Solve (2x^2-05x-3)(4x^2-10)quad=

$$( 2 x ^ { 2 } - 05 x - 3 ) ( 4 x ^ { 2 } - 10 ) \quad =$$

Evaluate

$$8x^{4}-32x^{2}+30$$

Solution Steps

Multiply 0 and 5 to get 0.

$$\left(2x^{2}-0x-3\right)\left(4x^{2}-10\right)$$

Anything times zero gives zero.

$$\left(2x^{2}-0-3\right)\left(4x^{2}-10\right)$$

Use the distributive property to multiply 2x^{2}-0-3 by 4x^{2}-10.

$$4\left(2x^{2}-0\right)x^{2}-10\left(2x^{2}-0\right)-12x^{2}+30$$

Multiply -1 and 0 to get 0.

$$4\left(2x^{2}+0\right)x^{2}-10\left(2x^{2}-0\right)-12x^{2}+30$$

Anything plus zero gives itself.

$$4\times 2x^{2}x^{2}-10\left(2x^{2}-0\right)-12x^{2}+30$$

Multiply 4 and 2 to get 8.

$$8x^{2}x^{2}-10\left(2x^{2}-0\right)-12x^{2}+30$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$8x^{4}-10\left(2x^{2}-0\right)-12x^{2}+30$$

Multiply -1 and 0 to get 0.

$$8x^{4}-10\times 2x^{2}-12x^{2}+30$$

Multiply -10 and 2 to get -20.

$$8x^{4}-20x^{2}-12x^{2}+30$$

Combine -20x^{2} and -12x^{2} to get -32x^{2}.

$$8x^{4}-32x^{2}+30$$

Expand

$$8x^{4}-32x^{2}+30$$

Solution Steps

Multiply 0 and 5 to get 0.

$$\left(2x^{2}-0x-3\right)\left(4x^{2}-10\right)$$

Anything times zero gives zero.

$$\left(2x^{2}-0-3\right)\left(4x^{2}-10\right)$$

Use the distributive property to multiply 2x^{2}-0-3 by 4x^{2}-10.

$$4\left(2x^{2}-0\right)x^{2}-10\left(2x^{2}-0\right)-12x^{2}+30$$

Multiply -1 and 0 to get 0.

$$4\left(2x^{2}+0\right)x^{2}-10\left(2x^{2}-0\right)-12x^{2}+30$$

Anything plus zero gives itself.

$$4\times 2x^{2}x^{2}-10\left(2x^{2}-0\right)-12x^{2}+30$$

Multiply 4 and 2 to get 8.

$$8x^{2}x^{2}-10\left(2x^{2}-0\right)-12x^{2}+30$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$8x^{4}-10\left(2x^{2}-0\right)-12x^{2}+30$$

Multiply -1 and 0 to get 0.

$$8x^{4}-10\times 2x^{2}-12x^{2}+30$$

Multiply -10 and 2 to get -20.

$$8x^{4}-20x^{2}-12x^{2}+30$$

Combine -20x^{2} and -12x^{2} to get -32x^{2}.

$$8x^{4}-32x^{2}+30$$