HELP! just started my first college math class and forgot how to FACTOR!?

Question

My last math class was 2 years ago in highschool and now i have forgotten all of it :( we started to factor today and i just could not remember how to do it. So if you can help me out and we be very greatful!

5x^2-x-6
( )( ) so i need to find a number that multiplies to get -6 and adds to get -x???? can you show me how to do this step by step?

7x^2-2x-5

Answer 1

Q. How to factor quadratics with lead coefficient other than zero
I got a really cool trick for this, but it's hard to write it here, limitations with the graphics and so forth. Let me try and write it out and get back with you...

Make a 4 by four box with positions for four numbers in it, like this:
|,,,,,,,, ,,,,,,,,,,|
|........ ..........|

In the top row, put the factors of the first term, that is, factors of 5x^2 the only possiblities are 1 and 5, so put those.
| 5... 1|
| ... ... |
In the bottom row, put the factors of the last term, that is, -6 Here's where you have to work in the signs. Since -6 is negative, one has to be negative and the other positive. You could try 2 and 3 but these won't work as I'm about to show. I'll cut to the chase and use -6 and 1

|5.....1|
|-6....1|

Ok, now why did I use -6 and 1 rather than 2 and 3? And why did I put the negative where I put it? Here's how I knew. I've done this before so I anticipated the next step which is ...
Multiply down the diagonals, add the results, and hopefully you shouold get the middle number. here, (5x1) + (-6x1) = 5 + (-6) = -1, which is the middle number (it's understood in front of the x) If you'd tried various arrangements of 2 and 3 you simply wouldn't have gotten that result.
Finally, read the coefficients fom the columns
(5x - 6)(x+1)
Now, suppose you had put the -6 and 1 in the bottom row in the opposite order? Then they wouldn't add up to -1 (try it!) and you'd know that you need to rearrange them. that's why I recommend doing this with a pencil rather than a pen. So there's still a certain amount of trial and error involved but not near as much as before.
Try the other one the same way.
Since 7 is prime, you can only go with 7 and 1. Put these in the top row.
|7.....1|
|.........|
Also 5 is prime so there's only so much you can do with that. But note that you're aiming for a small middle coeffieicnet, -2. So you want to arrange the numbers so that the products along the diagonals to be very close to each other, and opposite in sign, so as to add up to -2, That is, you wouldn't put the 5 opposite the 7 on the diagonal, because that would give a product of 35. Instead, put the 5 directly underneath the 7 so your products along the diagonals will have a shot at adding up to -2....
|7....1|
|5...-1|
(7x-1)(1x5) = -2
Note that if you'd put the negative on the 5 rather than on the 1 you would have come up with +2 as the product along the diagonals. Again, use pencil, not pen. (It's bad luck to do math with a pen anyhow, as I tell my students.)
Now just read the factors from the columns
(7x + 5)(x-1)

Now when your lead coefficient is 1, that's a lot easier, you just need to come up with two numbers that multiply together to make the last number and add together to make the middle number. With a lead coeffieicnt other than 1, however, most books tell you to do a tedious process of trial and error. And this puts people off, becuase they think math should be more systematic, it should not rely on trial and error. And they're right! That's why I like this trick. There's still a certain amount of trial and error, but a lot less, and it's far more systematic.
Good luck!

Answer 2

I sort of do what She-Nerd said. But I do it more in my head...

Just find the factors of the coefficient of the x^2, and the factors of the last term. Then try multiplying and adding the various terms together until you find the ones that will give you the coefficient of the x. I'm not quite sure how to explain that, but if you've done it before, hopefully you'll know what I'm talking about.

Anyway, I got (5x-6)(x+1) and (7x+5)(x-1). So if you just play around with the numbers to refresh your mind, it should come back to you.

Answer 3

FACTORISING
1) Take out the biggest number that goes into all the terms.
2) Take each letter in turn and take out the highest power (e.g. x,x^2 etc) that will go into EVERY term.
3) Open the brackets and fill in all the bits needed to reproduce each term.

e.g. factorise 15x^4y+20x^2y^3z-35^3yz^2

= 5x^2y(3x^2+4y^2z-7xz^2)

5 = biggest number that'll divide into 15,20 and 35
x^2 = highest powers of x and y that will go into all three terms
Z was not in all three terms so it can't come out as a common factor.

REMEMBER
1) The bits taken out and put at the front are the COMMON FACTORS
2) The bits inside the brackets are what's needed to get back to the original terms if you were to multiply the brackets out again.


1) Simplifying
5x2 + -1x + -6

Reorder the terms:
-6 + -1x + 5x2

Factor a trinomial.
(-1 + -1x)(6 + -5x)Final result:(-1 + -1x)(6 + -5x)


2) Simplifying
7x2 + -2x + -5

Reorder the terms:
-5 + -2x + 7x2

Factor a trinomial.
(-5 + -7x)(1 + -1x)Final result:(-5 + -7x)(1 + -1x)

Answer 4

5x^2 - x - 6 =5( x^2 - 1/5x - 6/5) =
5(x^2 - 1/5x + (1/10)^2 - (1/10)^2 - 6/5) =
5((x -1/10)^2 - (1/100 + 6/5)) =
5((x-1/10)^2 - (1/100 + 120/100))=
5((x-1/10)^2 - 121/100) =
5(x-1/10-11/10)(x - 1/10 + 11/10) =
5(x - 12/10)(x + 1) = (5x - 6)(x+1)

7x^2 - 2x - 5 =
7(x^2 - 2/7 x - 5/7) =
7(x^2 - 2/7 + (1/7)^2 - (1/7)^2 - 5/7) =
7((x-1/7)^2 - (1/49 + 35/49))=
7((x-1/7)^2 - 36/49) =
7(x-1/7 -6/7)(x-1/7+6/7) =
7(x-1)(x+5/7)= (x-1)(7x + 5)

Answer 5

and your in college ??



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