Solve (5-4)(3-a)-(a-4)^2=

$$( 5 - 4 ) ( 3 - a ) - ( a - 4 ) ^ { 2 } =$$

Evaluate

$$-a^{2}+7a-13$$

Solution Steps

Subtract 4 from 5 to get 1.

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

Use the distributive property to multiply 1 by 3-a.

$$3-a-\left(a-4\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-4\right)^{2}.

$$3-a-\left(a^{2}-8a+16\right)$$

To find the opposite of a^{2}-8a+16, find the opposite of each term.

$$3-a-a^{2}+8a-16$$

Combine -a and 8a to get 7a.

$$3+7a-a^{2}-16$$

Subtract 16 from 3 to get -13.

$$-13+7a-a^{2}$$

Expand

$$-a^{2}+7a-13$$

Solution Steps

Subtract 4 from 5 to get 1.

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

Use the distributive property to multiply 1 by 3-a.

$$3-a-\left(a-4\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-4\right)^{2}.

$$3-a-\left(a^{2}-8a+16\right)$$

To find the opposite of a^{2}-8a+16, find the opposite of each term.

$$3-a-a^{2}+8a-16$$

Combine -a and 8a to get 7a.

$$3+7a-a^{2}-16$$

Subtract 16 from 3 to get -13.

$$-13+7a-a^{2}$$