Answer:
x² - 16 ⇒ (x + 4)(x - 4)
(2x + 1)³ ⇒ 8x³ + 12x² + 6x + 1
(2x + 3y)² ⇒ 4x² + 12xy + 9y²
x³ + 8y³ ⇒ (x + 2y)(x² - 2xy + 4y²)
Explanation:
* Lets explain how to solve the problem
# x² - 16
∵ x² - 16 is a difference of two squares
- Its factorization is two brackets with same terms and different
middle signs
- To factorize it find the square root of each term
∵ √x² = x and √16 = 4
∴ The terms of each brackets are x and 4 and the bracket have
different middle signs
∴ x² - 16 = (x + 4)(x - 4)
* x² - 16 ⇒ (x + 4)(x - 4)
# (2x + 1)³
- To solve the bracket we will separate (2x + 1)³ to (2x + 1)(2x + 1)²
∵ (2x + 1)² = (2x)(2x) + 2(2x)(1) + (1)(1) = 4x² + 4x + 1
∴ (2x + 1)³ = (2x + 1)(4x² + 4x + 1)
∵ (2x + 1)(4x² + 4x + 1) = (2x)(4x²) + (2x)(4x) + (2x)(1) + (1)(4x²) + (1)(4x) + (1)(1)
∴ (2x + 1)(4x² + 4x + 1) = 8x³ + 8x² + 2x + 4x² + 4x + 1 ⇒ add like terms
∴ (2x + 1)(4x² + 4x + 1) = 8x³ + (8x² + 4x²) + (2x + 4x) + 1
∴ (2x + 1)(4x² + 4x + 1) = 8x³ + 12x² + 6x + 1
∴ (2x + 1)³ = 8x³ + 12x² + 6x + 1
* (2x + 1)³ ⇒ 8x³ + 12x² + 6x + 1
# (2x + 3y)²
∵ (2x + 3y)² = (2x)(2x) + 2(2x)(3y) + (3y)(3y)
∴ (2x + 3y)² = 4x² + 12xy + 9y²
* (2x + 3y)² ⇒ 4x² + 12xy + 9y²
# x³ + 8y³
∵ x³ + 8y³ is the sum of two cubes
- Its factorization is binomial and trinomial
- The binomial is cub root the two terms
∵ ∛x³ = x and ∛8y³ = 2y
∴ The binomial is (x + 2y)
- We will make the trinomial from the binomial
- The first term is (x)² = x²
- The second term is (x)(2y) = 2xy with opposite sign of the middle
sign in the binomial
- The third term is (2y)² = 4y²
∴ x³ + 8y³ = (x + 2y)(x² - 2xy + 4y²)
* x³ + 8y³ ⇒ (x + 2y)(x² - 2xy + 4y²)