Solve (5-2x)^2-(2x+5)*(2x-5)+10*(2x-5)=

$$( 5 - 2 x ) ^ { 2 } - ( 2 x + 5 ) \cdot ( 2 x - 5 ) + 10 \cdot ( 2 x - 5 ) =$$

Evaluate

$$0$$

Solution Steps

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

$$25-20x+4x^{2}-\left(2x+5\right)\left(2x-5\right)+10\left(2x-5\right)$$

Consider \left(2x+5\right)\left(2x-5\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.

$$25-20x+4x^{2}-\left(\left(2x\right)^{2}-25\right)+10\left(2x-5\right)$$

Expand \left(2x\right)^{2}.

$$25-20x+4x^{2}-\left(2^{2}x^{2}-25\right)+10\left(2x-5\right)$$

Calculate 2 to the power of 2 and get 4.

$$25-20x+4x^{2}-\left(4x^{2}-25\right)+10\left(2x-5\right)$$

To find the opposite of 4x^{2}-25, find the opposite of each term.

$$25-20x+4x^{2}-4x^{2}+25+10\left(2x-5\right)$$

Combine 4x^{2} and -4x^{2} to get 0.

$$25-20x+25+10\left(2x-5\right)$$

Add 25 and 25 to get 50.

$$50-20x+10\left(2x-5\right)$$

Use the distributive property to multiply 10 by 2x-5.

$$50-20x+20x-50$$

Combine -20x and 20x to get 0.

$$50-50$$

Subtract 50 from 50 to get 0.

$$0$$

Factor

$$0$$