Solve (x-2)^2+2(x-2)(y-3)+(y-3)^2

$$( x - 2 ) ^ { 2 } + 2 ( x - 2 ) ( y - 3 ) + ( y - 3 ) ^ { 2 }$$

Evaluate

$$\left(x+y-5\right)^{2}$$

Solution Steps

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

$$x^{2}-4x+4+2\left(x-2\right)\left(y-3\right)+\left(y-3\right)^{2}$$

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

$$x^{2}-4x+4+\left(2x-4\right)\left(y-3\right)+\left(y-3\right)^{2}$$

Use the distributive property to multiply 2x-4 by y-3.

$$x^{2}-4x+4+2xy-6x-4y+12+\left(y-3\right)^{2}$$

Combine -4x and -6x to get -10x.

$$x^{2}-10x+4+2xy-4y+12+\left(y-3\right)^{2}$$

Add 4 and 12 to get 16.

$$x^{2}-10x+16+2xy-4y+\left(y-3\right)^{2}$$

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

$$x^{2}-10x+16+2xy-4y+y^{2}-6y+9$$

Combine -4y and -6y to get -10y.

$$x^{2}-10x+16+2xy-10y+y^{2}+9$$

Add 16 and 9 to get 25.

$$x^{2}-10x+25+2xy-10y+y^{2}$$

Expand

$$x^{2}+2xy-10x+y^{2}-10y+25$$

Solution Steps

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

$$x^{2}-4x+4+2\left(x-2\right)\left(y-3\right)+\left(y-3\right)^{2}$$

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

$$x^{2}-4x+4+\left(2x-4\right)\left(y-3\right)+\left(y-3\right)^{2}$$

Use the distributive property to multiply 2x-4 by y-3.

$$x^{2}-4x+4+2xy-6x-4y+12+\left(y-3\right)^{2}$$

Combine -4x and -6x to get -10x.

$$x^{2}-10x+4+2xy-4y+12+\left(y-3\right)^{2}$$

Add 4 and 12 to get 16.

$$x^{2}-10x+16+2xy-4y+\left(y-3\right)^{2}$$

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

$$x^{2}-10x+16+2xy-4y+y^{2}-6y+9$$

Combine -4y and -6y to get -10y.

$$x^{2}-10x+16+2xy-10y+y^{2}+9$$

Add 16 and 9 to get 25.

$$x^{2}-10x+25+2xy-10y+y^{2}$$