Polynomials

2 exercises3 questions solvedFree 14-day trial ↓

Exercise 2.1Geometrical Meaning of the Zeroes of a Polynomial

Six graphs of polynomial functions y = p(x) are shown (figures i through vi). For each graph, determine the number of zeroes of p(x) by identifying where the curve intersects the x-axis.

Solution

The number of zeroes of p(x) = the number of points where the graph intersects (or touches) the x-axis. (i) The graph does not intersect the x-axis at all. Number of zeroes = 0 (ii) The graph intersects the x-axis at exactly 1 point. Number of zeroes = 1 (iii) The graph intersects the x-axis at 3 points. Number of zeroes = 3 (iv) The graph intersects the x-axis at 2 points. Number of zeroes = 2 (v) The graph intersects the x-axis at 4 points. Number of zeroes = 4 (vi) The graph intersects the x-axis at 3 points. Number of zeroes = 3 Key rule: Each intersection with the x-axis represents one zero. A polynomial of degree n has at most n zeroes.

The textbook is solved. The exam won't ask the textbook.

Board papers twist Polynomials into exam-pattern questions the exercises never show. Practice unlimited chapter-wise tests with step-wise solutions — verify your WhatsApp number and your free trial starts right here.

CBSE · CLASS 10

🎟️14-day free trial · code PRAC auto-applied

Mobile Number

+91

I agree to the Terms & Conditions and Privacy Policy, including receiving study updates on WhatsApp.

OTP will be sent to your WhatsApp

⭐ Trusted by 12,000+ students across India

All chapters: Class 10 Mathematics →Polynomials practice questions →

Exercise 2.2Relationship between Zeroes and Coefficients of a Polynomial

For each quadratic polynomial, find its zeroes and confirm that they satisfy both the sum-of-zeroes and product-of-zeroes relations with the coefficients: (i) x² − 2x − 8 (ii) 4s² − 4s + 1 (iii) 6x² − 3 − 7x (iv) 4u² + 8u (v) t² − 15 (vi) 3x² − x − 4

Solution

(i) x² − 2x − 8 = x² − 4x + 2x − 8 = x(x−4) + 2(x−4) = (x+2)(x−4) Zeroes: α = −2, β = 4 Verification: α+β = −2+4 = 2 = −(−2)/1 = −b/a ✓; αβ = (−2)(4) = −8 = −8/1 = c/a ✓ (ii) 4s² − 4s + 1 = (2s−1)² Zeroes: α = β = 1/2 Verification: α+β = 1 = 4/4 = −b/a ✓; αβ = 1/4 = 1/4 = c/a ✓ (iii) 6x² − 3 − 7x = 6x² − 7x − 3 = 6x² − 9x + 2x − 3 = 3x(2x−3) + 1(2x−3) = (3x+1)(2x−3) Zeroes: α = −1/3, β = 3/2 Verification: α+β = −1/3+3/2 = 7/6 = 7/6 = −b/a ✓; αβ = (−1/3)(3/2) = −1/2 = −3/6 = c/a ✓ (iv) 4u² + 8u = 4u(u + 2) Zeroes: α = 0, β = −2 Verification: α+β = −2 = −8/4 = −b/a ✓; αβ = 0 = 0/4 = c/a ✓ (v) t² − 15 = (t−√15)(t+√15) Zeroes: α = √15, β = −√15 Verification: α+β = 0 = 0/1 = −b/a ✓; αβ = −15 = −15/1 = c/a ✓ (vi) 3x² − x − 4 = 3x² − 4x + 3x − 4 = x(3x−4) + 1(3x−4) = (x+1)(3x−4) Zeroes: α = −1, β = 4/3 Verification: α+β = −1+4/3 = 1/3 = 1/3 = −b/a ✓; αβ = (−1)(4/3) = −4/3 = −4/3 = c/a ✓

Construct a quadratic polynomial for each pair of values, where the first value is the sum of zeroes and the second is the product of zeroes: (i) 1/4, −1 (ii) √2, 1/3 (iii) 0, √5 (iv) 1, 1 (v) −1/4, 1/4 (vi) 4, 1

Solution

For zeroes α, β: quadratic polynomial = x² − (α+β)x + αβ = x² − (sum)x + (product) (i) Sum = 1/4, Product = −1 p(x) = x² − (1/4)x + (−1) = x² − x/4 − 1 Multiplying by 4: 4x² − x − 4 (ii) Sum = √2, Product = 1/3 p(x) = x² − √2x + 1/3 Multiplying by 3: 3x² − 3√2x + 1 (iii) Sum = 0, Product = √5 p(x) = x² − 0·x + √5 = x² + √5 (iv) Sum = 1, Product = 1 p(x) = x² − x + 1 (v) Sum = −1/4, Product = 1/4 p(x) = x² − (−1/4)x + 1/4 = x² + x/4 + 1/4 Multiplying by 4: 4x² + x + 1 (vi) Sum = 4, Product = 1 p(x) = x² − 4x + 1