Solve -3/2y+1-(y+3/2)^2+(y-1/4)*(y+1/4)+frac{3}{2}y-frac{1}{16}

$$- \frac { 3 } { 2 } y + 1 - ( y + \frac { 3 } { 2 } ) ^ { 2 } + ( y - \frac { 1 } { 4 } ) \cdot ( y + \frac { 1 } { 4 } ) + \frac { 3 } { 2 } y - \frac { 1 } { 16 }$$

Evaluate

$$-3y-\frac{11}{8}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+\frac{3}{2}\right)^{2}.

$$-\frac{3}{2}y+1-\left(y^{2}+3y+\frac{9}{4}\right)+\left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\right)+\frac{3}{2}y-\frac{1}{16}$$

To find the opposite of y^{2}+3y+\frac{9}{4}, find the opposite of each term.

$$-\frac{3}{2}y+1-y^{2}-3y-\frac{9}{4}+\left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\right)+\frac{3}{2}y-\frac{1}{16}$$

Combine -\frac{3}{2}y and -3y to get -\frac{9}{2}y.

$$-\frac{9}{2}y+1-y^{2}-\frac{9}{4}+\left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\right)+\frac{3}{2}y-\frac{1}{16}$$

Subtract \frac{9}{4} from 1 to get -\frac{5}{4}.

$$-\frac{9}{2}y-\frac{5}{4}-y^{2}+\left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\right)+\frac{3}{2}y-\frac{1}{16}$$

Consider \left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\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 \frac{1}{4}.

$$-\frac{9}{2}y-\frac{5}{4}-y^{2}+y^{2}-\frac{1}{16}+\frac{3}{2}y-\frac{1}{16}$$

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

$$-\frac{9}{2}y-\frac{5}{4}-\frac{1}{16}+\frac{3}{2}y-\frac{1}{16}$$

Subtract \frac{1}{16} from -\frac{5}{4} to get -\frac{21}{16}.

$$-\frac{9}{2}y-\frac{21}{16}+\frac{3}{2}y-\frac{1}{16}$$

Combine -\frac{9}{2}y and \frac{3}{2}y to get -3y.

$$-3y-\frac{21}{16}-\frac{1}{16}$$

Subtract \frac{1}{16} from -\frac{21}{16} to get -\frac{11}{8}.

$$-3y-\frac{11}{8}$$

Expand

$$-3y-\frac{11}{8}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+\frac{3}{2}\right)^{2}.

$$-\frac{3}{2}y+1-\left(y^{2}+3y+\frac{9}{4}\right)+\left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\right)+\frac{3}{2}y-\frac{1}{16}$$

To find the opposite of y^{2}+3y+\frac{9}{4}, find the opposite of each term.

$$-\frac{3}{2}y+1-y^{2}-3y-\frac{9}{4}+\left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\right)+\frac{3}{2}y-\frac{1}{16}$$

Combine -\frac{3}{2}y and -3y to get -\frac{9}{2}y.

$$-\frac{9}{2}y+1-y^{2}-\frac{9}{4}+\left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\right)+\frac{3}{2}y-\frac{1}{16}$$

Subtract \frac{9}{4} from 1 to get -\frac{5}{4}.

$$-\frac{9}{2}y-\frac{5}{4}-y^{2}+\left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\right)+\frac{3}{2}y-\frac{1}{16}$$

Consider \left(y-\frac{1}{4}\right)\left(y+\frac{1}{4}\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 \frac{1}{4}.

$$-\frac{9}{2}y-\frac{5}{4}-y^{2}+y^{2}-\frac{1}{16}+\frac{3}{2}y-\frac{1}{16}$$

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

$$-\frac{9}{2}y-\frac{5}{4}-\frac{1}{16}+\frac{3}{2}y-\frac{1}{16}$$

Subtract \frac{1}{16} from -\frac{5}{4} to get -\frac{21}{16}.

$$-\frac{9}{2}y-\frac{21}{16}+\frac{3}{2}y-\frac{1}{16}$$

Combine -\frac{9}{2}y and \frac{3}{2}y to get -3y.

$$-3y-\frac{21}{16}-\frac{1}{16}$$

Subtract \frac{1}{16} from -\frac{21}{16} to get -\frac{11}{8}.

$$-3y-\frac{11}{8}$$