Solve [3(1/2-y)(1/2+y)-3(1/2-y)^2+(2y-1)^2+y(2y+4)](3y-1)

$$[ 3 ( \frac { 1 } { 2 } - y ) ( \frac { 1 } { 2 } + y ) - 3 ( \frac { 1 } { 2 } - y ) ^ { 2 } + ( 2 y - 1 ) ^ { 2 } + y ( 2 y + 4 ) ] ( 3 y - 1 )$$

Evaluate

$$9y^{2}-1$$

Solution Steps

Use the distributive property to multiply 3 by \frac{1}{2}-y.

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

Use the distributive property to multiply \frac{3}{2}-3y by \frac{1}{2}+y and combine like terms.

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

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

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

Use the distributive property to multiply -3 by \frac{1}{4}-y+y^{2}.

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

Subtract \frac{3}{4} from \frac{3}{4} to get 0.

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

Combine -3y^{2} and -3y^{2} to get -6y^{2}.

$$\left(-6y^{2}+3y+\left(2y-1\right)^{2}+y\left(2y+4\right)\right)\left(3y-1\right)$$

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

$$\left(-6y^{2}+3y+4y^{2}-4y+1+y\left(2y+4\right)\right)\left(3y-1\right)$$

Combine -6y^{2} and 4y^{2} to get -2y^{2}.

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

Combine 3y and -4y to get -y.

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

Use the distributive property to multiply y by 2y+4.

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

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

$$\left(-y+1+4y\right)\left(3y-1\right)$$

Combine -y and 4y to get 3y.

$$\left(3y+1\right)\left(3y-1\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(3y\right)^{2}-1$$

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

$$3^{2}y^{2}-1$$

Calculate 3 to the power of 2 and get 9.

$$9y^{2}-1$$

Expand

$$9y^{2}-1$$

Solution Steps

Use the distributive property to multiply 3 by \frac{1}{2}-y.

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

Use the distributive property to multiply \frac{3}{2}-3y by \frac{1}{2}+y and combine like terms.

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

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

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

Use the distributive property to multiply -3 by \frac{1}{4}-y+y^{2}.

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

Subtract \frac{3}{4} from \frac{3}{4} to get 0.

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

Combine -3y^{2} and -3y^{2} to get -6y^{2}.

$$\left(-6y^{2}+3y+\left(2y-1\right)^{2}+y\left(2y+4\right)\right)\left(3y-1\right)$$

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

$$\left(-6y^{2}+3y+4y^{2}-4y+1+y\left(2y+4\right)\right)\left(3y-1\right)$$

Combine -6y^{2} and 4y^{2} to get -2y^{2}.

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

Combine 3y and -4y to get -y.

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

Use the distributive property to multiply y by 2y+4.

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

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

$$\left(-y+1+4y\right)\left(3y-1\right)$$

Combine -y and 4y to get 3y.

$$\left(3y+1\right)\left(3y-1\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(3y\right)^{2}-1$$

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

$$3^{2}y^{2}-1$$

Calculate 3 to the power of 2 and get 9.

$$9y^{2}-1$$