Solve (2y-5)^2+(3y-5)(3y+5)+40y=

$$( 2 y - 5 ) ^ { 2 } + ( 3 y - 5 ) ( 3 y + 5 ) + 40 y =$$

Evaluate

$$y\left(13y+20\right)$$

Solution Steps

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

$$4y^{2}-20y+25+\left(3y-5\right)\left(3y+5\right)+40y$$

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

$$4y^{2}-20y+25+\left(3y\right)^{2}-25+40y$$

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

$$4y^{2}-20y+25+3^{2}y^{2}-25+40y$$

Calculate 3 to the power of 2 and get 9.

$$4y^{2}-20y+25+9y^{2}-25+40y$$

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

$$13y^{2}-20y+25-25+40y$$

Subtract 25 from 25 to get 0.

$$13y^{2}-20y+40y$$

Combine -20y and 40y to get 20y.

$$13y^{2}+20y$$

Expand

$$13y^{2}+20y$$

Solution Steps

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

$$4y^{2}-20y+25+\left(3y-5\right)\left(3y+5\right)+40y$$

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

$$4y^{2}-20y+25+\left(3y\right)^{2}-25+40y$$

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

$$4y^{2}-20y+25+3^{2}y^{2}-25+40y$$

Calculate 3 to the power of 2 and get 9.

$$4y^{2}-20y+25+9y^{2}-25+40y$$

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

$$13y^{2}-20y+25-25+40y$$

Subtract 25 from 25 to get 0.

$$13y^{2}-20y+40y$$

Combine -20y and 40y to get 20y.

$$13y^{2}+20y$$