Math Problem Statement

Factorization of algebraic expressions including: a^2 - 6ab + 9b^2 - 16c^2, x^4 + x^2 + 1, x^4 - 8x^2 + 4, 8x^3 - 27y^3 - 4x + 6y, x^4 - 8x^2 - 33 - 14y^2, a^4/b^4 + 8a^2/b^2 + 36

Solution

Let’s break down the factorizations in the image. The problems seem to focus on factoring different algebraic expressions. I’ll go step by step for each one:

6.1 Factorize a26ab+9b216c2a^2 - 6ab + 9b^2 - 16c^2

  • The expression a26ab+9b2a^2 - 6ab + 9b^2 is a perfect square trinomial, which can be written as (a3b)2(a - 3b)^2.
  • Now we have (a3b)216c2(a - 3b)^2 - 16c^2, which is a difference of squares: (A2B2)=(AB)(A+B)(A^2 - B^2) = (A - B)(A + B).
  • So, the factorization becomes: (a3b+4c)(a3b4c)(a - 3b + 4c)(a - 3b - 4c) Answer: (a3b+4c)(a3b4c)(a - 3b + 4c)(a - 3b - 4c).

6.2 Factorize x4+x2+1x^4 + x^2 + 1

  • We recognize this as a quadratic in disguise where u=x2u = x^2, so the expression becomes u2+u+1u^2 + u + 1.
  • This doesn't factor nicely over the reals, but we can factor it over complex numbers using roots of unity.
  • For simplicity, the factorization can be written as: (x2+x+1)(x2x+1)(x^2 + x + 1)(x^2 - x + 1) Answer: (x2+x+1)(x2x+1)(x^2 + x + 1)(x^2 - x + 1).

6.3 Factorize x48x2+4x^4 - 8x^2 + 4

  • Let u=x2u = x^2, so the expression becomes u28u+4u^2 - 8u + 4, which is a quadratic.
  • The quadratic formula gives:

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Math Problem Analysis

Mathematical Concepts

Algebra
Factorization
Difference of Squares
Quadratics

Formulas

Difference of Squares: A^2 - B^2 = (A - B)(A + B)
Quadratic Formula: ax^2 + bx + c
Cubic Factoring
Polynomial Factorization

Theorems

Difference of Squares
Quadratic Equation
Factoring Cubes

Suitable Grade Level

Grades 9-12