What is 3(-1-3)^2+4(3)-6(-1)+2 ?

The solution to 3(-1-3)^2+4(3)-6(-1)+2 is 68

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3(-1-3)^2+4(3)-6(-1)+2

3 (- 1 - 3)^{2} + 4 (3) - 6 (- 1) + 2

68

Solution steps

3 (- 1 - 3)^{2} + 4 (3) - 6 (- 1) + 2

Follow the PEMDAS order of operations

Calculate within parentheses

(- 1 - 3) : - 4

- 1 - 3

- 1 - 3 = - 4

= - 4

= 3 (- 4)^{2} + 4 (3) - 6 (- 1) + 2

Calculate exponents

(- 4)^{2} : 16

(- 4)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2} = 16

= 16

= 3 \cdot 16 + 4 (3) - 6 (- 1) + 2

Multiply and divide (left to right)

3 \cdot 16 : 48

3 \cdot 16

3 \cdot 16 = 48

= 48

= 48 + 4 (3) - 6 (- 1) + 2

Multiply and divide (left to right)

4 (3) : 12

4 (3)

Apply rule:

(a) = a

(3) = 3

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

= 48 + 12 - 6 (- 1) + 2

Multiply and divide (left to right)

6 (- 1) : - 6

6 (- 1)

Apply rule

a \cdot (- b) = - a \cdot b

6 (- 1) = - 6 \cdot 1 = - 6

= - 6

= 48 + 12 - (- 6) + 2

Add and subtract (left to right)

48 + 12 - (- 6) + 2 : 68

48 + 12 - (- 6) + 2

48 + 12 = 60

= 60 - (- 6) + 2

Apply rule

- (- a) = + a

- (- 6) = + 6

= 60 + 6 + 2

60 + 6 = 66

= 66 + 2

66 + 2 = 68

= 68

= 68

(33 - (3 + 7) - 8) \div 5

(4 + 3) \times 7

- 1^{2} - 3 (- 1)

6 (2 - 3^{2} - 5)

2 (13)^{3}