Solve 5T(1+1/4T)+(T^2+T-1)^2-(T+1)^3-T^2(T+1/2)

$$5 T ( 1 + \frac { 1 } { 4 } T ) + ( T ^ { 2 } + T - 1 ) ^ { 2 } - ( T + 1 ) ^ { 3 } - T ^ { 2 } ( T + \frac { 1 } { 2 } )$$

Evaluate

$$T^{4}-\frac{13T^{2}}{4}$$

Solution Steps

Use the distributive property to multiply 5T by 1+\frac{1}{4}T.

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

Square T^{2}+T-1.

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

Combine \frac{5}{4}T^{2} and -T^{2} to get \frac{1}{4}T^{2}.

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

Combine 5T and -2T to get 3T.

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

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(T+1\right)^{3}.

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

To find the opposite of T^{3}+3T^{2}+3T+1, find the opposite of each term.

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

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

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

Combine \frac{1}{4}T^{2} and -3T^{2} to get -\frac{11}{4}T^{2}.

$$3T-\frac{11}{4}T^{2}+T^{4}+T^{3}+1-3T-1-T^{2}\left(T+\frac{1}{2}\right)$$

Combine 3T and -3T to get 0.

$$-\frac{11}{4}T^{2}+T^{4}+T^{3}+1-1-T^{2}\left(T+\frac{1}{2}\right)$$

Subtract 1 from 1 to get 0.

$$-\frac{11}{4}T^{2}+T^{4}+T^{3}-T^{2}\left(T+\frac{1}{2}\right)$$

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

$$-\frac{11}{4}T^{2}+T^{4}+T^{3}-\left(T^{3}+\frac{1}{2}T^{2}\right)$$

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

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

Combine T^{3} and -T^{3} to get 0.

$$-\frac{11}{4}T^{2}+T^{4}-\frac{1}{2}T^{2}$$

Combine -\frac{11}{4}T^{2} and -\frac{1}{2}T^{2} to get -\frac{13}{4}T^{2}.

$$-\frac{13}{4}T^{2}+T^{4}$$

Expand

$$T^{4}-\frac{13T^{2}}{4}$$

Solution Steps

Use the distributive property to multiply 5T by 1+\frac{1}{4}T.

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

Square T^{2}+T-1.

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

Combine \frac{5}{4}T^{2} and -T^{2} to get \frac{1}{4}T^{2}.

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

Combine 5T and -2T to get 3T.

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

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(T+1\right)^{3}.

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

To find the opposite of T^{3}+3T^{2}+3T+1, find the opposite of each term.

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

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

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

Combine \frac{1}{4}T^{2} and -3T^{2} to get -\frac{11}{4}T^{2}.

$$3T-\frac{11}{4}T^{2}+T^{4}+T^{3}+1-3T-1-T^{2}\left(T+\frac{1}{2}\right)$$

Combine 3T and -3T to get 0.

$$-\frac{11}{4}T^{2}+T^{4}+T^{3}+1-1-T^{2}\left(T+\frac{1}{2}\right)$$

Subtract 1 from 1 to get 0.

$$-\frac{11}{4}T^{2}+T^{4}+T^{3}-T^{2}\left(T+\frac{1}{2}\right)$$

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

$$-\frac{11}{4}T^{2}+T^{4}+T^{3}-\left(T^{3}+\frac{1}{2}T^{2}\right)$$

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

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

Combine T^{3} and -T^{3} to get 0.

$$-\frac{11}{4}T^{2}+T^{4}-\frac{1}{2}T^{2}$$

Combine -\frac{11}{4}T^{2} and -\frac{1}{2}T^{2} to get -\frac{13}{4}T^{2}.

$$-\frac{13}{4}T^{2}+T^{4}$$