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Expand the following algebraic expressions: 1. 6(x+3) 2. -4(x+3) 3. x(x+2) 4. 20I(a+3b) + 4(2a+b) 5. 4(2b+1) - 3(6-2)

Mathematics

ScanSolve AI Answer

Step-by-step solution

11 steps

Here are the solutions to the expressions:

1.

  1. Distribute the 6 to each term inside the parenthesis.

    6(x+3)=6⋅x+6⋅36(x+3) = 6 \cdot x + 6 \cdot 3
  2. Perform the multiplications.

    6x+186x + 18

    The expanded expression is:

    6x + 18\boxed{\text{6x + 18}}

    2.

  3. Distribute the −4-4 to each term inside the parenthesis.

    −4(x+3)=−4⋅x+(−4)⋅3-4(x+3) = -4 \cdot x + (-4) \cdot 3
  4. Perform the multiplications.

    −4x−12-4x - 12

    The expanded expression is:

    -4x - 12\boxed{\text{-4x - 12}}

    3.

  5. Distribute the xx to each term inside the parenthesis.

    x(x+2)=x⋅x+x⋅2x(x+2) = x \cdot x + x \cdot 2
  6. Perform the multiplications.

    x2+2xx^2 + 2x

    The expanded expression is:

    \boxed{\text{x^2 + 2x}}

    4.

  7. Distribute $20I$ into the first parenthesis and 4 into the second parenthesis.

    20I(a+3b)+4(2a+b)=(20I⋅a+20I⋅3b)+(4⋅2a+4⋅b)20I(a+3b) + 4(2a+b) = (20I \cdot a + 20I \cdot 3b) + (4 \cdot 2a + 4 \cdot b)
  8. Perform the multiplications.

    20Ia+60Ib+8a+4b20Ia + 60Ib + 8a + 4b

    The expanded expression is:

    20Ia + 60Ib + 8a + 4b\boxed{\text{20Ia + 60Ib + 8a + 4b}}

    5.

  9. Distribute 4 into the first parenthesis and simplify the second parenthesis.

    4(2b+1)−3(6−2)=(4⋅2b+4⋅1)−3(4)4(2b+1) - 3(6-2) = (4 \cdot 2b + 4 \cdot 1) - 3(4)
  10. Perform the multiplications.

    8b+4−128b + 4 - 12
  11. Combine the constant terms.

    8b−88b - 8

    The expanded expression is:

    8b - 8\boxed{\text{8b - 8}}
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