Question

Anyone know an easy way to factor quadratic equations like these without graphing or quadratic formula?

Answer 1

\[14x^2+41x+15=0\]

Answer 2

@mathmate

Answer 3

@satellite73

Answer 4

Yeah, there is a method where you split the middle term.

Answer 5

@Seratul
Do you know how to factor integers? Like 24=2*2*2*3?

Answer 6

Yes.

Answer 7

Good!
The method work like the following:
Let's say we have
15x^2+19x+6 and we'd like to factorize (without using quadratic formula).

Answer 8

...where a=15, b=19, c=6
Then we have ac=15*6=90
And we look for two integers, m & n such that
m*n=90, m+n=19
so far so good?

Answer 9

So far so good :)

Answer 10

Can you find the two numbers m and n?

Answer 11

Oh, 10 and 9.

Answer 12

Good, now, let me figure out the next step! lol

Answer 13

Lmao. Take your time. I appreciate you helping me btw.

Answer 14

Then 9x and 10x would represent what you need to add together to get 19x.

Answer 15

Alright.

Answer 16

In this case, it is relatively easy, because 9=3*3, and 10=5*2

Answer 17

And 5 times 3 is 15?

Answer 18

yes, so it would be (5x + )(3x+ )
which you can fill easily.

Answer 19

How would I fill it?

Answer 20

(5x + p )(3x+ q )
You need p*3x=9x, and 5x+q=10x

Answer 21

Why would q have a variable if we are trying to get a whole number of 6?

Answer 22

Well, p*q \(should\) be 6 from what we have already done at the beginning.
You can solve p and find q from pq=6 also.

Answer 23

Oh. I guess I can just do trail and error, so would it be (5x+3)(3x+2)=0?

Answer 24

Exactly!

Answer 25

Thank you! Is it alright if I try to do 14x^2+41x+15=0 and just write my steps here?

Answer 26

It's just I think there is an even easier way with less guessing, but please do and work on this one, using the same steps.

Answer 27

So ac=14*15=210, and b=41
Did you find m and n?

Answer 28

Okay.
14*15=210
m*n=210
m+n=41
21+20=41
21=7,3
20=5,2,2
14= 7*2
So we can write (7x+q)(2x+p)=0

Answer 29

Oh, sorry. I went a little too quick.

Answer 30

Does this seem correct so far?

Answer 31

You need to first find m and n.
Your first try was 21+20, but the product is not 210.
You are better off trying the correct product (210) first, and I'll show you how to get closer to the sum of 41.

Answer 32

Can you give me any pair with m*n=210, don't worry about the sum.

Answer 33

Oh crap.
Alright, 21 and 10 should work.

Answer 34

Good, 21*10=210, first step satisfied. But 21+10=31....too small, right?

Answer 35

Unfortunately to small :(

Answer 36

When the sum is too small, you \(stretch\) the range, get the big number bigger, and try a smaller than the small.

Answer 37

Any tries?

Answer 38

I was gonna say 70 and 3, but that stretches a little too much.

Answer 39

So you know that 35<m<70, and 3<n<10.
Any more candidates?

Answer 40

5 and 42. Closer...

Answer 41

*21<m<70 lol

Answer 42

So you know which way to go?

Answer 43

Oh lol... Hmm, 35 and 6 :D

Answer 44

Exactly! Not too bad, right?

Answer 45

Next step.

Answer 46

It seems okay so far.

Answer 47

Okay, we break down 35 and 6.
35 - 7,5
6 - 3,2

Answer 48

I'll show you an easier way, the method came back to me.
Write (14x+35)(px+q)

Answer 49

May I ask where those numbers came from?

Answer 50

So that we know
14 comes from a, 35 comes from m.
(it could be 14 and n, but we'll try that later)

Answer 51

We don't need to do the p, q, actually.
We'll just write (14x+35).
Do you see a common factor?

Answer 52

7

Answer 53

Exactly, since the given expression does not have a common factor, so the 7 does not belong to the first factor (therefore the second).
Now we have
(2x+5)(7x+?)
right?

Answer 54

Ah, yes.

Answer 55

Can you finish it, before we try (14x+n)

Answer 56

It has to equal 15 so 3?

Answer 57

(2x+5)(7x+3)=0

Answer 58

Exactly, so we just factorized it to be (2x+5)(7x+3), check (especially the first few times using this method).
2x*7x=14x^2, 2x(3)+(5)(7x)=6x+35x=41x, 5*3=15, yay!

Answer 59

Now, what if we did
(14x+6)
can you do the rest?

Answer 60

Okay, so they have 2 in common.
So (7x+3)(2x+5)
Oh, its the same thing.

Answer 61

So this shows that you can just choose any of m or n to start, and get the same result.
Is everything clear?

Answer 62

Yes, thank god. Makes it so much clearer. Thank you very much. Do you have time for one more though? I want to try with a negative number.

Answer 63

Sure, let's do it!

Answer 64

:D
45x^2-74x-55

Answer 65

But remember to take care of the sign for m and n!

Answer 66

I want to see you factor it! :)

Answer 67

Do it step by step, don't jump!

Answer 68

Okay. Hopefully I don't mess up.
45*-55= -2475

Answer 69

Excellent.
Next step would be...

Answer 70

We need to figure out m and n

Answer 71

Good! Keep going!

Answer 72

Remember, try any product that gives -2475 and go from there.
Also, since the sum is -74, you expect the larger number to be negative.

Answer 73

m*n=-2475
m+n=-74
25-99=-74

Answer 74

Excellent! Keep the momentum there! Go, go ,go!

Answer 75

(45x+25x)
= (9x+5)(5x+q)

Answer 76

Now the last little step! Go, go, go!! You're almost there!!!!

Answer 77

5q=-55
q=-11
(9x+5)(5x-11)=0????

Answer 78

Yep, but do a FOIL anyway!

Answer 79

OKAY!!
9x*5x-99x+25x-55
45x^2-74x-55=0!!!! :DDDD

Answer 80

Well done! Cheer leaders dancing! :)

Answer 81

Lmao.
Thanks for sticking with me for the past hour lol. I extremely appreciate it. I was never so happy to answer the problem!

Answer 82

You sir are amazing!

Answer 83

No problem! I suggest you write up the method for future reference, or perhaps bookmark this post as well! If I forget (again), I'll ask you! lol

Answer 84

Of course. I'll probably be using this for the rest of algebra 2/trig.

Answer 85

Have a good day or night :)

Answer 86

Good! Talk to you another time! :)