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# Value of a Polynomial

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

 1 What do you Mean by Value of a Polynomial? 2 Thinking Out of the Box! 3 Solved Examples 4 Important Notes on Value of a Polynomial 5 Interactive Questions

## 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 $$P(a)$$

 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 $$x$$.

Think Tank
• 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.

Important Notes
• 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 $$x = 2$$

$P(2) = 2^3 - 2 \times 2^2 + 1\\[0.2cm] P(2) = 8 - 2 \times 4 + 1\\[0.2cm] P(2) = 8 - 8 + 1\\[0.2cm] P(2) = 1$

 $$\therefore$$ $$P(1) = 0$$ 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

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we at Cuemath believe in.

## 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 $$P(x) = 0$$

Download SOLVED Practice Questions of Value of a Polynomial for FREE
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Grade 9 | Questions Set 1
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