Solve (5-x)(5+x)+(x-5)^2+(2x-10)(x+3)

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

Evaluate

$$2\left(x-5\right)\left(x-2\right)$$

Solution Steps

Consider \left(5-x\right)\left(5+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$25-x^{2}+\left(x-5\right)^{2}+\left(2x-10\right)\left(x+3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$25-x^{2}+x^{2}-10x+25+\left(2x-10\right)\left(x+3\right)$$

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

$$25-10x+25+\left(2x-10\right)\left(x+3\right)$$

Add 25 and 25 to get 50.

$$50-10x+\left(2x-10\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-10 by x+3 and combine like terms.

$$50-10x+2x^{2}-4x-30$$

Combine -10x and -4x to get -14x.

$$50-14x+2x^{2}-30$$

Subtract 30 from 50 to get 20.

$$20-14x+2x^{2}$$

Expand

$$2x^{2}-14x+20$$

Solution Steps

Consider \left(5-x\right)\left(5+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$25-x^{2}+\left(x-5\right)^{2}+\left(2x-10\right)\left(x+3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$25-x^{2}+x^{2}-10x+25+\left(2x-10\right)\left(x+3\right)$$

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

$$25-10x+25+\left(2x-10\right)\left(x+3\right)$$

Add 25 and 25 to get 50.

$$50-10x+\left(2x-10\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-10 by x+3 and combine like terms.

$$50-10x+2x^{2}-4x-30$$

Combine -10x and -4x to get -14x.

$$50-14x+2x^{2}-30$$

Subtract 30 from 50 to get 20.

$$20-14x+2x^{2}$$