Solve =(4x+3)^2-2(5x-3)(x+1)-2x(3x+10)

$$= ( 4 x + 3 ) ^ { 2 } - 2 ( 5 x - 3 ) ( x + 1 ) - 2 x ( 3 x + 10 )$$

Evaluate

$$15$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9-2\left(5x-3\right)\left(x+1\right)-2x\left(3x+10\right)$$

Use the distributive property to multiply -2 by 5x-3.

$$16x^{2}+24x+9+\left(-10x+6\right)\left(x+1\right)-2x\left(3x+10\right)$$

Use the distributive property to multiply -10x+6 by x+1 and combine like terms.

$$16x^{2}+24x+9-10x^{2}-4x+6-2x\left(3x+10\right)$$

Combine 16x^{2} and -10x^{2} to get 6x^{2}.

$$6x^{2}+24x+9-4x+6-2x\left(3x+10\right)$$

Combine 24x and -4x to get 20x.

$$6x^{2}+20x+9+6-2x\left(3x+10\right)$$

Add 9 and 6 to get 15.

$$6x^{2}+20x+15-2x\left(3x+10\right)$$

Use the distributive property to multiply -2x by 3x+10.

$$6x^{2}+20x+15-6x^{2}-20x$$

Combine 6x^{2} and -6x^{2} to get 0.

$$20x+15-20x$$

Combine 20x and -20x to get 0.

$$15$$

Factor

$$3\times 5$$