# simplifying help

• Feb 25th 2011, 04:19 PM
maybnxtseasn
simplifying help
i havent taken an algebra class for sometime and im curious...how do i approach at simplifying the following denominator

3/(4x²+5x+1) so i can perform partial fraction integration
• Feb 25th 2011, 04:29 PM
Sudharaka
Quote:

Originally Posted by maybnxtseasn
i havent taken an algebra class for sometime and im curious...how do i approach at simplifying the following denominator

3/(4x²+5x+1) so i can perform partial fraction integration

Dear maybnxtseasn,

$\displaystyle \dfrac{3}{4x²+5x+1}$

$\displaystyle \dfrac{3}{(4x+1)(x+1)}$

Decomposing this into partial fractions;

$\displaystyle \dfrac{3}{4x²+5x+1}=\dfrac{4}{4x+1}-\dfrac{1}{x+1}$

Hope you can continue with the integration.
• Feb 25th 2011, 07:06 PM
maybnxtseasn
thanks but im curious as to the steps u use to simplify this?
• Feb 25th 2011, 07:15 PM
NOX Andrew
Are you asking how he knew $\displaystyle 4x^2 + 5x + 1 = (4x + 1)(x + 1)$? If so, one method is as follows:

The product of the coefficients of the first and last terms is $\displaystyle 4 x 1 = 4$.

$\displaystyle 1$ and $\displaystyle 4$ are factors of $\displaystyle 4$ (the product from the previous step) and sum to $\displaystyle 5$ (the coefficient of the middle term). So, write the middle term, $\displaystyle 5x$, as $\displaystyle 1x + 4x$: $\displaystyle 4x^2 + 1x + 4x + 1$.

Then, group each pair of terms together: $\displaystyle (4x^2 + 1x) + (4x + 1)$.

Perform any factoring on each group of terms: $\displaystyle x(4x + 1) + (4x + 1)$.

Now, factor $\displaystyle (4x + 1)$: $\displaystyle (4x + 1)(x + 1)$.
• Feb 26th 2011, 03:02 AM
HallsofIvy
Another way to do that (but much more complicated that just factoring directly) is to solve the equation $\displaystyle 4x^2+ 5x+ 1= 0$- not by factoring, of course, but by the quadratic formula:
[tex]x= \frac{-5\pm\sqrt{5^2- 4(4)(1)}}{2(4)}= \frac{-5\pm\sqrt{25- 16}}{8}= \frac{-5\pm\sqrt{9}}{8}= \frac{-5\pm3}{8}[tex]
so that $\displaystyle x= -\frac{1}{4}$ and $\displaystyle x= 1$

Now, we can say that [tex]4x^2+ 5x+ 1= 4(x+ \frac{1}{4})(x- 1)= (4x+ 1)(x- 1)[/quote]
(I put the "4" in front to make t he leading coefficient 4.)

That is, as I said, the hard way!
• Feb 26th 2011, 05:06 PM
Sudharaka
Quote:

Originally Posted by HallsofIvy
Another way to do that (but much more complicated that just factoring directly) is to solve the equation $\displaystyle 4x^2+ 5x+ 1= 0$- not by factoring, of course, but by the quadratic formula:
[tex]x= \frac{-5\pm\sqrt{5^2- 4(4)(1)}}{2(4)}= \frac{-5\pm\sqrt{25- 16}}{8}= \frac{-5\pm\sqrt{9}}{8}= \frac{-5\pm3}{8}[tex]
so that $\displaystyle x= -\frac{1}{4}$ and $\displaystyle x= 1$

Now, we can say that [tex]4x^2+ 5x+ 1= 4(x+ \frac{1}{4})(x- 1)= (4x+ 1)(x- 1)
(I put the "4" in front to make t he leading coefficient 4.)

That is, as I said, the hard way!

Dear HallsofIvy,

You new avatar make some of the text unreadable. I have seen this in your recent post a number of times and I tried the zoom in/out feature in the browser to eliminate the problem. But it was not helpful. Dont know if this is only a problem with my browser but anyhow I have included a screenshot so that you get an idea about how I see your post.