Solve (2y+1/5)^2-(y-1)^2

$$( 2 y + \frac { 1 } { 5 } ) ^ { 2 } - ( y - 1 ) ^ { 2 }$$

Evaluate

$$3y^{2}+\frac{14y}{5}-\frac{24}{25}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2y+\frac{1}{5}\right)^{2}.

$$4y^{2}+\frac{4}{5}y+\frac{1}{25}-\left(y-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$4y^{2}+\frac{4}{5}y+\frac{1}{25}-\left(y^{2}-2y+1\right)$$

To find the opposite of y^{2}-2y+1, find the opposite of each term.

$$4y^{2}+\frac{4}{5}y+\frac{1}{25}-y^{2}+2y-1$$

Combine 4y^{2} and -y^{2} to get 3y^{2}.

$$3y^{2}+\frac{4}{5}y+\frac{1}{25}+2y-1$$

Combine \frac{4}{5}y and 2y to get \frac{14}{5}y.

$$3y^{2}+\frac{14}{5}y+\frac{1}{25}-1$$

Subtract 1 from \frac{1}{25} to get -\frac{24}{25}.

$$3y^{2}+\frac{14}{5}y-\frac{24}{25}$$

Expand

$$3y^{2}+\frac{14y}{5}-\frac{24}{25}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2y+\frac{1}{5}\right)^{2}.

$$4y^{2}+\frac{4}{5}y+\frac{1}{25}-\left(y-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$4y^{2}+\frac{4}{5}y+\frac{1}{25}-\left(y^{2}-2y+1\right)$$

To find the opposite of y^{2}-2y+1, find the opposite of each term.

$$4y^{2}+\frac{4}{5}y+\frac{1}{25}-y^{2}+2y-1$$

Combine 4y^{2} and -y^{2} to get 3y^{2}.

$$3y^{2}+\frac{4}{5}y+\frac{1}{25}+2y-1$$

Combine \frac{4}{5}y and 2y to get \frac{14}{5}y.

$$3y^{2}+\frac{14}{5}y+\frac{1}{25}-1$$

Subtract 1 from \frac{1}{25} to get -\frac{24}{25}.

$$3y^{2}+\frac{14}{5}y-\frac{24}{25}$$