Class 9 · Maths

Polynomials

A polynomial is an algebraic expression in which the variable has only whole-number exponents. This chapter covers the degree and types of polynomials, zeroes of a polynomial, the remainder and factor theorems, factorisation and the standard algebraic identities.

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Exercise 2.1

Q1. Exercise 2.1 — Question 1

Is 4x² − 3x + 7 a polynomial in one variable? Write its degree.

In 4x² − 3x + 7 the only variable is x, and the powers of x are 2, 1 and 0 — all whole numbers. Since a polynomial in one variable needs a single variable with whole-number exponents, this expression qualifies. The highest power of x present is 2, and the degree is the highest power.

Formula: Degree = highest power of the variable

Final Answer: Yes, it is a polynomial in one variable (x) and its degree is 2.

Common Mistake: Calling an expression with a negative or fractional power (like x⁻¹ or √x) a polynomial.

Exam Tip: Check two things every time: one variable only, and whole-number powers.

Exercise 2.2

Q1. Exercise 2.2 — Question 1

Find the value of the polynomial p(x) = 5x − 4x² + 3 at x = 0 and at x = −1.

At x = 0: p(0) = 5(0) − 4(0)² + 3 = 0 − 0 + 3 = 3 At x = −1: p(−1) = 5(−1) − 4(−1)² + 3 = −5 − 4(1) + 3 = −5 − 4 + 3 = −6

Final Answer: p(0) = 3 and p(−1) = −6

Common Mistake: Writing (−1)² as −1. Squaring a negative number always gives a positive result.

Exam Tip: Put brackets around the substituted value before simplifying — it prevents sign errors.

Exercise 2.4

Q1. Exercise 2.4 — Question 1

Use the factor theorem to check whether (x + 1) is a factor of x³ + x² + x + 1.

By the factor theorem, (x + 1) is a factor of p(x) only if p(−1) = 0. Here p(x) = x³ + x² + x + 1. p(−1) = (−1)³ + (−1)² + (−1) + 1 = −1 + 1 − 1 + 1 = 0 Since the value is 0, the factor theorem confirms that (x + 1) divides the polynomial exactly.

Formula: Factor theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0

Final Answer: Yes — p(−1) = 0, so (x + 1) is a factor.

Common Mistake: Substituting x = +1 for the factor (x + 1). For (x + a) you must use x = −a.

Exam Tip: Write the zero of the divisor first (for x + 1 it is −1), then substitute.

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