Solve 3x(x+1)-(x-2)^2-6=(x+3)(x-5)+x*(x+9)+5

$$3 x ( x + 1 ) - ( x - 2 ) ^ { 2 } - 6 = ( x + 3 ) ( x - 5 ) + x \cdot ( x + 9 ) + 5$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

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

$$3x^{2}+3x-\left(x-2\right)^{2}-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

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

$$3x^{2}+3x-\left(x^{2}-4x+4\right)-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$3x^{2}+3x-x^{2}+4x-4-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

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

$$2x^{2}+3x+4x-4-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

Combine 3x and 4x to get 7x.

$$2x^{2}+7x-4-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

Subtract 6 from -4 to get -10.

$$2x^{2}+7x-10=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

Use the distributive property to multiply x+3 by x-5 and combine like terms.

$$2x^{2}+7x-10=x^{2}-2x-15+x\left(x+9\right)+5$$

Use the distributive property to multiply x by x+9.

$$2x^{2}+7x-10=x^{2}-2x-15+x^{2}+9x+5$$

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

$$2x^{2}+7x-10=2x^{2}-2x-15+9x+5$$

Combine -2x and 9x to get 7x.

$$2x^{2}+7x-10=2x^{2}+7x-15+5$$

Add -15 and 5 to get -10.

$$2x^{2}+7x-10=2x^{2}+7x-10$$

Subtract 2x^{2} from both sides.

$$2x^{2}+7x-10-2x^{2}=7x-10$$

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

$$7x-10=7x-10$$

Subtract 7x from both sides.

$$7x-10-7x=-10$$

Combine 7x and -7x to get 0.

$$-10=-10$$

Compare -10 and -10.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

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

$$3x^{2}+3x-\left(x-2\right)^{2}-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

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

$$3x^{2}+3x-\left(x^{2}-4x+4\right)-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$3x^{2}+3x-x^{2}+4x-4-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

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

$$2x^{2}+3x+4x-4-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

Combine 3x and 4x to get 7x.

$$2x^{2}+7x-4-6=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

Subtract 6 from -4 to get -10.

$$2x^{2}+7x-10=\left(x+3\right)\left(x-5\right)+x\left(x+9\right)+5$$

Use the distributive property to multiply x+3 by x-5 and combine like terms.

$$2x^{2}+7x-10=x^{2}-2x-15+x\left(x+9\right)+5$$

Use the distributive property to multiply x by x+9.

$$2x^{2}+7x-10=x^{2}-2x-15+x^{2}+9x+5$$

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

$$2x^{2}+7x-10=2x^{2}-2x-15+9x+5$$

Combine -2x and 9x to get 7x.

$$2x^{2}+7x-10=2x^{2}+7x-15+5$$

Add -15 and 5 to get -10.

$$2x^{2}+7x-10=2x^{2}+7x-10$$

Subtract 2x^{2} from both sides.

$$2x^{2}+7x-10-2x^{2}=7x-10$$

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

$$7x-10=7x-10$$

Subtract 7x from both sides.

$$7x-10-7x=-10$$

Combine 7x and -7x to get 0.

$$-10=-10$$

Compare -10 and -10.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$