How to factor large trinomials like 12x^2+47x+40

Question

I know the basics of factoring, and I know the add-to multiply-to method, and I know how to guess and check. I figured put the answer using a trinomial factoring calc online, but I'd like to know a method I could use for future reference. I needs halp. Thanks!

Answer 1

12x²+47x+40

 

multiply 12*40 using their prime factors

12*40=2x2x3x2x2x2x5

now try to group the factors such that the products of the factors add or subtract to give the middle term of 47

(in this case add because all terms are positive)

another thing to try is to express the middle term, in this case 47, as a sum (or a difference if you have negative signs) and then try to make these numbers with your factors

for example, 47=40+7 but we can't make the 7 by multiplying our factors

how about 47=32+15 ? look at the factors...

2x2x3x2x2x2x5=(2x2x2x2x2)x(3x5)=32x15

12=4x3 and 40=8x5 and the 8 multiplies the 4 while the 5 multiplies the 3

(4x   )(3x   )=(4x+5)(3x+8)

another trick to use...

when using guess and check for factoring 12 and 40, some possibilities can be eliminated

first make sure the polynomial is in standard form ax²+bx+c

a, b, and c can be even or odd

there are only 8 ways that this can happen: O for odd and E for even

OOO

EEE

EEO

OEE

EOE

OOE

EOO

OEO

in your example you have EOE

when this happens "a" has to be factored into ExO and "c" has to be factored into ExO as well

12=2x6 can't be used because they are both even

12=1x12 or 3x4 can be used

40=1x40 and 5x8 can be used but 2x20 and 4x10 cannot be used because both are even

this limits your possibilities

another example...

20x²-48x+27  is the EEO case whereby "a" must be factored into ExE

20 can only be factored into 2x10 because 1x20 and 4x5 have an odd number in each case and cannot be ued

2x10 and 3x9 and 2x9=18, 3x10=30, and 30+18=48, the middle term

(2x-3)(10x-9)

hints for the other cases...

OOO is irreducible over the set of integers (the only thing you can do is look for a common coefficient factor)

EEE, factor out a common coefficient number

EEO, already gave the hint

OEE, "c" must be factored as ExE

EOE, already gave hint

OOE, "c" must be factored as ExO

EOO, "a" must be factored as ExO

OEO, no trick for this case because a and c are both odd and have to be factored into OxO

try this example...

9x²+44x-60

OEE case where 60 has to be ExE

1x60, 3x20, 4x15, and 5x12 cannot be used

2x30 or 6x10 can be used and 9=1x9 or 3x3

9x60=3x3x2x2x3x5

take it from here..

Hope this helps.

Post more problems if you like and I'll explain them for you.

 

 

 

Answer 2

For a random trinomial your most reliable method is setting it equal to zero and using the quadratic formula. Before using it check a part of that formula called the discriminant. This will tell you a lot, such as if your trinomial even can be factored.

Answer 3

Hayna,

 

When factoring trinomials of the form ax² + bx + c, the first test you run is common factors.  Since there is none in this case, you should try factoring it by product of 2 binomials, or by the quadratic formula.

 

I wrote a simple program in my TI-83, which allows me to solve quadratic equations by inputting the values of a, b and c.  I run it and it neatly returned the following solutions: x₁ = -5/4  x₂ = -8/3 from these, I could figured it out the corresponding product of two binomials: (4x + 5)(3x + 8).  This is called working backwards and it is used when you know the answer, which was my case.  

 

Now, if you don't know the answer, then you may solve this problem by applying the quadratic formula.

 

Solving this equation by product of 2 binomials would be a good exercise if you would like to practice FOIL, given the 3 options you have to factor 12x² : (12x)(x), (4x)(3x) and (6x)(2x). 

 

It is a good exercise anyway you approach it.