Solve 3(2x-3)*(3x+2)-2(x+4)*(4x-3)+41x-10x^2

$$3 ( 2 x - 3 ) \cdot ( 3 x + 2 ) - 2 ( x + 4 ) \cdot ( 4 x - 3 ) + 41 x - 10 x ^ { 2 }$$

Evaluate

$$6$$

Solution Steps

Use the distributive property to multiply 3 by 2x-3.

$$\left(6x-9\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+41x-10x^{2}$$

Use the distributive property to multiply 6x-9 by 3x+2 and combine like terms.

$$18x^{2}-15x-18-2\left(x+4\right)\left(4x-3\right)+41x-10x^{2}$$

Use the distributive property to multiply -2 by x+4.

$$18x^{2}-15x-18+\left(-2x-8\right)\left(4x-3\right)+41x-10x^{2}$$

Use the distributive property to multiply -2x-8 by 4x-3 and combine like terms.

$$18x^{2}-15x-18-8x^{2}-26x+24+41x-10x^{2}$$

Combine 18x^{2} and -8x^{2} to get 10x^{2}.

$$10x^{2}-15x-18-26x+24+41x-10x^{2}$$

Combine -15x and -26x to get -41x.

$$10x^{2}-41x-18+24+41x-10x^{2}$$

Add -18 and 24 to get 6.

$$10x^{2}-41x+6+41x-10x^{2}$$

Combine -41x and 41x to get 0.

$$10x^{2}+6-10x^{2}$$

Combine 10x^{2} and -10x^{2} to get 0.

$$6$$

Factor

$$2\times 3$$