Answer:
the answer will be 8
Explanation:
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
�
2
+
2
�
−
8
z
2
+2z−8 .
Explanation
The equation is given to us:
(
�
2
−
4
)
(
�
−
3
)
(
�
+
2
)
(
�
2
+
�
−
12
)
(z
2
+z−12)
(z+2)
(z−3)
(z
2
−4)
We can solve it by dividing the equation into two parts, numerator and denominator,
Part-1
Numerator
Solving the numerator,
(
�
2
−
4
)
(
�
−
3
)
(z−3)
(z
2
−4)
Using the algebric identity, (a²-b²) = (a=b)(a-b),
(
�
2
−
4
)
(
�
−
3
)
=
(
�
2
−
2
2
)
(
�
−
3
)
=
(
�
+
2
)
(
�
−
2
)
)
(
�
−
3
)
(z−3)
(z
2
−4)
=
(z−3)
(z
2
−2
2
)
=
(z−3)
(z+2)(z−2))
Denominator
Now, Solving the denominator,
(
�
+
2
)
(
�
2
+
�
−
12
)
(z
2
+z−12)
(z+2)
by factorization of the denominator,
\begin{gathered}\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}\end{gathered}
Part-2
Now after putting numerator and denominator, together;\begin{gathered}\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}\end{gathered}
After canceling out,
(
�
+
2
)
(
�
−
2
)
)
(
�
−
3
)
×
(
�
+
4
)
(
�
−
3
)
(
�
+
2
)
=
(
�
−
2
)
)
1
×
(
�
+
4
)
1
=
(
�
−
2
)
×
(
�
+
4
)
=
�
2
+
4
�
−
2
�
−
8
=
�
2
+
2
�
−
8
(z−3)
(z+2)(z−2))
×
(z+2)
(z+4)(z−3)
=
1
(z−2))
×
1
(z+4)
=(z−2)×(z+4)
=z
2
+4z−2z−8
=z
2
+2z−8
Hence, the quotient in simplest form state is
�
2
+
2
�
−
8
z
2
+2z−8 .