Factoring trinomials

[Syringer]

Here's a method that I learned in HS when the term "a" in ax^2 + bx + c is not 1, requiring more steps..

1) Rewrite your equation as x^2 + bx + a*c
2) factor this equation..which should be easy since "a" is 1 here, so that you get (x+s)(x+t)
3) Rewrite (x+s)(x+t) as (x+s/a)(x+t/a) where a is the original "a" term in the polynomial
4) Simpifly the equation if necessary
5) Now move the denominator in the 2nd part of each part of the (..) and (..) terms to the front, and you're done!

Example: 10x^2 + 3x - 18..so "a" is 10
1) x^2 + 3x - 180
2) (x+15)(x-12) <-- hardest part
3) (x+15/10)(x-12/10) <--- divide each by 10 since a is 10
4) (x+3/2)(x-6/5) <--- simplify
5) (2x+3)(5x-6) <--- move denominator to front

It'll take some getting used to, but after a few tries you should get used to it, and it makes such equations much easier to deal with.

 

[yosuke188]

Originally posted by: 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!

It should actually be (x+32)(x+15)=0 since the answers are -32 and -15.

You set each term equal to zero, so x+32=0 and x+15=0... x=-32, -15

 

[rgwalt]

Originally posted by: 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!

When I use the quadratic formula, I get x = -1.6 or -0.75.

Putting them into the factored form gives (x + 1.6)(x + 0.75)

Multiplied out this gives x^2 + 2.35 + 1.2

Multiplying this equation by 20 gives 20x^2 + 47x + 24

So the factored form can be written as 20(x + 1.6)(x + 0.75)

That form is completely correct in a mathematical sense, but they are probably after a factorization that gives all whole numbers. If that is the case, then you can write down all the whole number factors of 20 and figure out which combination will give you whole number when multiplied with 1.6 and 0.75.

For instance, the factorization of 20 gives 1,20 2,10, and 4,5. 1,20 does not work, nor does 2,10 for obvious reasons. This leaves us with 4,5. Try multiplying 1.6 by 4 and 0.75 by 5. Does this give whole numbers? No, then reverse it. Does multiplying 1.6 by 5 and 0.75 by 4 give whole numbers? Yes! So, what we have is the following factorization:

(5x + 8)(4x + 3)

So, tips: Make sure you apply your quadratic formula correctly! The answers will give you the factors of the polynomial x^2 + (b/a)x + (c/a)x To get the factorization of ax^2 + bx + x, multiply the factorization by a, and distribute factors of a to get a factorization with all whole numbers.

Hopefully this isn't too confusing. Good luck!

R