Solve (a+y-2)(a+y+2)-(a-y)^2-4(ay-1)+1

$$( a + y - 2 ) ( a + y + 2 ) - ( a - y ) ^ { 2 } - 4 ( a y - 1 ) + 1$$

Evaluate

$$1$$

Solution Steps

Consider \left(a+y-2\right)\left(a+y+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=a+y and b=2. Square 2.

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

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

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

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

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

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

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

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

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

Combine 2ay and 2ay to get 4ay.

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

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

$$4ay-4-4\left(ay-1\right)+1$$

Use the distributive property to multiply -4 by ay-1.

$$4ay-4-4ay+4+1$$

Combine 4ay and -4ay to get 0.

$$-4+4+1$$

Add -4 and 4 to get 0.

$$1$$

Factor

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