Solve (5t-1)^2+(2t-2)(-t+4)

$$( 5 t - 1 ) ^ { 2 } + ( 2 t - 2 ) ( - t + 4 )$$

Evaluate

$$23t^{2}-7$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5t-1\right)^{2}.

$$25t^{2}-10t+1+\left(2t-2\right)\left(-t+4\right)$$

Use the distributive property to multiply 2t-2 by -t+4.

$$25t^{2}-10t+1+2t\left(-t\right)+8t-2\left(-t\right)-8$$

Multiply -2 and -1 to get 2.

$$25t^{2}-10t+1+2t\left(-t\right)+8t+2t-8$$

Combine 8t and 2t to get 10t.

$$25t^{2}-10t+1+2t\left(-t\right)+10t-8$$

Combine -10t and 10t to get 0.

$$25t^{2}+1+2t\left(-t\right)-8$$

Subtract 8 from 1 to get -7.

$$25t^{2}-7+2t\left(-t\right)$$

Multiply t and t to get t^{2}.

$$25t^{2}-7+2t^{2}\left(-1\right)$$

Multiply 2 and -1 to get -2.

$$25t^{2}-7-2t^{2}$$

Combine 25t^{2} and -2t^{2} to get 23t^{2}.

$$23t^{2}-7$$

Expand

$$23t^{2}-7$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5t-1\right)^{2}.

$$25t^{2}-10t+1+\left(2t-2\right)\left(-t+4\right)$$

Use the distributive property to multiply 2t-2 by -t+4.

$$25t^{2}-10t+1+2t\left(-t\right)+8t-2\left(-t\right)-8$$

Multiply -2 and -1 to get 2.

$$25t^{2}-10t+1+2t\left(-t\right)+8t+2t-8$$

Combine 8t and 2t to get 10t.

$$25t^{2}-10t+1+2t\left(-t\right)+10t-8$$

Combine -10t and 10t to get 0.

$$25t^{2}+1+2t\left(-t\right)-8$$

Subtract 8 from 1 to get -7.

$$25t^{2}-7+2t\left(-t\right)$$

Multiply t and t to get t^{2}.

$$25t^{2}-7+2t^{2}\left(-1\right)$$

Multiply 2 and -1 to get -2.

$$25t^{2}-7-2t^{2}$$

Combine 25t^{2} and -2t^{2} to get 23t^{2}.

$$23t^{2}-7$$