Factoring trinomials

[Cabages]

I have a big math test Thursday, and all of my math assignments are due tommorow. I need to do about 4 of them, but everything builds off factoring and these trinomials, and I just dont get them!

Could anyone explain how to do this problem, and then hopefully I will catch on to the rest?

10x² + 3x - 18



Thanks. This really gets me out of a bind.

Edit: I went to some tutoring today. The guy said to get factors of the first and last number, 10 and 18, and then times them or add them together or something.

I kinda see how you could get 5x - 6)(2x + 3) out of doing it that way, but I dont know really how to get to that point. Is there any other way to find out?

BTW I have no TI-89

 

[sash1]

(5x - 6)(2x + 3)

edit: oops I think I got the signs wrong

 

[DrPizza]

would it help if I referred to factoring as "un-FOILing"?

 

[txrandom]

Put this into your Ti-89:

solve(10x^2+3x-18=0,x)

 

[txrandom]

It's (5x-6)(2x+3) btw.

 

[rgwalt]

Do you know the quadratic formula? If so, then use it.

R

 

[Toastedlightly]

there is some rule... Ax^2 + bx +c = 0

AC or something.

 

[DrPizza]

Originally posted by: sash1
(5x - 6)(2x + 3)

Gee, nice explanation. Congrats on just giving him the answer.


cabages, because it's -18, one of the binomials is going to have a (-) and the other is going to have a (+).

Your possibilities for the first term are 1x, 2x, 5x, and 10x (all the factors of 10x^2)
Your possibilities for the second term are 1,2,3,6,9,18 (all the factors of 18)

So, your possibilities are:
(x+1)(10x-18)
(x+2)(10x-9)
(x+3)(10x-6)
(x+6)(10x-3)
(x+9)(10x-2)
(x+18)(10x-1)

(2x+1)(5x-18)
(2x+2)(5x-9)
(2x+3)(5x-6)
(2x+6)(5x-3)
(2x+9)(5x-2)
(2x+18)(5x-1)

(5x+1)(2x-18)
(5x+2)(2x-9)
(5x+3)(2x-6)
(5x+6)(2x-3)
(5x+9)(2x-2)
(5x+18)(2x-1)

(10x+1)(x-18)
(10x+2)(x-9)
(10x+3)(x-6)
(10x+6)(x-3)
(10x+9)(x-2)
(10x+18)(x-1)

To see which one works, simply start guessing and checking these by using FOIL. It's a pain in the neck though, so here are some hints:

Since you can't factor a GCF of 2 at the beginning, factors such as (10x+2) won't happen.

 

[Skunkwourk]

I suck at math but maybe this helps.
I can say one thing you should look at is the 10 and the 18 because you know that the numbers your answer will have to equal those two when multiplied together.

Edit: Dr Pizza is smarter than I am. Im terrible at explaining math.

 

[RapidSnail]

Just curios, what grade are you in?

 

[thegimp03]

Man, looking at this stuff brings me back to Pre-Algebra. I've forgotten almost all of it.

 

[Cabages]

Originally posted by: rgwalt
Do you know the quadratic formula? If so, then use it.

R

I had a math teacher try to teach me it that way last year, hes a nut though.

So I plug in the numbers to the corresponding positions in the quadratic formula, and the two resulting answers go in these empty positions -> (x+ )(x+ )?

 

[Mrvile]

This should be between 8th and 10th grade.

 

[DrPizza]

Originally posted by: txrandom
Put this into your Ti-89:

solve(10x^2+3x-18=0,x)

exactly why I presume that most people who use TI-89's don't know math; they only know how to use their calculator. I've seen kids come out of high school who supposedly had an A average in math who failed calc I in college because they didn't understand enough math and the mean prof refused to let them use a calculator.

 

[Cabages]

Originally posted by: RapidSnail
Just curios, what grade are you in?

Im a senior, skipped pre-calculus to this class, which is CE Algebra, basically college algebra.

My teacher last year though couldnt teach, and you got half your grade from your notebook and it was way to easy to cheat on tests.

 

[DrPizza]

I should also point out that there are algorithms for picking the correct coefficients rather quickly; but it's most useful to understand the guess and check method first. There are some other hints for the guess and check method though:

You don't have to "FOIL" each one; you know the "F" and the "L" parts are going to check, so just check the "OI" part.
If you get the right number but wrong sign on the "OI" part, just change both of the signs in your factors.
i.e. (5x + 6)(2x - 3) the OI part gives -15x +12x = -3x. You wanted +3x, so simply change it to:
(5x - 6)(2x + 3)

If you know the quadratic formula, and you set your trinomial equal to zero, when you use the formula and solve for x, you get roots of 6/5 and -3/2. From this, you can figure out what the factors were supposed to be.

 

[Howard]

Damn it, I used to know a pretty easy method.

 

[Oscar1613]

Originally posted by: Cabages

Originally posted by: rgwalt
Do you know the quadratic formula? If so, then use it.

R

I had a math teacher try to teach me it that way last year, hes a nut though.

So I plug in the numbers to the corresponding positions in the quadratic formula, and the two resulting answers go in these empty positions -> (x+ )(x+ )?

yes

x = (-b +/- sqrt(b^2 - 4*a*c))/(2*a)

where a*x^2 + b*x + c = 0

edit: actually the result would be (x - )(x - )

 

[DrPizza]

Also, here's a classic: 80x^2-241x+48

The pairs that multiply to 80 are 1 and 80, 2 and 40, 4 and 20, 5 and 16, 8 and 10.
The pairs that multiply to 48 are 1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8.

You end up with 100 possibilities (when you consider the signs)

However, consider (2x+?)(40x+??) When you Foil, the 2x*?? and ?*40x will both give you an even number. When you combine both even numbers, you will have another even number. Thus, it's impossible to have an even pair. Thus, you can eliminate 2 and 40, 4 and 20, 8 and 10. Likewise, you don't have to consider 2 and 24, 4 and 12, 6 and 8.

This leave 1 and 80 OR 5 and 16 to use for the x's, and it leaves 1 and 48 OR 3 and 16 for the constants.

I'm going to guess that it's the 5x and 16x, and 3 and 16.
You CAN'T have (16x+16), because then there would be a GCF
So, the only combinations possible are (16x+3)(5x-16) OR (16x-3)(5x+16).
The first FOIL gives -241x for the middle term, thus the second factorization is correct.

This trick only works when the middle term is an odd number.

edit in bold; there are 100 possibilities; I thought I had changed that when I added in "when you consider the signs"; it was 50 possibilities, plus you have to consider the signs +then- or -then+

 

[sash1]

Originally posted by: DrPizza

Originally posted by: sash1
(5x - 6)(2x + 3)

Gee, nice explanation. Congrats on just giving him the answer.

hey, dude, thanks! i try my best. not all of us have a doctorate you know

 

[RapidSnail]

Originally posted by: Cabages

Originally posted by: RapidSnail
Just curios, what grade are you in?

Im a senior, skipped pre-calculus to this class, which is CE Algebra, basically college algebra.

My teacher last year though couldnt teach, and you got half your grade from your notebook and it was way to easy to cheat on tests.

:thumbsup:

 

[RapidSnail]

Originally posted by: DrPizza
Also, here's a classic: 80x^2-241x+48

The pairs that multiply to 80 are 1 and 80, 2 and 40, 4 and 20, 5 and 16, 8 and 10.
The pairs that multiply to 48 are 1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8.

You end up with 50 possibilities (when you consider the signs)

However, consider (2x+?)(40x+??) When you Foil, the 2x*?? and ?*40x will both give you an even number. When you combine both even numbers, you will have another even number. Thus, it's impossible to have an even pair. Thus, you can eliminate 2 and 40, 4 and 20, 8 and 10. Likewise, you don't have to consider 2 and 24, 4 and 12, 6 and 8.

This leave 1 and 80 OR 5 and 16 to use for the x's, and it leaves 1 and 48 OR 3 and 16 for the constants.

I'm going to guess that it's the 5x and 16x, and 3 and 16.
You CAN'T have (16x+16), because then there would be a GCF
So, the only combinations possible are (16x+3)(5x-16) OR (16x-3)(5x+16).
The first FOIL gives -241x for the middle term, thus the second factorization is correct.

This trick only works when the middle term is an odd number.

Good stuff Pizza.

 

[DrPizza]

Also, here are straightforward steps for factoring a trinomial when the coefficient on the x^2 term is 1:

x^2 (sign1) bx (sign 2) c (consider b and c to be positive numbers; sign simply meaning the sign in front of them.
Work from right to left. List the factors of c, that (sign 2) up to b, put (sign 1) on the bigger number.

example:

x^2+13x-48
list the factors of 48: 1,48; 2,24; 3,16; 4,12; 6,8
which ones subtract to give you 13? Ans: 3 and 16
(x 3)(x 16)
Now, the sign in front of the 13 goes no the bigger number.
(x 3)(x+16)
The only step that really takes thought is putting in the last sign; since it was negative 48 that we factored, we know that the signs must be opposite signs; therefore it's - in front of the 3.
(x-3)(x+16)

 

[Cabages]

I like this Quadratic Formula thing.

Here is another problem from my assignment:

20k² + 47k + 24

I plugged it into the quadratic formula and got -32 and -15

I tried plugging them in like this (x-32)(x-15)

I FOILed it out and got x²-47x+480 so the middle one is right (except for the - sign, but the end one isnt.

And I honestly dont know what is going on here: "where a*x^2 + b*x + c = 0"

I can tell I missed something from the past posts.

I just got an email back from my teacher, and everything isnt due until THURSDAY. I am going to go to a private tutor at the Sylvan Center tommorow, I guess, because I still dont know the rule of squareds or cubes.

Thanks for all the help!

 

[chuckywang]

I used the quadratic formual as well and got -32/20 and -15/20.