Question

How do you add or subtract $\dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}?$

Answer 1

Hint: We do cross multiplication. Using the polynomial multiplication, we multiply the terms. We may use the polynomial addition, polynomial subtraction and polynomial division, if necessary.

Complete step by step solution:
We are asked to subtract a polynomial fraction from another polynomial fraction, $\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}$
Since we have to find the difference of the two polynomial fractions $\dfrac{x+6}{5x+10}$ and $\dfrac{x-2}{4x+8},$ we see if the denominators are the same.
In the given problem, the denominators of the polynomial fractions are different.
So, we have to do cross multiplication of the terms as follows,
$\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=\dfrac{\left( x+6 \right)\left( 4x+8 \right)-\left( 5x+10 \right)\left( x-2 \right)}{\left( 5x+10 \right)\left( 4x+8 \right)}.......\left( 1 \right)$
We are going to find each of the products on the left-hand side of the above equation.
Let us take the first product, $\left( x+6 \right)\left( 4x+8 \right).$
We use polynomial multiplication.
We multiply each term of the second polynomial with each term of the first polynomial, regardless of the power of the variables.
We get,
$\Rightarrow \left( x+6 \right)\left( 4x+8 \right)=4{{x}^{2}}+8x+24x+48$
Using the polynomial addition in the above equation to get,
$\Rightarrow \left( x+6 \right)\left( 4x+8 \right)=4{{x}^{2}}+32x+48.......\left( 2 \right)$
Let us take the next product, $\left( 5x+10 \right)\left( x-2 \right).$
By polynomial multiplication we will find the following step,
$\Rightarrow \left( 5x+10 \right)\left( x-2 \right)=5{{x}^{2}}-10x+10x-20.$
We multiplied all the terms of the first polynomial with all the terms of the second polynomial.
Now we add or subtract the terms with variables having the same exponent.
We get,
$\Rightarrow \left( 5x+10 \right)\left( x-2 \right)=5{{x}^{2}}-20.......\left( 3 \right)$
Similarly, we take another product, $\left( 5x+10 \right)\left( 4x+8 \right).$
The polynomial multiplication will give us the following,
$\Rightarrow \left( 5x+10 \right)\left( 4x+8 \right)=20{{x}^{2}}+40x+40x+80.$
Using the rule of the polynomial addition,
$\Rightarrow \left( 5x+10 \right)\left( 4x+8 \right)=20{{x}^{2}}+80x+80.......\left( 4 \right)$
We are substituting the equations \[\left( 2 \right),\left( 3 \right)\] and $\left( 4 \right)$ in the equation $\left( 1 \right),$
  $\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=\dfrac{4{{x}^{2}}+32x+48-\left( 5{{x}^{2}}-20 \right)}{20{{x}^{2}}+80x+80}$
We open the bracket and the use polynomial addition and subtraction,
 \[\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=\dfrac{4{{x}^{2}}+32x+48-5{{x}^{2}}+20}{20{{x}^{2}}+80x+80}\]
And we get,
\[\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=\dfrac{-{{x}^{2}}+32x+68}{20{{x}^{2}}+80x+80}\]
So, we will get,
\[\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=\dfrac{-{{x}^{2}}+32x+68}{20\left( {{x}^{2}}+4x+4 \right)}\]
That is,
\[\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=\dfrac{-{{x}^{2}}+32x+68}{20{{\left( x+2 \right)}^{2}}}\]
We are multiplying both the numerator and the denominator of the Right-hand side with $-1,$ \[\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=\dfrac{{{x}^{2}}-32x-68}{-20{{\left( x+2 \right)}^{2}}}\]
Now we will get,
\[\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=-\dfrac{{{x}^{2}}-32x-68}{20{{\left( x+2 \right)}^{2}}}.\]
And from this we will find,
\[\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=-\dfrac{\left( x-34 \right)\left( x+2 \right)}{20{{\left( x+2 \right)}^{2}}},\] Since ${{x}^{2}}-32x-68=\left( x-34 \right)\left( x+2 \right).$
Cancelling $x+2$ from numerator and denominator,
\[\Rightarrow \dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=-\dfrac{\left( x-34 \right)}{20\left( x+2 \right)}\]
Hence the difference is \[\dfrac{x+6}{5x+10}-\dfrac{x-2}{4x+8}=-\dfrac{\left( x-34 \right)}{20\left( x+2 \right)}=\dfrac{34-x}{20\left( x+2 \right)}.\]

Note: In polynomial addition, we add the coefficients of the variables with the same power to make the sum polynomial of the same degree as of the summand polynomial with highest degree. Similar is the case for polynomial subtraction. In polynomial multiplication, we multiply the coefficients of all the variables as well as the variables, regardless of the power.