Polynomials – Solved Problems with Explanation
Polynomials are algebraic expressions consisting of variables and coefficients. They form the foundation for higher algebra and are widely used in mathematics, science, and competitive examinations.
In this article, we will solve important polynomial problems step by step, covering definitions, operations, factorization, and evaluation.
What is a Polynomial?
A polynomial in one variable x is an expression of the form:
anxn + an−1xn−1 + … + a1x + a0
- Degree: Highest power of the variable
- Coefficient: Numerical factor of a term
- Constant term: Term without a variable
Solved Polynomial Problems
Problem 1: Find the degree of the polynomial 5x³ − 4x² + 7x − 9
Solution:
The highest power of x is 3.
Degree = 3
Problem 2: Add the polynomials (3x² + 4x − 5) and (2x² − x + 7)
Solution:
(3x² + 4x − 5) + (2x² − x + 7)
= (3x² + 2x²) + (4x − x) + (−5 + 7)
= 5x² + 3x + 2
Sum = 5x² + 3x + 2
Problem 3: Subtract (4x² − 3x + 6) from (7x² + 2x − 1)
Solution:
(7x² + 2x − 1) − (4x² − 3x + 6)
= 7x² + 2x − 1 − 4x² + 3x − 6
= 3x² + 5x − 7
Difference = 3x² + 5x − 7
Problem 4: Multiply (x + 2)(x + 5)
Solution:
(x + 2)(x + 5)
= x(x + 5) + 2(x + 5)
= x² + 5x + 2x + 10
= x² + 7x + 10
Product = x² + 7x + 10
Problem 5: Factorize x² − 9x + 20
Solution:
Product = 20, Sum = −9
Numbers are −4 and −5
x² − 4x − 5x + 20 = 0
(x − 4)(x − 5)
Factorized form = (x − 4)(x − 5)
Problem 6: Find the value of 2x³ − 3x² + x − 5 when x = 2
Solution:
Substitute x = 2
2(2³) − 3(2²) + 2 − 5
= 16 − 12 + 2 − 5
= 1
Value = 1
Problem 7: Check whether x − 1 is a factor of x³ − 1
Solution (Factor Theorem):
Let f(x) = x³ − 1
f(1) = 1 − 1 = 0
Since f(1) = 0, x − 1 is a factor.
Problem 8: Use identity to expand (x − 3)²
Solution:
(x − 3)² = x² − 6x + 9
Expanded form = x² − 6x + 9
Exam Tips
✔ Always arrange polynomial in descending order
✔ Use identities to save time
✔ Apply factor theorem for quick checks
Common Mistakes
- Sign errors while adding or subtracting
- Incorrect degree identification
- Skipping brackets during multiplication
- Wrong substitution during evaluation
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