Value of a Polynomial
In this mini-lesson, we will learn about the value of a polynomial by understanding its meaning, and we will also understand how to find the value of a polynomial expression.
We know that a polynomial contains variables, constant terms, and operators.
A polynomial is a mathematical expression written in the form:
\[a_0x^n + a_1x^{n-1} + a_2x^{n-2}...........+ a_nx^0\]
The above expression is also called "polynomials in the standard form."
Where \(a_0, a_1, a_2.........a_n\) are constants and \(n\) is a natural number.
If we change the value of the variable, the value of the polynomial also changes.
For example, if the value of \(x\) in the polynomial \(P(x) = x + 1\) changed from \(x = 1\) to \(x = 2\) the value of \(P(x)\) also changes from \(2\) to \(3\)
Let's consider the graph of the function \(y = x^2\)
We can clearly see that the graph is taking different values for different values of \(x\).
Now, let's explore the value of a polynomial in more detail.
Lesson Plan
What do you Mean by Value of a Polynomial?
The value of a polynomial means the value a polynomial takes if we substitute the variables with any number.
For example, \(P(x)\) is a polynomial; the value \(P(x)\) takes at any \(x = a\) is \(P(a)\)
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Value of \(P(x)\) at \(x = a\) is \(P(a)\) |
To find the value of \(P(x)\) at \(x = a\), we need to replace \(x\) with \(a\) in the polynomial.
Let's consider the graph of the function \(y = x^3 - 6x^2 + 11x - 6\)
Now, we can clearly observe that the graph of \(y = x^3 - 6x + 11x - 6\) is taking different values at different values of \(x\).
At \(x = 1\), \(y = 0\)
At \(x = 1.5\), \(y = 0.375\)
At \(x = 2\), \(y = 0\)
At \(x = 2.5\), \(y = -0.375\)
Enter the polynomial and explore the graph of the polynomial to check for the value of \(y\) at any value of \(x\).
- If a polynomial goes to negative infinity when \(x\) goes to negative infinity, and positive infinity when \(x\) goes to positive infinity, will this polynomial ever take a zero value at some \(x\)? If it can take, then can we find the number of values of \(x\) at which the polynomial takes zero value? Think about it.
How to Find the Value of a Polynomial Expression?
To find the value of a polynomial at a point \(x = a\), we only need to replace \(x\) with \(a\) in the equation of the polynomial.
So, if \(P(x)\) is a polynomial, the value \(P(x)\) takes at any \(x = a\) is \(P(a)\).
Let's consider an example.
If \(P(x) = x^2 + x + 1\), then the value of \(P(x)\) at different values of \(x\) can be calculated as follows:
At \(x = 0\), it means that we have to find the value of \(P(0)\).
So, replace \(x\) with \(0\).
\(P(0) = 0^2 + 0 + 1 = 0 + 0 + 1 = 1\)
At \(x = 1\), it means that we have to find the value of \(P(1)\).
So, replace \(x\) with \(1\).
\(P(1) = 1^2 + 1 + 1 = 1 + 1 + 1 = 3\)
At \(x = 2\), it means that we have to find the value of \(P(2)\).
So, replace \(x\) with \(2\).
\(P(2) = 2^2 + 2 + 1 = 4 + 2 + 1 = 7\)
At \(x = -1\), it means that we have to find the value of \(P(-1)\).
So, replace \(x\) with \(-1\).
\(P(-1) = {(-1)}^2 + (-1) + 1 = 1 - 1 + 1 = 1\)
Let's explore the value of polynomial expressions at different values of \(x\) with the help of a table.
| Polynomial expression | at \(x = 0\) | at \(x = 1\) | at \(x = 2\) |
| \(x + 7\) | \(7\) | \(8\) | \(9\) |
| \(3x - 2\) | \(-2\) | \(1\) | \(4\) |
| \(x^2 - 5x + 12\) | \(12\) | \(8\) | \(6\) |
| \(x^4 + 2x^3 - x^2 + 11x + 1\) | \(1\) | \(14\) | \(51\) |
The simulation shown below is the value of polynomial calculator.
Enter the polynomial and the value of \(x\) for which you want to find the value of the polynomial.
- The value of polynomial \(P(x)\) at \(x = a\) is \(P(a)\).
- A polynomial can take multiple values for the value of \(x\).
- If a polynomial takes the value 0 at any value of \(x\), then \(x\) is called the zero of the polynomial.
Solved Examples
| Example 1 |
Help John find the value of the polynomial \(P(x) = x^3 - 2x^2 + 1\) at \(x = 1\) and \(x = 2\).
Solution
Given the polynomial \(P(x) = x^3 - 2x^2 + 1\).
The value of \(P(x) = x^3 - 2x^2 + 1\) at \(x = 1\) and \(x = 2\) is:
At \(x = 1\)
\[P(1) = 1^3 - 2 \times 1^2 + 1\\[0.2cm]
P(1) = 1 - 2 \times 1 + 1\\[0.2cm]
P(1) = 1 - 2 + 1\\[0.2cm]
P(1) = 0\]
at \(x = 2\)
\[P(2) = 2^3 - 2 \times 2^2 + 1\\[0.2cm]
P(2) = 8 - 2 \times 4 + 1\\[0.2cm]
P(2) = 8 - 8 + 1\\[0.2cm]
P(2) = 1\]
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\(\therefore\) \(P(1) = 0\) and \(P(2) = 1\) |
| Example 2 |
Mathew finds out that the polynomial \(P(x) = x^3 - 3x^2 - x + 3\) has \(3\) as one of its zeros.
Can you check whether he is right?
Solution
If \(3\) is the zero of the polynomial, then \(P(3) = 0\)
Put \(x = 3\) in the polynomial.
\[\begin{align} P(3) &= 3^3 - 3 \times 3^2 - 3 + 3 \\[0.2cm]
P(3) &= 27 - 3 \times 9 - 3 + 3 \\[0.2cm]
P(3) &= 27 - 27 - 3 + 3 \\[0.2cm]
P(3) &= 0 \end{align}\]
\(3\) is the zero of polynomial \(P(x) = x^3 - 3x^2 - x + 3\)
| \(\therefore\) Mathew is right. |
| Example 3 |
If the polynomial \(P(x) = x^4 - kx^3 + 6x^2 - x + 3\) has the value \(9\) at \(x = 2\), then what is the value of \(k\)?
Solution
If polynomial \(P(x) = x^4 - kx^3 + 6x^2 - x + 3\) has the value \(9\) at \(x = 2\), then \(P(2) = 9\)
Substitute \(x\) with 2 in the expression of polynomial
\[\begin{align} P(2) &= 2^4 - k \times 2^3 + 6 \times 2^2 - 2 + 3 \\[0.2cm]
P(2) &= 16 - k \times 8 + 6 \times 4 - 2 + 3 \\[0.2cm]
P(2) &= 16 - 8k + 24 - 2 + 3 \\[0.2cm]
P(2) &= 41 - 8k \\[0.2cm]
9 &= 41 - 8k \\[0.2cm]
8k &= 41 - 9 \\[0.2cm]
8k &= 32 \\[0.2cm]
k &= 4 \end{align}\]
| \(\therefore\) k = 4 |
Interactive Questions
Here are a few activities for you to practice.
Select/Type your answer and click the "Check Answer" button to see the result.
Let's Summarize
The mini-lesson targeted the fascinating concept of the value of a polynomial. The math journey around the value of a polynomial starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
About Cuemath
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Frequently Asked Questions (FAQs)
1. How do you find the value of a polynomial?
The value of a polynomial can be calculated by substituting the variable with any number or constant.
Value of polynomial example for \(3x - 2\) at \(x = 2\) is:
\(3 \times 2 - 2 = 6 - 2 = 4\)
2. What do you mean by the value of a polynomial?
The value of the polynomial \(P(x)\) at \(x = a\) is \(P(a)\).
\(P(a)\) is the value of the polynomial at \(x = a\).
3. What is the value of zero in a polynomial?
The value of zero in a polynomial \(P(x)\) means the value of the polynomial at \(x = 0\)
It means \(P(0)\).
For example the value of zero in a polynomial \(P(x) = 2x^4 - 7x^2 + 9\) is \(P(0) = 2 \times 0^4 - 7 \times 0^2 + 9 = 0 + 0 + 9 = 9\)
4. How do you find the value of k in any polynomial?
The value of \(k\) in a polynomial can be obtained by putting the value of the variable inside the polynomial and equating the obtained polynomial with its value at that point.
For example, if the polynomial \(P(x) = x^3 - kx + 1\) takes the value 5 at \(x = 1\), then the value of \(k\) can be obtained by solving \(P(1) = 5\)
\(P(1) = 1^3 - k \times 1 + 1 = 5\)
\(1 - k + 1 = 5 \\[0.2cm]
-k + 2 = 5 \\[0.2cm]
-k = 3 \\[0.2cm]
k = -3 \)
5. What is the form of the polynomial equation?
If \(P(x)\) is a polynomial, then a polynomial equation can be obtained by equating \(P(x)\) to some constant or another equation.
For example, if \(P(x) = x^2 - 7x + 5\), then \(P(x)\) can be converted to a polynomial equation by equating \(P(x) = 0\)
Hence, \(x^2 - 7x + 5 = 0\) is a polynomial equation.
6. Define the value of a variable at a point.
The value of a variable \(x\) at a point \((2, 3)\) is the value of the x-coordinate of the point.
7. What is a polynomial value?
The polynomial value \(P(a)\) is the value of polynomial \(P(x)\) at \(x = a\).
8. Is zero a polynomial function?
Yes, 0 is a polynomial function; it is a constant polynomial.
The polynomial can be represented as \(P(x) = 0\)