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pre - algebra : multiplying polynomials name date period 1. $3n(6n - 1)…

Question

pre - algebra : multiplying polynomials
name
date
period

  1. $3n(6n - 1)$
  2. $3m(8m - 1)$
  3. $5k(4k - 7)$
  4. $2(7b - 8)$
  5. $5n(4n + 6)$
  6. $3(6n + 5)$
  7. $2b(3b - 3)$
  8. $8(8n - 2)$
  9. $3(8p + 5)$
  10. $4b(4b - 6)$
  11. $3(3n - 4)$
  12. $2(6x - 7)$
  13. $3(5n - 2)$
  14. $5(3x - 1)$
  15. $6x^2(3x - 6)$
  16. $7(8x + 2)$
  17. $4r(r + 1)$
  18. $2x^2(5x + 3)$
  19. $5p(6p - 6)$
  20. $3v(3v + 4)$

Explanation:

Response

Let's solve problem 1: \( 3n(6n - 1) \)

Step 1: Apply the distributive property (multiplication over subtraction)

The distributive property states that \( a(b - c) = ab - ac \). Here, \( a = 3n \), \( b = 6n \), and \( c = 1 \). So we multiply \( 3n \) by each term inside the parentheses.
\( 3n \times 6n - 3n \times 1 \)

Step 2: Simplify each product

For the first product: \( 3n \times 6n = 18n^2 \) (using the rule \( a^m \times a^n = a^{m + n} \), here \( n \times n = n^{1 + 1}=n^2 \) and \( 3\times6 = 18 \))
For the second product: \( 3n \times 1 = 3n \)

So putting it together, we get \( 18n^2 - 3n \)

Step 1: Apply the distributive property

Using \( a(b - c)=ab - ac \) with \( a = 3m \), \( b = 8m \), \( c = 1 \)
\( 3m\times8m - 3m\times1 \)

Step 2: Simplify each product

First product: \( 3m\times8m = 24m^2 \) ( \( 3\times8 = 24 \) and \( m\times m=m^2 \))
Second product: \( 3m\times1 = 3m \)

So the result is \( 24m^2 - 3m \)

Step 1: Apply the distributive property

\( 5k\times4k - 5k\times7 \) (using \( a(b - c)=ab - ac \) with \( a = 5k \), \( b = 4k \), \( c = 7 \))

Step 2: Simplify each product

First product: \( 5k\times4k = 20k^2 \) ( \( 5\times4 = 20 \) and \( k\times k = k^2 \))
Second product: \( 5k\times7 = 35k \)

So we have \( 20k^2 - 35k \)

Answer:

\( 18n^2 - 3n \)

Let's solve problem 2: \( 3m(8m - 1) \)