Question

4(6a + 2b^2) - 1/3 (27b^2 - 15a)
if a = -2 and b = 3

Answer 1

`4(6a + 2b^2) - 1/3 (27b^2 - 15a)`

a = -2

b = 3

`4\left(6\left(-2\right)+2^2\right)-1/3\left(27\left(3\right)^2-15\left(-2\right)\right)`

Follow the PEMDAS order of operations:-

• Parentheses

• Exponents

• Multiplication or Division

• Addition or Subtraction

`4\left(6\left(-2\right)+2^2\right)`

calculate within Parentheses

`6\left(-2\right)+2^2`

`2^(2) =4`

`6(-2)=-12`

Add:-

`-12+4=-8`

`4\left(-8\right)-1/3\left(27\left(3\right)^2-15\left(-2\right)\right)`

calculate within Parentheses:-

`27\left(3\right)^2-15\left(-2\right)`

`(3)^(2) =9`

`27*9=243`

`243-15\left(-2\right)`

`15(-2)=-30`

`243-\left(-30\right)`

`=273`

`4\left(-8\right)-1/3\cdot \:273`

Multiply:-

`4\left(-8\right)=-32`

`-32-1/3\cdot \:273`

`1/3*273=91`

`=-32-91`

`=-123`

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