Question

For all \[h\ne \pm 4\] , simplify \[\dfrac{3{{h}^{2}}-9h-12}{16-{{h}^{2}}}\] .
(A) \[\dfrac{-3\left( h+1 \right)}{\left( h-4 \right)}\]
(B) \[\dfrac{3\left( h-1 \right)}{\left( h-4 \right)}\]
(C) \[\dfrac{-3\left( h+1 \right)}{\left( h+4 \right)}\]
(D) \[\dfrac{\left( h+1 \right)}{3\left( h+4 \right)}\]
(E) \[\dfrac{3\left( h-1 \right)}{\left( h+4 \right)}\]

Answer 1

Hint: The given expression is in fraction form, the numerator and denominator are equal to \[\left( 3{{h}^{2}}-9h-12 \right)\] and \[\left( 16-{{h}^{2}} \right)\] respectively. Now, factorize the numerator and simplify it. Then, use the formula \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] to simplify the denominator \[\left( 16-{{h}^{2}} \right)\] .
Now, replace \[\left( 3{{h}^{2}}-9h-12 \right)\] and \[\left( 16-{{h}^{2}} \right)\] in the expression \[\dfrac{3{{h}^{2}}-9h-12}{16-{{h}^{2}}}\] by the simplified form of the numerator and the denominator. Then, solve it further.

Complete step-by-step answer:
According to the question, we have one expression and we have to simplify that expression.
\[\dfrac{3{{h}^{2}}-9h-12}{16-{{h}^{2}}}\] ………………………………(1)
The given expression is in the fraction form. We know that every fraction has a numerator and denominator.
The numerator of the given expression = \[\left( 3{{h}^{2}}-9h-12 \right)\] ………………………(2)
The denominator of the given expression = \[\left( 16-{{h}^{2}} \right)\] …………………………………(3)
We need to simplify the numerator and denominator of the given expression.
Now, factoring the numerator, we get
\[\begin{align}
  & \left( 3{{h}^{2}}-9h-12 \right) \\
 & =3{{h}^{2}}-12h+3h-12 \\
 & =3h\left( h-4 \right)+3\left( h-4 \right) \\
\end{align}\]
\[=\left( 3h+3 \right)\left( h-4 \right)\]
\[=3\left( h+1 \right)\left( h-4 \right)\]
So, \[\left( 3{{h}^{2}}-9h-12 \right)=3\left( h+1 \right)\left( h-4 \right)\] …………………………….(4)
We know the formula, \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] …………………………(5)
From equation (3), we have the denominator of the given expression, \[\left( 16-{{h}^{2}} \right)\] .
Now, simplifying the denominator,
\[\left( 16-{{h}^{2}} \right)\]
\[={{4}^{2}}-{{h}^{2}}\] ……………………..(6)
Bow, transforming equation (6) using the formula shown in equation (5), we get
\[={{4}^{2}}-{{h}^{2}}\]
\[=\left( 4+h \right)\left( 4-h \right)\]
So, \[\left( 16-{{h}^{2}} \right)=\left( 4+h \right)\left( 4-h \right)\] ………………………………..(7)
From equation (1), we have the expression, \[\dfrac{3{{h}^{2}}-9h-12}{16-{{h}^{2}}}\] .
From equation (4), we have the simplified form of the numerator, \[\left( 3{{h}^{2}}-9h-12 \right)=3\left( h+1 \right)\left( h-4 \right)\] .
Now, replacing the numerator \[\left( 3{{h}^{2}}-9h-12 \right)\] by \[3\left( h+1 \right)\left( h-4 \right)\] in the expression \[\dfrac{3{{h}^{2}}-9h-12}{16-{{h}^{2}}}\] , we get
\[\dfrac{3{{h}^{2}}-9h-12}{16-{{h}^{2}}}\]
\[=\dfrac{3\left( h+1 \right)\left( h-4 \right)}{16-{{h}^{2}}}\] …………………………(8)
From equation (7), we have the simplified form of the denominator, \[\left( 16-{{h}^{2}} \right)=\left( 4+h \right)\left( 4-h \right)\] .
Now, replacing the denominator \[\left( 16-{{h}^{2}} \right)\] by \[\left( 4+h \right)\left( 4-h \right)\] in equation (8), we get
\[=\dfrac{3\left( h+1 \right)\left( h-4 \right)}{\left( 4+h \right)\left( 4-h \right)}\]
\[=\dfrac{3\left( h+1 \right)\left( h-4 \right)}{-\left( 4+h \right)\left( h-4 \right)}\] …………………………………(9)
In the above equation, we can observe that the term \[\left( h-4 \right)\] is in the numerator and the denominator.
Now, on canceling the term \[\left( h-4 \right)\] in the numerator and the denominator, we get
\[\begin{align}
  & =\dfrac{3\left( h+1 \right)}{-\left( 4+h \right)} \\
 & =\dfrac{-3\left( h+1 \right)}{\left( 4+h \right)} \\
\end{align}\]
So, \[\dfrac{3{{h}^{2}}-9h-12}{16-{{h}^{2}}}=\dfrac{-3\left( h+1 \right)}{\left( 4+h \right)}\] .
Therefore, the correct option is (C).

Note: In this question, one might get confused about what to do with the information, \[h\ne \pm 4\] . Here, this information is very useful because it defines our expression. When \[h=\pm 4\] then the denominator expression \[\left( 16-{{h}^{2}} \right)\] equals zero due to which our whole expression \[\dfrac{3{{h}^{2}}-9h-12}{16-{{h}^{2}}}\] becomes undefined. So, to define the expression, we must have \[h=\pm 4\] .