Question

@mathmate

Answer 1

Factor x^2+x-20

Answer 2

What you need to do to factor
ax^2+bx+c
is to find two numbers m, n so that
m*n = ac
m+n=b
If a=1, then life is a little easier, because
we do mn=c
m+n=b
Here
a=1, b=1,c=-20

Answer 3

There you go, can you write down the answer?

Answer 4

(x+5)(x-4) :)

Answer 5

Great! Well done.
Let try another one.

Answer 6

x^2+x-12

Answer 7

I like your diagram,makes life easier!

Answer 8

im not sure :(

Answer 9

First find factors of 12 that have a difference of 1.
From
1/12
2/6
3/4
...

Answer 10

Worry about the sign later.

Answer 11

(x+4)(x-3)

Answer 12

Bingo! You got it again!
Now try this
Factor 6x^2+7x-20

Answer 13

It will be a little harder, but you need to find m, n first.

Answer 14

O_O i dont like these!

Answer 15

Can you find m,n?
Try
1,120 (difference =119, no good)
2,60
3,40
4,30 (diff=26, still no good)
...

Answer 16

BTW, if c is negative, you look for the difference, if c is positive, you look for the sum.

Answer 17

there are no factors to this :/

Answer 18

to continue the list above,
5,24
6,20
8,15
10,12
...
do you find m,n

Answer 19

k, will continue later.

Answer 20

Nothing goes into those numbers T_T

Answer 21

If you don't see the numbers that way, you can always use the Quadratic Formula.

\(\sf\LARGE x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

where \(\sf\Large ax^2+bx+c\)

and then whatever you get as the value of x, lets say x=1, x=3

then we know that if it was factored it would become:
x = 1
-1 -1
x - 1 = 0

x = 3
-3 -3
x - 3 = 0

and we know that

(x-1)(x-3) was the factored form of the equation.

Answer 22

8*15=120, 8-15=-7,
so now we have to work on the sign.
Since b=+7, we will have to put

to get 15-8=+7.

Now that we have got m,n, then all we need to do is to split the x term into two using 15x-8x instead of 7x:
\(6x^2+15x-8x-20\) and factorize by grouping
=\(6x^2+15x-8x-20\) now extract the - sign in the group -8x - 20
=\( (6x^2+15x) - (8x+20)\) now extract the common factors 3x and -4 from each group
=\( 3x(2x+5) - 4(2x+5)\) now extract the common factor 2x+5
=\( (3x-4)(2x+5)\) and we're done

Answer 23

got it can we do another one like that? :)

Answer 24

Sure! coming up.
Factor \(30x^2+23x−14\)

Answer 25

O_O

Answer 26

Now
m n difference
1 420 419 (too big, will skip a few)
10 42 32
12 35 23
14 30 16
...
Which one would you choose for m,n?

Answer 27

With some huge equations like that, if it is too timely to list all the factors and find its roots, the quadratic equation is always an option :)

Answer 28

@Thesmarterone i need to learn this way though XD
Mm its 12 and 35 right?

Answer 29

@TheSmartOne practice makes perfect. If we can do 420, 36 would be a piece of cake.

Answer 30

exactly! so can you complete the diagram?

Answer 31

next class bbl

Answer 32

Sometimes you can try to check if the equation follows one of these patterns:

\(\sf (a+b)^2 = a^2+2ab+b^2\)

\(\sf (a-b)^2 = a^2-2ab+b^2\)

And there are many ways to solve one equation, you want to find the way that is easiest for you unless your teacher wants you to solve it one particular way :P

Answer 33

@TheSmartOne very true.
I was trying to help her learn one method. Once she gets it, then we introduce shorter ones, like the quad. formula, standard forms. If we give her 4 methods to start, it is easy to get confused. That's also why I chose one that does not belong to the perfect squares, difference of two squares, etc. They will come later.
The grouping one is the one that is most difficult for most students.

Answer 34

@mathmate That is true, and I believe you are doing a great job teaching her! :D

Answer 35

Thank you!
It takes more than one... she is a great and focused student!

Answer 36

Ok :) ready to learn >_< ok i know the quadratic formula way but this is what i am currently doing :P ok so can we start with a fresh one ? im confused now :(

Answer 37

@mathmate

Answer 38

where were we?
Factor 30x^2+23x−14.

Answer 39

and:

Answer 40

and
Now
m n difference
1 420 419 (too big, will skip a few)
10 42 32
12 35 23
14 30 16
...
so 12 and 35 are our choices of m,n needed to complete the circle.

Answer 41

Ok so it would be like this :
\[(30x^2+35x)(12x-14)\]
for the first one we divide by 5 we get
5x(6x+7)
for the 2nd one we divide by 2 we get...
2(6x+7)
final answer
(6x+7)(5x+2)

Answer 42

wait i messed up >_<

Answer 43

its -12...