Solve F[t+1]=t+1^3-5[t+1]^2-4(t+1)+20

$$F [ t + 1 ] = t + 1 ^ { 3 } - 5 [ t + 1 ] ^ { 2 } - 4 ( t + 1 ) + 20$$

Evaluate

$$12-13t-5t^{2}$$

Solution Steps

Calculate 1 to the power of 3 and get 1.

$$t+1-5\left(t+1\right)^{2}-4\left(t+1\right)+20$$

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

$$t+1-5\left(t^{2}+2t+1\right)-4\left(t+1\right)+20$$

Use the distributive property to multiply -5 by t^{2}+2t+1.

$$t+1-5t^{2}-10t-5-4\left(t+1\right)+20$$

Combine t and -10t to get -9t.

$$-9t+1-5t^{2}-5-4\left(t+1\right)+20$$

Subtract 5 from 1 to get -4.

$$-9t-4-5t^{2}-4\left(t+1\right)+20$$

Use the distributive property to multiply -4 by t+1.

$$-9t-4-5t^{2}-4t-4+20$$

Combine -9t and -4t to get -13t.

$$-13t-4-5t^{2}-4+20$$

Subtract 4 from -4 to get -8.

$$-13t-8-5t^{2}+20$$

Add -8 and 20 to get 12.

$$-13t+12-5t^{2}$$

Expand

$$12-13t-5t^{2}$$

Solution Steps

Calculate 1 to the power of 3 and get 1.

$$t+1-5\left(t+1\right)^{2}-4\left(t+1\right)+20$$

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

$$t+1-5\left(t^{2}+2t+1\right)-4\left(t+1\right)+20$$

Use the distributive property to multiply -5 by t^{2}+2t+1.

$$t+1-5t^{2}-10t-5-4\left(t+1\right)+20$$

Combine t and -10t to get -9t.

$$-9t+1-5t^{2}-5-4\left(t+1\right)+20$$

Subtract 5 from 1 to get -4.

$$-9t-4-5t^{2}-4\left(t+1\right)+20$$

Use the distributive property to multiply -4 by t+1.

$$-9t-4-5t^{2}-4t-4+20$$

Combine -9t and -4t to get -13t.

$$-13t-4-5t^{2}-4+20$$

Subtract 4 from -4 to get -8.

$$-13t-8-5t^{2}+20$$

Add -8 and 20 to get 12.

$$-13t+12-5t^{2}$$