#### jramirez23

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I am especially having problems with the following equation:

4x[sup:krjvv7l5]2[/sup:krjvv7l5] + 27x + 35.

I am extremely clueless; I have even tried some websites and I still need help.

- Thread starter jramirez23
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I am especially having problems with the following equation:

4x[sup:krjvv7l5]2[/sup:krjvv7l5] + 27x + 35.

I am extremely clueless; I have even tried some websites and I still need help.

The possible factors for the first term are either 2x*2x or x*4x. So we have as possibilities...

(2x + ___)(2x + ___) or (x + ___)(4x + ___) or (4x + ___)(x + ___)

The possible factors for the last term are 1 and 35 or -1 and -35 or 5 and 7 or -5 and -7. So, start plugging in various possibilities until you get the correct combination that when the binomials are multiplied together you get the original expression. For instance, I'll try my first listed possibility.

(2x + 1)(2x + 35) or (x + 1)(4x + 35) or (4x + 1)(x + 35)

(2x + 1)(2x + 35) = 4x[sup:2b51imu2]2[/sup:2b51imu2] + 77x + 35 <--- Nope!

(x + 1)(4x + 35) = 4x[sup:2b51imu2]2[/sup:2b51imu2] + 39x + 35 <--- Nope!

(4x + 1)(x + 35) = 4x[sup:2b51imu2]2[/sup:2b51imu2] + 141x + 35 <--- Nope!

Now, you can continue.

Here's a method I always use for factoring trinomials of the form ax[sup:33tqulyc]2[/sup:33tqulyc] + bx + c where "a" is something other than 1.jramirez23 said:I am having a very difficult time learning how to factor ax[sup:33tqulyc]2[/sup:33tqulyc] + bx + c.

I am especially having problems with the following equation:

4x[sup:33tqulyc]2[/sup:33tqulyc] + 27x + 35.

I am extremely clueless; I have even tried some websites and I still need help.

Multiply "a" * "c". In your example, multiply 4 * 35, to get 140.

Now, look for two numbers whose product is 140, and whose SUM is "b," or 27 in your example.

7 and 20 will work, since 7*20 = 140, and 7 + 20 = 27.

Rewrite the middle term as 7x + 20x:

4x[sup:33tqulyc]2[/sup:33tqulyc] + 7x + 20x + 35

Factor by grouping. Remove a common factor of x from the first two terms, and a common factor of 5 from the last two terms:

x(4x + 7) + 5(4x + 7)

Remove the common factor of (4x + 7):

(4x + 7)(x + 5)

There's the factorization you're looking for. This may sound complicated, but it really isn't. It takes MUCH longer to explain it than it does to

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That's not trial and error?Mrspi said:Now, look for two numbers whose product is 140, and whose SUM is "b," or 27 in your example.

7 and 20 will work, since 7*20 = 140, and 7 + 20 = 27.

It isn't trial-and-error in the same way as Loren's method (which, by the way, is the one I learned in high school myself, and used for years).tkhunny said:That's not trial and error?Mrspi said:Now, look for two numbers whose product is 140, and whose SUM is "b," or 27 in your example.

7 and 20 will work, since 7*20 = 140, and 7 + 20 = 27.

Just my opinion, of course. The opinions of others may vary!

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Thanks for helping! I found a good way that I think best suits me.

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The most direct way is to use the quadratic equation (which is derived by completing the square.jramirez23 said:Thanks for helping! I found a good way that I think best suits me.

\(\displaystyle A\cdot x^2\, + \,B\cdot x \,+ \,C\, = \, A\cdot\, (x\, - \frac{-B\,+\,\sqrt{B^2\,-\,4\cdot A \cdot C}}{2\cdot A})\, \cdot \, (x\, - \frac{-B\,-\,\sqrt{B^2\,-\,4\cdot A \cdot C}}{2\cdot A})\)

Remember this formula - you will meet this "guy" many more time.