Question

How do you simplify \[\dfrac{{{x}^{2}}-3x-10}{{{x}^{2}}-8x+15}.\dfrac{{{x}^{2}}+5x-6}{{{x}^{2}}+4x+4}\]?

Answer 1

Hint: We have four quadratic equations in the denominator and the numerator. We factor them using the vanishing method by considering $x=a=5$ . We then eliminate the common factors and get the simplified form of the expression.

Complete step by step solution:
We have been given the division of four quadratic equations. We need to find the factor of the equations.
The four quadratic equations are \[{{x}^{2}}-3x-10\], \[{{x}^{2}}-8x+15\], \[{{x}^{2}}+5x-6\], \[{{x}^{2}}+4x+4\].
 We use vanishing methods to solve them.
We find the value of $x=a$ for which the function $f\left( x \right)={{x}^{2}}-3x-10=0$.
We take $x=a=5$. We can see $f\left( 5 \right)={{5}^{2}}-3\times 5-10=25-15-10=0$.
So, the root of the $f\left( x \right)={{x}^{2}}-3x-10$ will be the function $\left( x-5 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Therefore, the term $\left( x-5 \right)$ is a factor of the polynomial \[{{x}^{2}}-3x-10\].
So, \[{{x}^{2}}-3x-10=\left( x-5 \right)\left( x+2 \right)\].
For $g\left( x \right)={{x}^{2}}-8x+15$, we take $x=5$. We can see $f\left( 5 \right)={{5}^{2}}-8\times 5+15=25-40+15=0$.
So, the root of the $g\left( x \right)={{x}^{2}}-8x+15$ will be the function $\left( x-5 \right)$.
Therefore, \[{{x}^{2}}-8x+15=\left( x-5 \right)\left( x-3 \right)\].
For $h\left( x \right)={{x}^{2}}+5x-6$, we take $x=a=1$. We can see $f\left( 1 \right)={{1}^{2}}+5\times 1-6=1+5-6=0$.
So, the root of the $g\left( x \right)={{x}^{2}}+5x-6$ will be the function $\left( x-1 \right)$.
Therefore, \[{{x}^{2}}+5x-6=\left( x-1 \right)\left( x+6 \right)\].
For $k\left( x \right)={{x}^{2}}+4x+4$, we form square and get \[{{x}^{2}}+4x+4={{\left( x+2 \right)}^{2}}\]
Now we put the factorised values for the two equations.
We get \[\dfrac{{{x}^{2}}-3x-10}{{{x}^{2}}-8x+15}.\dfrac{{{x}^{2}}+5x-6}{{{x}^{2}}+4x+4}=\dfrac{\left( x-5 \right)\left( x+2 \right)}{\left( x-5 \right)\left( x-3 \right)}.\dfrac{\left( x-1 \right)\left( x+6 \right)}{{{\left( x+2 \right)}^{2}}}\].
Now we can see in the denominator and the numerator the common factor is \[\left( x-5 \right)\left( x+2 \right)\].
We can eliminate the factor from both denominator and the numerator.
So, \[\dfrac{{{x}^{2}}-3x-10}{{{x}^{2}}-8x+15}.\dfrac{{{x}^{2}}+5x-6}{{{x}^{2}}+4x+4}=\dfrac{\left( x-5 \right)\left( x+2 \right)}{\left( x-5 \right)\left( x-3 \right)}.\dfrac{\left( x-1 \right)\left( x+6 \right)}{{{\left( x+2 \right)}^{2}}}=\dfrac{\left( x-1 \right)\left( x+6 \right)}{\left( x-3 \right)\left( x+2 \right)}\]. The only condition being $x\ne -2,5$.

Therefore, the simplified form of \[\dfrac{{{x}^{2}}-3x-10}{{{x}^{2}}-8x+15}.\dfrac{{{x}^{2}}+5x-6}{{{x}^{2}}+4x+4}\] is \[\dfrac{\left( x-1 \right)\left( x+6 \right)}{\left( x-3 \right)\left( x+2 \right)}\].

Note: We can also factorise the quadratic equations using grouping method.
In case of \[{{x}^{2}}-3x-10\], we break the middle term $-3x$ into two parts of $-5x$ and $2x$.
So, ${{x}^{2}}-3x-10={{x}^{2}}-5x+2x-10$.
Factorising we get
$\begin{align}
  & {{x}^{2}}-3x-10 \\
 & ={{x}^{2}}-5x+2x-10 \\
 & =x\left( x-5 \right)+2\left( x-5 \right) \\
 & =\left( x-5 \right)\left( x+2 \right) \\
\end{align}$
Similarly, \[{{x}^{2}}-8x+15={{x}^{2}}-5x-3x+15=\left( x-5 \right)\left( x-3 \right)\].
\[{{x}^{2}}+5x-6={{x}^{2}}+6x-x-6=\left( x-1 \right)\left( x+6 \right)\] and \[{{x}^{2}}+4x+4={{x}^{2}}+2x+2x+4={{\left( x+2 \right)}^{2}}\].