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natta225 [31]
3 years ago
6

Drag the tiles to the correct boxes to complete the pairs.

Mathematics
1 answer:
andreev551 [17]3 years ago
4 0

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)

Step-by-step 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)

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Given:

The multiples of 2 form a sequence as follows: 2, 4, 6, 8, 10, 12, 14, 16, ...

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