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

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

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

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

1

Solution steps

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

Follow the PEMDAS order of operations

Calculate exponents

(1)^{2} : 1

(1)^{2}

(1)^{2} = 1

= 1

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

Multiply and divide (left to right)

3 \cdot 1 : 3

3 \cdot 1

3 \cdot 1 = 3

= 3

= 3 - 4 (1) + 2

Multiply and divide (left to right)

4 (1) : 4

4 (1)

Apply rule:

(a) = a

(1) = 1

= 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 4

= 3 - 4 + 2

Add and subtract (left to right)

3 - 4 + 2 : 1

3 - 4 + 2

3 - 4 = - 1

= - 1 + 2

- 1 + 2 = 1

= 1

= 1

- 3 (3)^{4} + 20

- (- (- 3 - 20) + 5 \times (4 - 8) + (- 6 + 4) \times 2)

42 \div (- 6) \times 8

\frac{4 - 3 \div 4}{2 + 1 \div 2}

5 + (9 \div 8) \times (5 \times 5 - 6) - 8