# simplifying help

• February 25th 2011, 05:19 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
• February 25th 2011, 05:29 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,

$\dfrac{3}{4x²+5x+1}$

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

Decomposing this into partial fractions;

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

Hope you can continue with the integration.
• February 25th 2011, 08:06 PM
maybnxtseasn
thanks but im curious as to the steps u use to simplify this?
• February 25th 2011, 08:15 PM
NOX Andrew
Are you asking how he knew $4x^2 + 5x + 1 = (4x + 1)(x + 1)$? If so, one method is as follows:

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

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

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

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

Now, factor $(4x + 1)$: $(4x + 1)(x + 1)$.
• February 26th 2011, 04:02 AM
HallsofIvy
Another way to do that (but much more complicated that just factoring directly) is to solve the equation $4x^2+ 5x+ 1= 0$- 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 $x= -\frac{1}{4}$ and $x= 1$

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!
• February 26th 2011, 06:06 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 $4x^2+ 5x+ 1= 0$- 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 $x= -\frac{1}{4}$ and $x= 1$

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.