Solve (2x-5)^2+(3x-5)(3x+5)+60x

$$( 2 x - 5 ) ^ { 2 } + ( 3 x - 5 ) ( 3 x + 5 ) + 60 x$$

Evaluate

$$x\left(13x+40\right)$$

Solution Steps

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

$$4x^{2}-20x+25+\left(3x-5\right)\left(3x+5\right)+60x$$

Consider \left(3x-5\right)\left(3x+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.

$$4x^{2}-20x+25+\left(3x\right)^{2}-25+60x$$

Expand \left(3x\right)^{2}.

$$4x^{2}-20x+25+3^{2}x^{2}-25+60x$$

Calculate 3 to the power of 2 and get 9.

$$4x^{2}-20x+25+9x^{2}-25+60x$$

Combine 4x^{2} and 9x^{2} to get 13x^{2}.

$$13x^{2}-20x+25-25+60x$$

Subtract 25 from 25 to get 0.

$$13x^{2}-20x+60x$$

Combine -20x and 60x to get 40x.

$$13x^{2}+40x$$

Expand

$$13x^{2}+40x$$

Solution Steps

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

$$4x^{2}-20x+25+\left(3x-5\right)\left(3x+5\right)+60x$$

Consider \left(3x-5\right)\left(3x+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.

$$4x^{2}-20x+25+\left(3x\right)^{2}-25+60x$$

Expand \left(3x\right)^{2}.

$$4x^{2}-20x+25+3^{2}x^{2}-25+60x$$

Calculate 3 to the power of 2 and get 9.

$$4x^{2}-20x+25+9x^{2}-25+60x$$

Combine 4x^{2} and 9x^{2} to get 13x^{2}.

$$13x^{2}-20x+25-25+60x$$

Subtract 25 from 25 to get 0.

$$13x^{2}-20x+60x$$

Combine -20x and 60x to get 40x.

$$13x^{2}+40x$$