Solve (z-x)(z+x)-(9-x)(I+x)

$$( z - x ) ( z + x ) - ( 9 - x ) ( I + x )$$

Evaluate

$$Ix-9x+z^{2}-9I$$

Solution Steps

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

$$z^{2}-x^{2}-\left(9-x\right)\left(I+x\right)$$

Apply the distributive property by multiplying each term of 9-x by each term of I+x.

$$z^{2}-x^{2}-\left(9I+9x-xI-x^{2}\right)$$

To find the opposite of 9I+9x-xI-x^{2}, find the opposite of each term.

$$z^{2}-x^{2}-9I-9x-\left(-xI\right)-\left(-x^{2}\right)$$

The opposite of -xI is xI.

$$z^{2}-x^{2}-9I-9x+xI-\left(-x^{2}\right)$$

The opposite of -x^{2} is x^{2}.

$$z^{2}-x^{2}-9I-9x+xI+x^{2}$$

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

$$z^{2}-9I-9x+xI$$

Expand

$$Ix-9x+z^{2}-9I$$

Solution Steps

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

$$z^{2}-x^{2}-\left(9-x\right)\left(I+x\right)$$

Apply the distributive property by multiplying each term of 9-x by each term of I+x.

$$z^{2}-x^{2}-\left(9I+9x-xI-x^{2}\right)$$

To find the opposite of 9I+9x-xI-x^{2}, find the opposite of each term.

$$z^{2}-x^{2}-9I-9x-\left(-xI\right)-\left(-x^{2}\right)$$

The opposite of -xI is xI.

$$z^{2}-x^{2}-9I-9x+xI-\left(-x^{2}\right)$$

The opposite of -x^{2} is x^{2}.

$$z^{2}-x^{2}-9I-9x+xI+x^{2}$$

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

$$z^{2}-9I-9x+xI$$