Simplify: (136xy^5)/(3xy^2) × (5+3x)^8/(5+3x)^6y

Let's start with multiplying the two fractions in the problem using the rule for fraction multiplication, which states that we multiply numerator by numerator and denominator by denominator while keeping the fraction line:

$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

But before we perform the fraction multiplication, let's note:

Important Note-

Notice that in both fractions of the multiplication there are expressions that are binomials squared.

In this case, we're referring to-$(5+3x)^8$and $(5+3x)^6$.

We'll treat these expressions simply as terms squared. This means that we won't open the parentheses and we'll treat both expressions exactly as we treat terms:

$a^8$and$a^6$

Now let's return to the problem and continue from where we left off:

$$Let's apply the rule for fraction multiplication mentioned above in the problem and perform the multiplication between the fractions:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}=\frac{136xy^5(5+3x)^8}{3xy^2y(5+3x)^6}=\frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6}$

In the first stage we performed the multiplication between the fractions using the above rule, then we simplified the expression in the fraction's denominator using the distributive property of multiplication, and the law of exponents for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

We applied in the final stage to the fraction's denominator that we got.

Now we'll use the above rule for fraction multiplication again, but in the opposite direction in order to present the resulting fraction as a multiplication of fractions where each fraction contains only numbers or terms with identical bases:

$\frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6}=\frac{136}{3}\cdot\frac{x}{x}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}\cdot1\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}$

Additionally in the second stage we applied the fact that dividing any number by itself gives 1.

We can now continue and simplify the expression using the law of exponents for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

Let's apply the above law to the last expression that we got:

$\frac{136}{3}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}y^{5-3}(5+3x)^{8-6}=\frac{136}{3}y^2(5+3x)^2=45\frac{1}{3}\cdot y^2(5+3x)^2$

In the first stage we applied the above law of exponents, treating the binomial as noted above, and then simplified the resulting expression, in the final stage we converted the improper fraction we got to a mixed number.

Let's summarize the solution to the problem:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}= \frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6} =45\frac{1}{3}\cdot y^2(5+3x)^2$

Therefore the correct answer is answer B.

Another Important Note:

In solving the problem above we detailed the steps to the solution, and used fraction multiplication in both directions multiple times and the above law of exponents,

We could have shortened the process, applied the distributive property of multiplication, and performed directly both the application of the above law of exponents and the numerical reduction to get directly the last line that we got:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}=45\frac{1}{3}\cdot y^2(5+3x)^2$

(meaning we could have skipped the part where we presented the fraction as a multiplication of fractions and even the initial fraction multiplication we performed and gone straight to reducing the fractions)

However, it should be emphasized that this quick solution method is conditional on the fact that between all terms in the numerator and denominator of each fraction in the problem, and also between the fractions themselves, multiplication is performed, meaning, that we can put it into a single fraction line like we did at the beginning and we can apply the distributive property and present as fraction multiplication as above, etc., this is a point worth noting, since not every problem we encounter will meet all the conditions mentioned here in this note.