Solve (-x+2)*(5x+3)*(2x-4)-3x*(x+1)=

$$(-x+2) \times (5x+3) \times (2x-4)-3x \times (x+1)=$$

Evaluate

$$-10x^{3}+31x^{2}-19x-24$$

Solution Steps

Apply the distributive property by multiplying each term of -x+2 by each term of 5x+3.

$$\left(5\left(-x\right)x+3\left(-x\right)+10x+6\right)\left(2x-4\right)-3x\left(x+1\right)$$

Apply the distributive property by multiplying each term of 5\left(-x\right)x+3\left(-x\right)+10x+6 by each term of 2x-4.

$$10\left(-x\right)x^{2}-20\left(-x\right)x+6\left(-x\right)x-12\left(-x\right)+20x^{2}-40x+12x-24-3x\left(x+1\right)$$

Multiply -20 and -1 to get 20.

$$10\left(-x\right)x^{2}+20xx+6\left(-x\right)x-12\left(-x\right)+20x^{2}-40x+12x-24-3x\left(x+1\right)$$

Multiply x and x to get x^{2}.

$$10\left(-x\right)x^{2}+20x^{2}+6\left(-x\right)x-12\left(-x\right)+20x^{2}-40x+12x-24-3x\left(x+1\right)$$

Multiply -12 and -1 to get 12.

$$10\left(-x\right)x^{2}+20x^{2}+6\left(-x\right)x+12x+20x^{2}-40x+12x-24-3x\left(x+1\right)$$

Combine 20x^{2} and 20x^{2} to get 40x^{2}.

$$10\left(-x\right)x^{2}+40x^{2}+6\left(-x\right)x+12x-40x+12x-24-3x\left(x+1\right)$$

Combine 12x and -40x to get -28x.

$$10\left(-x\right)x^{2}+40x^{2}+6\left(-x\right)x-28x+12x-24-3x\left(x+1\right)$$

Combine -28x and 12x to get -16x.

$$10\left(-x\right)x^{2}+40x^{2}+6\left(-x\right)x-16x-24-3x\left(x+1\right)$$

Multiply 10 and -1 to get -10.

$$-10xx^{2}+40x^{2}+6\left(-1\right)xx-16x-24-3x\left(x+1\right)$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$-10x^{3}+40x^{2}+6\left(-1\right)xx-16x-24-3x\left(x+1\right)$$

Multiply x and x to get x^{2}.

$$-10x^{3}+40x^{2}+6\left(-1\right)x^{2}-16x-24-3x\left(x+1\right)$$

Multiply 6 and -1 to get -6.

$$-10x^{3}+40x^{2}-6x^{2}-16x-24-3x\left(x+1\right)$$

Combine 40x^{2} and -6x^{2} to get 34x^{2}.

$$-10x^{3}+34x^{2}-16x-24-3x\left(x+1\right)$$

Use the distributive property to multiply -3x by x+1.

$$-10x^{3}+34x^{2}-16x-24-3x^{2}-3x$$

Combine 34x^{2} and -3x^{2} to get 31x^{2}.

$$-10x^{3}+31x^{2}-16x-24-3x$$

Combine -16x and -3x to get -19x.

$$-10x^{3}+31x^{2}-19x-24$$

Expand

$$-10x^{3}+31x^{2}-19x-24$$

Solution Steps

Apply the distributive property by multiplying each term of -x+2 by each term of 5x+3.

$$\left(5\left(-x\right)x+3\left(-x\right)+10x+6\right)\left(2x-4\right)-3x\left(x+1\right)$$

Apply the distributive property by multiplying each term of 5\left(-x\right)x+3\left(-x\right)+10x+6 by each term of 2x-4.

$$10\left(-x\right)x^{2}-20\left(-x\right)x+6\left(-x\right)x-12\left(-x\right)+20x^{2}-40x+12x-24-3x\left(x+1\right)$$

Multiply -20 and -1 to get 20.

$$10\left(-x\right)x^{2}+20xx+6\left(-x\right)x-12\left(-x\right)+20x^{2}-40x+12x-24-3x\left(x+1\right)$$

Multiply x and x to get x^{2}.

$$10\left(-x\right)x^{2}+20x^{2}+6\left(-x\right)x-12\left(-x\right)+20x^{2}-40x+12x-24-3x\left(x+1\right)$$

Multiply -12 and -1 to get 12.

$$10\left(-x\right)x^{2}+20x^{2}+6\left(-x\right)x+12x+20x^{2}-40x+12x-24-3x\left(x+1\right)$$

Combine 20x^{2} and 20x^{2} to get 40x^{2}.

$$10\left(-x\right)x^{2}+40x^{2}+6\left(-x\right)x+12x-40x+12x-24-3x\left(x+1\right)$$

Combine 12x and -40x to get -28x.

$$10\left(-x\right)x^{2}+40x^{2}+6\left(-x\right)x-28x+12x-24-3x\left(x+1\right)$$

Combine -28x and 12x to get -16x.

$$10\left(-x\right)x^{2}+40x^{2}+6\left(-x\right)x-16x-24-3x\left(x+1\right)$$

Multiply 10 and -1 to get -10.

$$-10xx^{2}+40x^{2}+6\left(-1\right)xx-16x-24-3x\left(x+1\right)$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$-10x^{3}+40x^{2}+6\left(-1\right)xx-16x-24-3x\left(x+1\right)$$

Multiply x and x to get x^{2}.

$$-10x^{3}+40x^{2}+6\left(-1\right)x^{2}-16x-24-3x\left(x+1\right)$$

Multiply 6 and -1 to get -6.

$$-10x^{3}+40x^{2}-6x^{2}-16x-24-3x\left(x+1\right)$$

Combine 40x^{2} and -6x^{2} to get 34x^{2}.

$$-10x^{3}+34x^{2}-16x-24-3x\left(x+1\right)$$

Use the distributive property to multiply -3x by x+1.

$$-10x^{3}+34x^{2}-16x-24-3x^{2}-3x$$

Combine 34x^{2} and -3x^{2} to get 31x^{2}.

$$-10x^{3}+31x^{2}-16x-24-3x$$

Combine -16x and -3x to get -19x.

$$-10x^{3}+31x^{2}-19x-24$$