What is 2(-4)^2+8(-4) ?

The solution to 2(-4)^2+8(-4) is 0

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2(-4)^2+8(-4)

2 (- 4)^{2} + 8 (- 4)

0

Solution steps

2 (- 4)^{2} + 8 (- 4)

Follow the PEMDAS order of operations

Calculate exponents

(- 4)^{2} : 16

(- 4)^{2}

Apply exponent rule:

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

n

is even

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

= 16

= 2 \cdot 16 + 8 (- 4)

Multiply and divide (left to right)

2 \cdot 16 : 32

2 \cdot 16

2 \cdot 16 = 32

= 32

= 32 + 8 (- 4)

Multiply and divide (left to right)

8 (- 4) : - 32

8 (- 4)

Apply rule

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

8 (- 4) = - 8 \cdot 4 = - 32

= - 32

= 32 - 32

Add and subtract (left to right)

32 - 32 : 0

32 - 32

32 - 32 = 0

= 0

= 0

\frac{32}{- 8 - 4 \cdot ( - 2 ) + 5}

- 16 \times 0 + 5

\frac{180 ( 6 - 2 )}{6}

(- 7 + 6)^{2}

- 1 (4) + (- 1) (4)