Introduction
In this article, we will be analyzing and solving problems from the Mathematics textbook for Class 8. Specifically, we will be focusing on exercise problems 51, 52, 53, 54, 55, 56, 57, and 58 from pages 24 and 25. These problems involve the factorization of polynomials and solving equations. Let’s dive into each problem one by one.
Problem 51: Polynomial Factorization
a) The polynomial x³ – 2x² + x can be factored as x(x² – 2x + 1), which further simplifies to x(x – 1)².
b) The polynomial 2x² + 4x + 2 – 2y² can be factored as 2(x² + 2x + 1) – 2y², which simplifies to 2[(x + 1)² – y²]. Further simplification gives us 2(x + 1 – y)(x + 1 + y).
c) The polynomial 2xy – x² – y² + 16 can be simplified as 16 – (x² – 2xy + y²), which is equivalent to 16 – (x – y)². Further simplification gives us (4 – x + y)(4 + x – y).
Problem 52: Divisibility Proof
To prove that (5n + 2)² – 4 is divisible by 5 for all integer values of n, we can express the given expression as (5n + 2)² – 2². Simplifying further, we get (5n + 2 – 2)(5n + 2 + 2), which is equal to 5n(5n + 4). Since 5 is a factor of 5n, it follows that 5n(5n + 4) is divisible by 5 for all integer values of n.
Problem 53: Polynomial Factorization
a) The polynomial x² – 3x + 2 can be factored as (x – 1)(x – 2), by splitting the middle term.
b) The polynomial x² + x – 6 can be factored as (x + 3)(x – 2).
c) The polynomial x² + 5x + 6 can be factored as (x + 2)(x + 3).
Problem 54: Polynomial Factorization
a) The polynomial x³ + 2x²y + xy² – 9x can be factored as x(x² + 2xy + y² – 9), which further simplifies to x[(x + y)² – 9]. Continuing with the factorization, we get x(x + y – 3)(x + y + 3).
b) The polynomial 2x – 2y – x² + 2xy – y² can be rewritten as (2x – 2y) – (x² – 2xy + y²), which simplifies to 2(x – y) – (x – y)². Further simplification gives us (x – y)(2 – x + y).
c) The polynomial x⁴ – 2x² can be factored as x²(x² – (√2)²), which further simplifies to x²(x – √2)(x + √2).
Problem 55: Solving Equations
a) To find the values of x that satisfy the equation x³ – 1/4x = 0, we can factor it as x(x² – (1/2)²) = 0. This leads to three possible solutions: x = 0, x = -1/2, and x = 1/2.
b) For the equation (2x – 1)² – (x + 3)² = 0, we can factor it as [(2x – 1) – (x + 3)][(2x – 1) + (x + 3)] = 0. Simplifying further gives us (x – 4)(3x + 2) = 0, which has two possible solutions: x = 4 and x = -2/3.
c) The equation x²(x – 3) + 12 – 4x = 0 can be rewritten as x²(x – 3) – 4(x – 3) = 0. Further factorization gives us (x – 3)(x² – 2²) = 0, which can be simplified to (x – 3)(x – 2)(x + 2) = 0. This equation has three possible solutions: x = 3, x = 2, and x = -2.
Problem 56: Evaluating Polynomials
a) Evaluating the polynomial x² + 1/2x + 1/16 at x = 49.75 gives us (49.75 + 1/4)² = (49.75 + 0.25)² = 50² = 2500.
b) Evaluating the polynomial x² – y² – 2y – 1 at x = 93 and y = 6 gives us (93 – 6 – 1)(93 + 6 + 1) = 86 * 100 = 8600.
Problem 57: Polynomial Factorization
a) The polynomial x² – 4x + 3 can be factored as (x – 1)(x – 3).
b) The polynomial x² + 5x + 4 can be factored as (x + 4)(x + 1).
c) The polynomial x² – x – 6 can be factored as (x + 2)(x – 3).
d) The polynomial x⁴ + 4 can be factored as (x² + 2)² – (2x)², which simplifies to (x² + 2 – 2x)(x² + 2 + 2x).
Problem 58: Divisibility Proof
To prove that n³ – n is divisible by 6 for all integer values of n, we can express the given expression as n(n² – 1) = n(n – 1)(n + 1). Since n, n – 1, and n + 1 form three consecutive integers, it follows that their product is divisible by both 3 and 2. Therefore, n³ – n is divisible by 6 for all integer values of n.
For more math exercises and solutions, visit Kienthucykhoa.com.
Conclusion
In this article, we analyzed and solved exercise problems from the Class 8 Mathematics textbook. We learned about polynomial factorization and solving equations. Additionally, we explored proofs involving divisibility. These concepts are essential for developing a strong foundation in Mathematics. Keep practicing and exploring more exercises to enhance your problem-solving skills.
