Create the Sum of two Fractions.

[gijas]

Split this fraction up into two fractions with different denominators.


(3x^2 + 7x + 8)
(x^2 + 3x + 2)


I understand you have to factor here. So the denominator would be..

(x + 1)(x + 2)

But the numerator I think is prime and that is where I get confused on what to do next.

 

[tkhunny]

Why do you care if the numerator cannot be factored?

You know the combined denominator looks like this: (x+1)(x+2)

Try some long division to get \(\displaystyle 3 + \frac{A}{(x+1)(x+2)}\)

This MUST have come from something that looked like this \(\displaystyle 3 + \frac{B}{x+1} + \frac{C}{x+2}\)

Find A, then B and C.

 

[mmm4444bot]

With this particular ratio, one method is to use long-hand polynomial division. Try it, and see what happens. :cool:

 

[mmm4444bot]

Ahh, I see that tk and I are thinking on similar lines. I'm kinda lazy, so I would use the form

3/1 + A/(x^2 + 3x + 2)

It seems to match the instructions.

 

[gijas]

Why do you care if the numerator cannot be factored?

You know the combined denominator looks like this: (x+1)(x+2)

Try some long division to get \(\displaystyle 3 + \frac{A}{(x+1)(x+2)}\)

This MUST have come from something that looked like this \(\displaystyle 3 + \frac{B}{x+1} + \frac{C}{x+2}\)

Find A, then B and C.

I am dividing 3/(3x^2 + 7x + 8) for A then factoring the result to arrive at B and C?

 

[mmm4444bot]

I am dividing 3/(3x^2 + 7x + 8)

That is not correct.

Divide 3x^2 + 7x + 8 by x^2 + 3x + 2 using long-hand polynomial division.

In our posts, the symbol A represents the remainder from this division. (It's a linear polynomial.)

We are using the symbol A, versus telling you the answer, to give you a chance to work the exercise yourself.

Have you learned how to divide one polynomial by another using long division?

 

[gijas]

That is not correct.

Divide 3x^2 + 7x + 8 by x^2 + 3x + 2 using long-hand polynomial division.

In our posts, the symbol A represents the remainder from this division. (It's a linear polynomial.)

We are using the symbol A, versus telling you the answer, to give you a chance to work the exercise yourself.

Have you learned how to divide one polynomial by another using long division?

Yes, I understand. I will let you know what I come up with.

 

[soroban]

Hello, gijas!

Split this fraction into two fractions with different denominators.

. . \(\displaystyle \dfrac{3x^2+7x+8}{x^2+3x+2}\)

We want: .\(\displaystyle \dfrac{3x^2 + 7x + 8}{(x+1)(x+2)} \;=\;\dfrac{ax+b}{x+1} + \dfrac{cx+d}{x+2}\)


Multiply by \(\displaystyle (x+1)(x+2):\)

. . \(\displaystyle 3x^2 + 7x + 8 \;=\;(ax+b)(x+2) + (cx+d)(x+1)\)

. . \(\displaystyle 3x^2+7x+8 \;=\;ax^2 + 2ax + bx + 2b + cx^2 + cx + dx + d \)

. . \(\displaystyle 3x^2+7x+8 \;=\;(a+c)x^2 + (2a + b+c+d)x + (2b+d) \)

Equate coefficients: .\(\displaystyle \begin{Bmatrix}a+c &=& 3 & [1] \\ 2a+b+c+d &=& 7 & [2] \\ 2b+d &=& 8 & [3]\end{Bmatrix}\)

From [1]: .\(\displaystyle c \:=\:3-a\)
From [3]: .\(\displaystyle d \:=\:8-2b\)


Substitute into [2]: .\(\displaystyle 2a + b + (3-a) + (8-2b) \:=\:7 \quad\Rightarrow\quad b \:=\:a+4\)

Then: .\(\displaystyle d \:=\:8 - 2b \:=\:8 - 2(a+4) \quad\Rightarrow\quad d \:=\:-2a\)

We have: .\(\displaystyle \begin{Bmatrix}a &=& a \\ b &=& a+4 \\ c &=& 3-a \\ d &=& -2a \end{Bmatrix}\)


On the right, replace \(\displaystyle a\) with a parameter \(\displaystyle t\):

. . \(\displaystyle \begin{Bmatrix}a &=& t \\ b &=& t+4 \\ c &=& 3-t \\ d &=& -2t\end{Bmatrix}\)


We have an infinite number of solutions:

.. \(\displaystyle \dfrac{3x^2+7x+8}{(x+1)(x+2)} \;=\;\dfrac{tx + (t+4)}{x+1} + \dfrac{(3-t)x - 2t}{x+2} \;\;\text{ for any real number }t\)

 

[lookagain]

Split this fraction up into two fractions with different denominators.


(3x^2 + 7x + 8)
(x^2 + 3x + 2)

Here are a couple of ways, where each is an algebraic fraction,
and the denominators are different:



\(\displaystyle = \ \dfrac{3x^2 + 6x + x + 8}{(x + 1)(x + 2)}\)


\(\displaystyle = \dfrac{3x^2 + 6x}{(x + 1)(x + 2)} \ + \ \dfrac{x + 8}{(x + 1)(x + 2)}\)


\(\displaystyle = \dfrac{3x(x + 2)}{(x + 1)(x + 2)} \ + \ \dfrac{x + 8}{x^2 + 3x + 2}\)


\(\displaystyle = \dfrac{3x}{x + 1} \ + \ \dfrac{x + 8}{x^2 + 3x + 2}\)


- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

\(\displaystyle Or:\)


\(\displaystyle = \ \dfrac{3x^2 - x + 8x + 8}{(x + 1)(x + 2)}\)


\(\displaystyle = \ \dfrac{3x^2 - x}{(x + 1)(x + 2)} \ + \ \dfrac{8x + 8}{(x + 1)(x + 2)}\)


\(\displaystyle = \ \dfrac{3x^2 - x}{x^2 + 3x + 2} \ + \ \dfrac{8(x + 1)}{(x + 1)(x + 2)}\)


\(\displaystyle = \ \dfrac{3x^2 - x}{x^2 + 3x + 2} \ + \ \dfrac{8}{x + 2}\)