sign gives one value and negative sign gives other value.
By separating the above equation we get
\[\Rightarrow u=0,-2\]
Therefore the zeros of the given polynomial are 0, -2.
Let us check whether the zeros we got are correct or not.
We have \[p\left( u \right)=4{{u}^{2}}+8u\]
Let us substitute \[u=0\] we get
\[\begin{align}
  & \Rightarrow p\left( 0 \right)=4{{\left( 0 \right)}^{2}}+8\left( 0 \right) \\
 & \Rightarrow p\left( 0 \right)=0 \\
\end{align}\]
Now let us substitute \[u=-2\] in above equation we get
\[\begin{align}
  & \Rightarrow p\left( -2 \right)=4{{\left( -2 \right)}^{2}}+8\left( -2 \right) \\
 & \Rightarrow p\left( -2 \right)=16-16=0 \\
\end{align}\]
Here, we can see that when we substitute \[u=0,-2\] the polynomial becomes ‘0’. So we can say that the zeros we got are correct.
Finally, we can conclude that \[p\left( u \right)=4{{u}^{2}}+8u\] has ‘0’ and ‘-2’ are the zeros.