Algebra Multiplication with Fractions

[Jason76]

\(\displaystyle (2x - 3)^{2}(5x^{2} + 2)^{3}[\frac{4}{(2x - 3)} + \frac{30x}{(5x^{2} + 2)}] \)

How does this become:

\(\displaystyle 4(2x - 3)(5x^{2} + 2)^{3} + 30x(2x - 3)^{2}(5x^{2} + 2)^{2}\) ?

Please just give me a few hints.

I'm thinking that the linear terms on the left (taken as a whole) are multiplied to each fraction on the right.

 

[daon2]

\(\displaystyle (2x - 3)^{2}(5x^{2} + 2)^{3}[\frac{4}{(2x - 3)} + \frac{30x}{(5x^{2} + 2)}] \)

How does this become:

\(\displaystyle 4(2x - 3)(5x^{2} + 2)^{3} + 30x(2x - 3)^{2}(5x^{2} + 2)^{2}\) ?

Please just give me a few hints.

I'm thinking that the linear terms on the left (taken as a whole) are multiplied to each fraction on the right.

Let's try this: Let \(\displaystyle A=(2x - 3)^{2}(5x^{2} + 2)^{3}\)

Then distribute: \(\displaystyle A[\frac{4}{(2x - 3)} + \frac{30x}{(5x^{2} + 2)}] = \frac{(A)4}{(2x - 3)} + \frac{(A)30x}{(5x^{2} + 2)}]\)

Now replace A again and simplify.

 

[Deleted member 4993]

\(\displaystyle (2x - 3)^{2}(5x^{2} + 2)^{3}[\frac{4}{(2x - 3)} + \frac{30x}{(5x^{2} + 2)}] \)

How does this become:

\(\displaystyle 4(2x - 3)(5x^{2} + 2)^{3} + 30x(2x - 3)^{2}(5x^{2} + 2)^{2}\) ?

Please just give me a few hints.

I'm thinking that the linear terms on the left (taken as a whole) are multiplied to each fraction on the right.

it is simiar to:

B²* C³ * (A/B + D/C)

= B²* C³ * A/B + B²* C³ * D/C

= B* C³ * A + B²* C² * D

 

[girodd]

yes, that's the distributive law, a(b+c) = ab + ac, no matter what a, b, c are. In this case, a is the product of the two polynomials sitting next to the parentheses containing the sum of the the two rational expressions.

 

[Jason76]

Thanks for the help guys. Those formulas for dealing with the problem look right.