Question

How do you simplify$\dfrac{{{x}^{2}}-6x+9}{81-{{x}^{4}}}$?

Answer 1

Hint: To simplify the above problem we will factorize polynomial in Numerator and Denominator. To factorize numerators we require to find two numbers which multiply to give +9 and at the same time sum to give -6, the coefficient of the x term. And in denominator it is a difference of two perfect square so we will apply this property ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$.then we make necessary calculations in Numerator and Denominator and will get required answer.

Complete step by step solution:
Given equation is $\dfrac{{{x}^{2}}-6x+9}{81-{{x}^{4}}}$
Numerator-
We factorize polynomials in numerator. To factorize numerators we require to find two numbers which multiply to give +9 and at the same time sum to give -6 , the coefficient of the x term.
The numbers -3 and -3 which multiply to give +9 and at the same time only pair to give sum -6 and the coefficient of x term.
$\Rightarrow \dfrac{{{x}^{2}}-3x-3x+9}{81-{{x}^{4}}}$
Now we take x common from first two term and -3 from the last two term
$\Rightarrow \dfrac{x\left( x-3 \right)-3\left( x-3 \right)}{81-{{x}^{4}}}$
Now we take (x-3) common from both the term
So now
$\Rightarrow \dfrac{\left( x-3 \right)\left( x-3 \right)}{81-{{x}^{4}}}$…………..(1)
By taking common -1 from both the term so, now Equation (1) can be rewritten as
$\Rightarrow \dfrac{\left( 3-x \right)\left( 3-x \right)}{81-{{x}^{4}}}$

Denominator-
Now in denominator part we have a difference of square
In denominator 81 can be written as ${{3}^{4}}$ so,
$\Rightarrow \dfrac{\left( 3-x \right)\left( 3-x \right)}{{{3}^{4}}-{{x}^{4}}}$
Now we will apply $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ and according to this property we can write
$\Rightarrow \dfrac{\left( 3-x \right)\left( 3-x \right)}{{{\left( {{3}^{2}} \right)}^{2}}-{{\left( {{x}^{2}} \right)}^{2}}}$
That means we can break ${{3}^{4}}$ into ${{\left( {{3}^{2}} \right)}^{2}}$and ${{x}^{4}}$ into ${{\left( {{x}^{2}} \right)}^{2}}$
Now
$\Rightarrow \dfrac{\left( 3-x \right)\left( 3-x \right)}{\left( {{3}^{2}}-{{x}^{2}} \right)\left( {{3}^{2}}+{{x}^{2}} \right)}$
Now again we will apply same property that is $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$in to $\left( {{3}^{2}}-{{x}^{2}} \right)$so it becomes
$\left( 3-x \right)\left( 3+x \right)$
Therefore
\[\Rightarrow \dfrac{\left( 3-x \right)\left( 3-x \right)}{\left( 3-x \right)\left( 3+x \right)\left( {{3}^{2}}+{{x}^{2}} \right)}\]
Now the same term will be cancel, out from numerator and denominator that is \[\left( 3-x \right)\] will be cancel from numerator and denominator
\[\Rightarrow \dfrac{\left( 3-x \right)}{\left( 3+x \right)\left( {{3}^{2}}+{{x}^{2}} \right)}\]
Now \[{{3}^{2}}\] can be written as 9 and equation can be written as
\[\Rightarrow \dfrac{\left( 3-x \right)}{\left( 3+x \right)\left( 9+{{x}^{2}} \right)}\]
Now the Final answer of the given equation is that
\[\dfrac{\left( 3-x \right)}{\left( 3+x \right)\left( 9+{{x}^{2}} \right)}\]

Note: During solving the above problem calculation mistakes can happen so please take care of it .
And alternative method to solve the numerator is that \[{{x}^{2}}-6x+9\] is complete square of the type
\[{{\left( a-b \right)}^{2}}=\left( {{a}^{2}}-2ab+{{b}^{2}} \right)\] As
\[{{x}^{2}}-6x+9\]= \[{{\left( x \right)}^{2}}-2\times x\times 3+{{3}^{2}}\]
 Hence,
\[{{x}^{2}}-6x+9\]= \[{{\left( x-3 \right)}^{2}}\]
This can be rewritten as
\[9-6x+{{x}^{2}}\]= \[{{\left( 3-x \right)}^{2}}\]
So this is another method to solve the numerator of the given equation.