Question

What is the quotient in simplest form state any restrictions on the variable z^2-4/z-3/z+2/z^2+z-12?

Answer 1

The expression is:


       (z^2 - 4)
       _______
      
          z - 3
 _______________

          z + 2
     ___________

       z^2 + z - 12


You can factor the numerator z^2 - 4 = (z + 2) (z - 2)

And the denominatior z^2 + z + 12 = (z + 4)(z - 3)


That permits to write the quotient as:


        (z + 2)(z - 2)
       ___________
      
             z - 3
  _______________

             z + 2
     ___________

       (z + 4)(z - 3)


Now you can multiplicate the numerator of the numerator times the denominator of the denominator, and multtiplicate the denominator of the numerator times the numerator of the denominator to obtain:


(z + 2)(z - 2)(z + 4)(z - 3)
____________________

       (z - 3)((z + 2)


Cancel the factors (z - 3) and (z + 2) because they are in both the numerator and the denominator =>

(z - 2)(z + 4) = z^2 + 2z - 8


The restrictions are that none of the cancelled factors can be 0, so z ≠ 3 and z ≠- 4.

Answer 2

To solve the problem we will divide the problem into two parts and then solve them separately later on by joining the two parts and further simplification we will get our solution.

The quotient in simplest form state is $\bold{z^2+2z-8}$.

Explanation

The equation is given to us:

•  $\dfrac{\dfrac{(z^2-4)}{(z-3)} }{\dfrac{(z+2)}{(z^2+z-12)} }$

We can solve it by dividing the equation into two parts, numerator and denominator,

Part-1

Numerator

Solving the numerator,

$\dfrac{(z^2-4)}{(z-3)}$

Using the algebric identity, (a²-b²) = (a=b)(a-b),

$\begin{aligned}&\ \ \ \ \dfrac{(z^2-4)}{(z-3)}\\\\&=\dfrac{(z^2-2^2)}{(z-3)}\\&=\dfrac{(z+2)(z-2))}{(z-3)}\\\end{aligned}$

Denominator

Now, Solving the denominator,

$\dfrac{(z+2)}{(z^2+z-12)}$

by factorization of the denominator,

$\begin{aligned}&\ \ \ \ \ \dfrac{(z+2)}{(z^2+z-12)}\\\\&=\dfrac{(z+2)}{(z^2+4z-3z-12)}}\\\\&=\dfrac{(z+2)}{z(z+4)-3(z+4)}}\\\\&=\dfrac{(z+2)}{(z+4)(z-3)}}\\\\\end{aligned}$

Part-2

Now after putting numerator and denominator, together;$\begin{aligned}& \dfrac{\dfrac{(z+2)(z-2))}{(z-3)}}{\dfrac{(z+2)}{(z+4)(z-3)}}} \\\\&\dfrac{(z+2)(z-2))}{(z-3)} \times \dfrac{(z+4)(z-3)}{(z+2)}\end{aligned}$

After canceling out,

$\begin{aligned}&\dfrac{(z+2)(z-2))}{(z-3)} \times \dfrac{(z+4)(z-3)}{(z+2)}\\\\&=\dfrac{(z-2))}{1} \times \dfrac{(z+4)}{1}\\\\&= (z-2)\times (z+4)\\&=z^2 +4z -2z -8\\&=z^2 +2z -8\end{aligned}$

Hence, the quotient in simplest form state is $\bold{z^2+2z-8}$.

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