Value of a Polynomial

Q: What is the value of a polynomial?

The value of a polynomial p(x) at x = a, written as p(a), is the numerical result obtained by substituting a for x in the polynomial and simplifying. It is a fixed number.

Q: How do you find the value of a polynomial at a point?

Replace every x in the polynomial with the given value. Calculate each term (powers first, then multiply by coefficients). Add all terms with correct signs.

Q: What does p(0) represent?

p(0) equals the constant term of the polynomial. When x = 0, all terms with x become zero, leaving only the constant. On the graph, p(0) is the y-intercept.

Q: What does p(1) represent?

p(1) equals the sum of all coefficients of the polynomial. When x = 1, every power of x equals 1, so each term reduces to just its coefficient.

Q: What is a zero of a polynomial?

A zero (or root) of a polynomial p(x) is a value a such that p(a) = 0. On the graph of y = p(x), zeroes are the x-intercepts (where the graph crosses the x-axis).

Q: How is the value of a polynomial related to the Remainder Theorem?

The Remainder Theorem states that when p(x) is divided by (x - a), the remainder equals p(a). So evaluating p(a) is equivalent to finding the remainder of the division.

Q: Can a polynomial be evaluated at a negative number?

Yes. Substitute the negative number carefully, using parentheses. For example, if x = -3, then x squared = (-3) squared = 9, not -9.

Q: Is this topic in the CBSE Class 9 syllabus?

Yes. The value of a polynomial is covered in the CBSE Class 9 Mathematics syllabus under the Polynomials chapter. It forms the basis for understanding zeroes, the Remainder Theorem, and the Factor Theorem.

Q: Can a polynomial be evaluated at a fraction?

Yes. Substitute the fraction for x and simplify each term. For example, if p(x) = 2x squared and x = 1/2, then p(1/2) = 2(1/4) = 1/2.

Q: How many zeroes can a polynomial have?

A polynomial of degree n can have at most n zeroes. A linear polynomial has exactly 1 zero, a quadratic has at most 2, a cubic has at most 3, and so on.

Class 9Polynomials

The value of a polynomial at a given point is the numerical result obtained by substituting a specific number for the variable in the polynomial expression. This is one of the foundational concepts in the study of polynomials in Class 9.

If p(x) is a polynomial in variable x, then p(a) denotes the value of the polynomial when x is replaced by a. The process of finding p(a) is called evaluating the polynomial at x = a.

This concept is directly connected to finding zeroes of a polynomial (values where p(a) = 0), the Remainder Theorem (the remainder when p(x) is divided by (x − a) equals p(a)), and the Factor Theorem (if p(a) = 0, then (x − a) is a factor of p(x)).

Evaluating polynomials is also the basis for graphing polynomial functions. By computing p(x) at several values of x, we obtain coordinate points (x, p(x)) that can be plotted on a Cartesian plane to visualise the shape of the polynomial curve. The x-intercepts of this graph correspond to the zeroes of the polynomial.

What is Value of a Polynomial?

Definition: The value of a polynomial p(x) at x = a is denoted by p(a) and is obtained by replacing every occurrence of x in the polynomial with the number a, then simplifying.

p(a) = value obtained by substituting x = a in p(x)

Notation:

p(x) = the polynomial expression in variable x
p(a) = the value of the polynomial at x = a
p(0) = the value at x = 0 (equals the constant term)
p(1) = the value at x = 1 (equals the sum of all coefficients)

Important:

p(a) is a number, not an expression. It is a fixed numerical value.
If p(a) = 0, then a is called a zero (or root) of the polynomial.
By the Remainder Theorem, p(a) equals the remainder when p(x) is divided by (x − a).
This process works for any polynomial of any degree and for any value of a (including fractions, negatives, and irrationals).

Value of a Polynomial Formula

Key Results:

1. Evaluating p(a):

For a polynomial p(x) = a_nxⁿ + a_n−1xⁿ⁻¹ + … + a₁x + a₀:

p(a) = a_naⁿ + a_n−1aⁿ⁻¹ + … + a₁a + a₀

2. Special values:

p(0) = a₀ (the constant term)
p(1) = a_n + a_n−1 + … + a₁ + a₀ (sum of all coefficients)
p(−1) = alternating signs: a_n(−1)ⁿ + a_n−1(−1)ⁿ⁻¹ + …

3. Connection to Remainder Theorem:

When p(x) is divided by (x − a), the remainder = p(a)

4. Connection to Factor Theorem:

If p(a) = 0, then (x − a) is a factor of p(x).
If p(a) ≠ 0, then (x − a) is NOT a factor of p(x).

Derivation and Proof

Steps to evaluate a polynomial at a given point:

Step-by-step method:

Write the polynomial p(x) clearly with all terms.
Substitute the given value (say x = a) for every x in the polynomial.
Calculate each term separately:
Find the power of a first, then multiply by the coefficient.
Add all terms (keeping track of signs).
Simplify to get the final numerical value.

Example walkthrough:

Evaluate p(x) = 2x³ − 3x² + x − 5 at x = 2.

p(2) = 2(2)³ − 3(2)² + (2) − 5
= 2(8) − 3(4) + 2 − 5
= 16 − 12 + 2 − 5
= 1

Tips for accuracy:

Be careful with negative values: (−2)² = 4 but −2² = −4.
For fractions, simplify each term before adding.
Always verify by substituting back or checking with a different method.
Use parentheses around negative numbers to avoid sign errors: write (−3)³ = −27, not −3³ which could be misread.

Horner’s Method (efficient evaluation):

For higher-degree polynomials, rewrite in nested form to reduce calculations:

p(x) = 2x³ − 3x² + x − 5 can be written as:
p(x) = ((2x − 3)x + 1)x − 5
At x = 2: ((2 × 2 − 3) × 2 + 1) × 2 − 5 = (1 × 2 + 1) × 2 − 5 = 3 × 2 − 5 = 6 − 5 = 1 ✓

This method reduces the number of multiplications and is used in computer algorithms for polynomial evaluation.

Connection to polynomial graphs:

Each evaluation p(a) gives a point (a, p(a)) on the graph of y = p(x).
By evaluating at several values, we can plot points and sketch the graph.
Where p(a) = 0, the graph crosses (or touches) the x-axis.
Where p(a) > 0, the graph is above the x-axis.
Where p(a) < 0, the graph is below the x-axis.

Types and Properties

Types of problems involving value of a polynomial:

1. Direct evaluation

Given p(x) and a value of x, find p(a).
Straightforward substitution and simplification.

2. Finding zeroes

Find values of x for which p(x) = 0.
Evaluate p(x) at different values to check if the result is zero.

3. Verifying the Remainder Theorem

Divide p(x) by (x − a) using long division.
Check that the remainder equals p(a).

4. Checking if (x − a) is a factor

Evaluate p(a). If p(a) = 0, then (x − a) is a factor.
This is the Factor Theorem.

5. Evaluating at special values

p(0) gives the constant term.
p(1) gives the sum of coefficients.
p(−1) gives alternating sum of coefficients.

6. Reverse problems

Given p(a) = some value, find unknown coefficients in the polynomial.

Solved Examples

Example 1: Example 1: Evaluating a quadratic polynomial

Problem: If p(x) = x² − 5x + 6, find p(3).

Solution:

Substituting x = 3:

p(3) = (3)² − 5(3) + 6
= 9 − 15 + 6
= 0

Since p(3) = 0, x = 3 is a zero of p(x).

Answer: p(3) = 0.

Example 2: Example 2: Evaluating a cubic polynomial

Problem: If p(x) = 2x³ + 3x² − 4x + 1, find p(−1).

Solution:

Substituting x = −1:

p(−1) = 2(−1)³ + 3(−1)² − 4(−1) + 1
= 2(−1) + 3(1) − 4(−1) + 1
= −2 + 3 + 4 + 1
= 6

Answer: p(−1) = 6.

Example 3: Example 3: Evaluating at x = 0

Problem: If p(x) = 4x³ − 7x² + 2x − 9, find p(0).

Solution:

Substituting x = 0:

p(0) = 4(0)³ − 7(0)² + 2(0) − 9
= 0 − 0 + 0 − 9
= −9

Note: p(0) always equals the constant term of the polynomial.

Answer: p(0) = −9.

Example 4: Example 4: Evaluating at a fraction

Problem: If p(x) = 2x² − x + 1, find p(1/2).

Solution:

Substituting x = 1/2:

p(1/2) = 2(1/2)² − (1/2) + 1
= 2(1/4) − 1/2 + 1
= 1/2 − 1/2 + 1
= 1

Answer: p(1/2) = 1.

Example 5: Example 5: Checking if a value is a zero

Problem: Is x = 2 a zero of p(x) = x³ − 4x² + x + 6?

Solution:

Substituting x = 2:

p(2) = (2)³ − 4(2)² + (2) + 6
= 8 − 16 + 2 + 6
= 0

Since p(2) = 0, x = 2 IS a zero of p(x).

By the Factor Theorem, (x − 2) is a factor of p(x).

Answer: Yes, x = 2 is a zero of p(x).

Example 6: Example 6: Evaluating at a negative value

Problem: If p(t) = t² + 4t − 12, find p(−6) and p(2).

Solution:

Finding p(−6):

p(−6) = (−6)² + 4(−6) − 12
= 36 − 24 − 12
= 0

Finding p(2):

p(2) = (2)² + 4(2) − 12
= 4 + 8 − 12
= 0

Both −6 and 2 are zeroes of p(t).

Verification: t² + 4t − 12 = (t + 6)(t − 2) ✓

Answer: p(−6) = 0, p(2) = 0.

Example 7: Example 7: Sum of coefficients using p(1)

Problem: If p(x) = 3x³ − 5x² + 7x − 2, find the sum of all coefficients.

Solution:

Sum of all coefficients = p(1):

p(1) = 3(1)³ − 5(1)² + 7(1) − 2
= 3 − 5 + 7 − 2
= 3

Verification: Coefficients are 3, −5, 7, −2. Sum = 3 − 5 + 7 − 2 = 3 ✓

Answer: Sum of coefficients = 3.

Example 8: Example 8: Finding unknown coefficient

Problem: If p(x) = x² + kx − 6 and p(2) = 0, find k.

Solution:

Substituting x = 2 and setting p(2) = 0:

p(2) = (2)² + k(2) − 6 = 0
4 + 2k − 6 = 0
2k − 2 = 0
2k = 2
k = 1

Verification: p(x) = x² + x − 6 = (x + 3)(x − 2). p(2) = 0 ✓

Answer: k = 1.

Example 9: Example 9: Evaluating a linear polynomial

Problem: If p(x) = 3x − 7, find p(0), p(1), p(7/3), and p(−2).

Solution:

p(0):

= 3(0) − 7 = −7

p(1):

= 3(1) − 7 = −4

p(7/3):

= 3(7/3) − 7 = 7 − 7 = 0

p(−2):

= 3(−2) − 7 = −6 − 7 = −13

Note: x = 7/3 is the zero of p(x) since p(7/3) = 0.

Answer: p(0) = −7, p(1) = −4, p(7/3) = 0, p(−2) = −13.

Example 10: Example 10: Verifying the Remainder Theorem

Problem: Find the remainder when p(x) = x³ − 2x² + 3x − 1 is divided by (x − 1) using (a) the Remainder Theorem and (b) long division.

Solution:

(a) Remainder Theorem:

Remainder = p(1) = (1)³ − 2(1)² + 3(1) − 1
= 1 − 2 + 3 − 1 = 1

(b) Long division:

Dividing x³ − 2x² + 3x − 1 by (x − 1):
Quotient = x² − x + 2, Remainder = 1

Verification: (x − 1)(x² − x + 2) + 1 = x³ − x² + 2x − x² + x − 2 + 1 = x³ − 2x² + 3x − 1 ✓

Answer: Remainder = 1 (both methods agree).

Real-World Applications

Applications of evaluating polynomials:

Finding zeroes: Evaluating p(a) at various values of a helps identify zeroes of the polynomial, which are the x-intercepts of the graph y = p(x).
Remainder Theorem: The value p(a) equals the remainder when p(x) is divided by (x − a). This avoids lengthy polynomial division.
Factor Theorem: If p(a) = 0, then (x − a) is a factor. This is the primary method for factorising polynomials of degree 3 and above.
Graphing polynomials: Evaluating p(x) at several points gives coordinates (x, p(x)) that can be plotted to sketch the graph.
Physics and engineering: Polynomial models of real-world quantities (distance, velocity, cost) are evaluated at specific times or inputs to predict outcomes.
Economics: Cost functions, revenue functions, and profit functions are polynomials evaluated at specific production levels.

Key Points to Remember

The value of a polynomial p(x) at x = a is denoted p(a) and found by substituting a for x.
p(a) is a number, not an expression.
p(0) equals the constant term of the polynomial.
p(1) equals the sum of all coefficients.
If p(a) = 0, then a is called a zero (or root) of the polynomial.
By the Remainder Theorem: the remainder when p(x) is divided by (x − a) equals p(a).
By the Factor Theorem: if p(a) = 0, then (x − a) is a factor of p(x).
Be careful with negative substitutions: (−2)² = 4 but −(2²) = −4.
Evaluating at fractions requires careful arithmetic with denominators.
This concept connects to graphing: each evaluation gives a point on the curve y = p(x).

Practice Problems

If p(x) = 3x squared - 7x + 2, find p(2), p(0), and p(-1).
If p(x) = x cubed - 6x squared + 11x - 6, verify that p(1) = 0, p(2) = 0, and p(3) = 0.
Find the value of p(x) = 5x squared - 3x + 7 at x = -2.
If p(x) = x squared + bx - 8 and p(4) = 0, find b.
Find p(0) and p(1) for the polynomial p(x) = 4x cubed - 2x squared + x - 3. What do these values represent?
Is x = -3 a zero of p(x) = x cubed + 27? Find p(-3).
If p(x) = 2x squared - 5x + k and p(3) = 5, find the value of k.
Find the remainder when x cubed - 3x squared + 3x - 1 is divided by (x - 1) using the Remainder Theorem.

Frequently Asked Questions

Q1. What is the value of a polynomial?

The value of a polynomial p(x) at x = a, written as p(a), is the numerical result obtained by substituting a for x in the polynomial and simplifying. It is a fixed number.

Q2. How do you find the value of a polynomial at a point?

Replace every x in the polynomial with the given value. Calculate each term (powers first, then multiply by coefficients). Add all terms with correct signs.

Q3. What does p(0) represent?

p(0) equals the constant term of the polynomial. When x = 0, all terms with x become zero, leaving only the constant. On the graph, p(0) is the y-intercept.

Q4. What does p(1) represent?

p(1) equals the sum of all coefficients of the polynomial. When x = 1, every power of x equals 1, so each term reduces to just its coefficient.

Q5. What is a zero of a polynomial?

A zero (or root) of a polynomial p(x) is a value a such that p(a) = 0. On the graph of y = p(x), zeroes are the x-intercepts (where the graph crosses the x-axis).

Q6. How is the value of a polynomial related to the Remainder Theorem?

The Remainder Theorem states that when p(x) is divided by (x - a), the remainder equals p(a). So evaluating p(a) is equivalent to finding the remainder of the division.

Q7. Can a polynomial be evaluated at a negative number?

Yes. Substitute the negative number carefully, using parentheses. For example, if x = -3, then x squared = (-3) squared = 9, not -9.

Q8. Is this topic in the CBSE Class 9 syllabus?

Yes. The value of a polynomial is covered in the CBSE Class 9 Mathematics syllabus under the Polynomials chapter. It forms the basis for understanding zeroes, the Remainder Theorem, and the Factor Theorem.

Q9. Can a polynomial be evaluated at a fraction?

Yes. Substitute the fraction for x and simplify each term. For example, if p(x) = 2x squared and x = 1/2, then p(1/2) = 2(1/4) = 1/2.

Q10. How many zeroes can a polynomial have?

A polynomial of degree n can have at most n zeroes. A linear polynomial has exactly 1 zero, a quadratic has at most 2, a cubic has at most 3, and so on.

Value of a Polynomial

What is Value of a Polynomial?

Value of a Polynomial Formula

Derivation and Proof

Types and Properties

Solved Examples

Example 1: Example 1: Evaluating a quadratic polynomial

Example 2: Example 2: Evaluating a cubic polynomial

Example 3: Example 3: Evaluating at x = 0

Example 4: Example 4: Evaluating at a fraction

Example 5: Example 5: Checking if a value is a zero

Example 6: Example 6: Evaluating at a negative value

Example 7: Example 7: Sum of coefficients using p(1)

Example 8: Example 8: Finding unknown coefficient

Example 9: Example 9: Evaluating a linear polynomial

Example 10: Example 10: Verifying the Remainder Theorem

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. What is the value of a polynomial?

Q2. How do you find the value of a polynomial at a point?

Q3. What does p(0) represent?

Q4. What does p(1) represent?

Q5. What is a zero of a polynomial?

Q6. How is the value of a polynomial related to the Remainder Theorem?

Q7. Can a polynomial be evaluated at a negative number?

Q8. Is this topic in the CBSE Class 9 syllabus?

Q9. Can a polynomial be evaluated at a fraction?

Q10. How many zeroes can a polynomial have?

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