Solve 1/2(2y^{2}+1)^{2}-2(2y^{2}+1)(2y^{2}-1)-[1/2(2y^2+1)-1]^2-(2y^2+1/2-y)^2

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

Evaluate

$$2+y+4y^{3}-11y^{4}$$

Solution Steps

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

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

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

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

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

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

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

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

To find the opposite of y^{4}-y^{2}+\frac{1}{4}, find the opposite of each term.

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

Square 2y^{2}+\frac{1}{2}-y.

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

To find the opposite of 4y^{4}-4y^{3}+3y^{2}-y+\frac{1}{4}, find the opposite of each term.

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

Combine -y^{4} and -4y^{4} to get -5y^{4}.

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

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

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

Subtract \frac{1}{4} from -\frac{1}{4} to get -\frac{1}{2}.

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

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

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

Use the distributive property to multiply -4y^{2}-2 by 2y^{2}-1 and combine like terms.

$$2y^{4}+2y^{2}+\frac{1}{2}-8y^{4}+2-5y^{4}-2y^{2}-\frac{1}{2}+4y^{3}+y$$

Combine 2y^{4} and -8y^{4} to get -6y^{4}.

$$-6y^{4}+2y^{2}+\frac{1}{2}+2-5y^{4}-2y^{2}-\frac{1}{2}+4y^{3}+y$$

Add \frac{1}{2} and 2 to get \frac{5}{2}.

$$-6y^{4}+2y^{2}+\frac{5}{2}-5y^{4}-2y^{2}-\frac{1}{2}+4y^{3}+y$$

Combine -6y^{4} and -5y^{4} to get -11y^{4}.

$$-11y^{4}+2y^{2}+\frac{5}{2}-2y^{2}-\frac{1}{2}+4y^{3}+y$$

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

$$-11y^{4}+\frac{5}{2}-\frac{1}{2}+4y^{3}+y$$

Subtract \frac{1}{2} from \frac{5}{2} to get 2.

$$-11y^{4}+2+4y^{3}+y$$

Expand

$$2+y+4y^{3}-11y^{4}$$

Solution Steps

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

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

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

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

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

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

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

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

To find the opposite of y^{4}-y^{2}+\frac{1}{4}, find the opposite of each term.

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

Square 2y^{2}+\frac{1}{2}-y.

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

To find the opposite of 4y^{4}-4y^{3}+3y^{2}-y+\frac{1}{4}, find the opposite of each term.

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

Combine -y^{4} and -4y^{4} to get -5y^{4}.

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

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

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

Subtract \frac{1}{4} from -\frac{1}{4} to get -\frac{1}{2}.

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

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

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

Use the distributive property to multiply -4y^{2}-2 by 2y^{2}-1 and combine like terms.

$$2y^{4}+2y^{2}+\frac{1}{2}-8y^{4}+2-5y^{4}-2y^{2}-\frac{1}{2}+4y^{3}+y$$

Combine 2y^{4} and -8y^{4} to get -6y^{4}.

$$-6y^{4}+2y^{2}+\frac{1}{2}+2-5y^{4}-2y^{2}-\frac{1}{2}+4y^{3}+y$$

Add \frac{1}{2} and 2 to get \frac{5}{2}.

$$-6y^{4}+2y^{2}+\frac{5}{2}-5y^{4}-2y^{2}-\frac{1}{2}+4y^{3}+y$$

Combine -6y^{4} and -5y^{4} to get -11y^{4}.

$$-11y^{4}+2y^{2}+\frac{5}{2}-2y^{2}-\frac{1}{2}+4y^{3}+y$$

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

$$-11y^{4}+\frac{5}{2}-\frac{1}{2}+4y^{3}+y$$

Subtract \frac{1}{2} from \frac{5}{2} to get 2.

$$-11y^{4}+2+4y^{3}+y$$