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Verify whether the following are zeroes of the polynomial indicated against them
\[P(x)=3x+1,x=\dfrac{1}{3}\]
\[P(x)=lx+m,x=\dfrac{m}{l}\]

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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Hint: Given a function \[y=f(x)\], we can find the ordered pairs for the function by substituting \[x=a\] in the equation of the function. By substituting \[x=a\] we get \[y=f(a)\], we should know that the \[\left( a,f(a) \right)\] is the coordinate of that lies on the graph of \[y=f(x)\]. If at \[x=a,f(a)=0\], then a is the root of the given function.

Complete step by step solution:
We are given the first function as \[P(x)=3x+1\] we need to find the functional value at \[x=\dfrac{1}{3}\]. To find this we substitute \[x=\dfrac{1}{3}\] in the equation of the function.
\[\Rightarrow P\left( \dfrac{1}{3} \right)=3\times \dfrac{1}{3}+1\]
Simplifying the above expression, we get
\[\Rightarrow P\left( \dfrac{1}{3} \right)=2\]
As \[P\left( \dfrac{1}{3} \right)\ne 0\] , \[x=\dfrac{1}{3}\] is not the root of the given polynomial function.
The next polynomial function is \[P(x)=lx+m\], we need to find the functional value at \[x=\dfrac{m}{l}\]. To do this we substitute \[x=\dfrac{m}{l}\] in the equation of the function,
\[\Rightarrow P(x)=l\left( \dfrac{m}{l} \right)+m\]
Simplifying the above expression, we get
\[\Rightarrow P(x)=2m\]
As we don’t know the value of m, we can not be sure whether \[x=\dfrac{m}{l}\] is zero or not.

Note:
We can find any number of ordered pairs for a given function by the same method. But we should be careful about one thing. We should not substitute values for which the function becomes undefined or values that are not part of the domain of function.
For example, for the function, \[\sqrt{x}\] we cannot substitute negative values as they are not in the domain of function. Also, for the function, \[\dfrac{1}{x}\] we cannot substitute \[x=0\] as the function becomes undefined for this value.
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