Question

How do you simplify $\dfrac{10{{h}^{2}}-17h+3}{3h-3}\cdot \dfrac{h-1}{10{{h}^{2}}-17h+3}$?

Answer 1

Hint: In the above question, we are given an expression to simplify, which is written as $\dfrac{10{{h}^{2}}-17h+3}{3h-3}\cdot \dfrac{h-1}{10{{h}^{2}}-17h+3}$. As we can see, it is basically a product of two fractions of the polynomials. For simplifying, first we need to determine the domain for this expression, that is, the values of the variable $h$ which are allowed. We can determine the domain by putting all the denominator terms equal to zero, and the obtained values of $h$ are to be discarded from the real number set. On equating the denominators to zero, we will get a linear and a quadratic equation. The linear equation can be solved by applying basic algebraic operations, while the quadratic equation will be solved using the quadratic formula given by $h=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step by step solution:
Let us write the expression given in the above question as
$\Rightarrow p\left( h \right)=\dfrac{10{{h}^{2}}-17h+3}{3h-3}\cdot \dfrac{h-1}{10{{h}^{2}}-17h+3}$
The given expression contains fractions. We know that the denominator of a fraction can never be equal to zero. Therefore, putting the first denominator term equal to zero, we have
\[\Rightarrow 3h-3=0\]
Adding \[3\] both sides
$\begin{align}
  & \Rightarrow 3h-3+3=0+3 \\
 & \Rightarrow 3h=3 \\
\end{align}$
Dividing by \[3\] we get
$\begin{align}
  & \Rightarrow \dfrac{3h}{3}=\dfrac{3}{3} \\
 & \Rightarrow h=1 \\
\end{align}$
Now, we put the second denominator term equal to zero to get
$\Rightarrow 10{{h}^{2}}-17h+3=0$
The above equation can be solved by using the quadratic formula given by
$\Rightarrow h=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Putting $a=10$, $b=-17$ and $c=3$ we get
\[\begin{align}
  & \Rightarrow h=\dfrac{-\left( -17 \right)\pm \sqrt{{{\left( -17 \right)}^{2}}-4\left( 10 \right)\left( 3 \right)}}{2\left( 10 \right)} \\
 & \Rightarrow h=\dfrac{17\pm \sqrt{289-120}}{20} \\
 & \Rightarrow h=\dfrac{17\pm \sqrt{169}}{20} \\
\end{align}\]
Putting \[\sqrt{169}=13\] we get
\[\begin{align}
  & \Rightarrow h=\dfrac{17\pm 13}{20} \\
 & \Rightarrow h=\dfrac{17+13}{20},h=\dfrac{17-13}{20} \\
 & \Rightarrow h=\dfrac{30}{20},h=\dfrac{4}{20} \\
 & \Rightarrow h=\dfrac{3}{2},h=\dfrac{1}{5} \\
\end{align}\]
Thus, for h equal to $\dfrac{3}{2}$, $\dfrac{1}{5}$ and $1$, the denominators are equal to zero. So these are not allowed. Therefore, the domain for the given expression is obtained as $h\in R-\left\{ \dfrac{1}{5},1,\dfrac{3}{2} \right\}$.
Now, we again consider the given expression.
$\Rightarrow p\left( h \right)=\dfrac{10{{h}^{2}}-17h+3}{3h-3}\cdot \dfrac{h-1}{10{{h}^{2}}-17h+3}$
The quadratic factor $10{{h}^{2}}-17h+3$ can be cancelled to get
$\begin{align}
  & \Rightarrow p\left( h \right)=\dfrac{1}{3h-3}\cdot \dfrac{h-1}{1} \\
 & \Rightarrow p\left( h \right)=\dfrac{h-1}{3h-3} \\
\end{align}$
Taking $3$ common from the denominator, we get
\[\Rightarrow p\left( h \right)=\dfrac{h-1}{3\left( h-1 \right)}\]
Cancelling \[\left( h-1 \right)\] we finally get
\[\Rightarrow p\left( h \right)=\dfrac{1}{3}\]

Hence, the given expression is simplified as \[\dfrac{1}{3}\] for $h\in R-\left\{ \dfrac{1}{5},1,\dfrac{3}{2} \right\}$.

Note: The given expression seems to be very easy since it contains common factors in the numerator and the denominator, which can be easily cancelled to get the final expression. But we must not forget to determine the domain for the given expression, since for the values discarded from the real set in the above solution, the denominator of the given expression becomes zero. Therefore, the given expression will not be defined for these values.