Question

can someone help me please
factor completely.
3x^2+21x+30
@563blackghost

Answer 1

show work pleasl

Answer 2

do you see that 3,21,30 all share a common factor?

Answer 3

yes

Answer 4

and what common factor is that?

Answer 5

3

Answer 6

very awesome
can you tell me what you get after dividing the following numbers 3,21, and 30 by 3?

Answer 7

1

Answer 8

3/3=1
21/3=?
30/3=?

Answer 9

21/3=7
30/3=10

Answer 10

so you should see that we have the following so far
\[3x^2+21x+30 \\ 3(1x^2+7x+10) \\ 3(x^2+7x+10)\]

(we are not done)

Answer 11

does this seem okay so far?

Answer 12

yes

Answer 13

okay now let's look at just the x^2+7x+10 for now

Answer 14

we will bring the 3 down later

Answer 15

so looking at \[x^2+7x+10\]
we need to come up with 2 numbers that multiply to be 10 while also adds up to be 7

Answer 16

if the coefficient of x^2 was something other that 1 say a
I would have said we need to come up with 2 numbers that multiply to be a*10 while adds up to be 7

that is if we had something like ax^2+7x+10
where a is given as some number

Answer 17

can you think of two numbers that multiply to be 10 while they also add up to be 7?

Answer 18

2 and 5

Answer 19

@freckles

Answer 20

right 2*5 is 10 while 2+5=7
so you can do this the short cut way since the coefficient of x^2 is 1

which is if you have x^2+bx+c
and you find two numbers m and n such that m*n=c while m+n=b
then you can say x^2+bx+c is the same as (x+m)(x+n)
since when multiplying this out you get x^2+mx+nx+mn
=x^2+(m+n)x+mn=x^2+bx+c


so all you need to do when you find m and n such that m*n=b while m+n=c
is write x^2+bx+c as (x+m)(x+n)


or if you want to do it the longer way the way that works for if you have something other than 1 as the coefficient of x^2
then you would do something like this
ax^2+bx+c
look for two numbers that multiply to be a*c and add up to b
say that numbers are m and n
so you replace bx with mx+nx
ax^2+mx+nx+c
and factor by grouping

Answer 21

if you don't feel like reading all of that right now then just read this part...
"
so all you need to do when you find m and n such that m*n=b while m+n=c
is write x^2+bx+c as (x+m)(x+n)
"

Answer 22

so using what I just said rewrite x^2+7x+10

Answer 23

okay i rewright

Answer 24

and remember your m and n are 2 an 5 correct?

Answer 25

yes

Answer 26

so you know you are going to write x^2+7x+10 as (x+ m )(x+ n )
where m and n are 2 and 5
it doesn't matter if you replace m with 5 or 2 just the vice versa for n

Answer 27

this is hard

this is i have so far
3x^2+21x+30
3(1x^2+7x+10)
3(x^2+7x+10)
x^2+7x+10
(x+m)(x+n)

Answer 28

you found your values for m and n already....

Answer 29

no thats the hard part

Answer 30

x^2+bx+c
in factored form is (x+m)(x+n)
if m*n is c and m+n is b

x^2+7x+10
in factored form is .....
since 2*5=10 and 2+5=7

what goes in place of the dots

Answer 31

X

Answer 32

no x^2+7x+10 is not X

Answer 33

you already found m and n
you just need to replace m and n

Answer 34

remember you said 2*5=10 and 2+5=7
and in my general thing I have above m*n=c and m+n=b
and my factored form for x^2+bx+c was (x+m)(x+n)

Answer 35

you found m and n to be 2 and 5

Answer 36

you just need to replace m and n now in the factored form for the quadratic I wrote above

Answer 37

1x^2

Answer 38

are you reading what I wrote?

Answer 39

yes

Answer 40

The factored form for x^2+bx+c is (x+m)(x+n)
where m*n=c and m+n=b

You have the factored form for x^2+7x+10 is .......
where 2*5=10 and 2+5=7


just replace the m and n in my factored form for the quadratic with your m and n which is 2 and 5
and you will have your factored form for your quadratic

Answer 41

so 2 and 5 is the replacement right

Answer 42

example
the factored form for x^2+5x+6 is (x+2)(x+3)
since 2*3=6 and 2+3=5

another example
the factored form for x^2-5x+6 is (x-2)(x-3)
since -2(-3)=6 and -2+(-3)=-5

Answer 43

another example
the factored form for x^2-x-6 is (x-3)(x+2)
since -3(2)=6 and -3+2=-1

Answer 44

in general
The factored form for x^2+bx+c is (x+m)(x+n)
if m*n=c and m+n=b

Answer 45

your problem:
the factored form for x^2+7x+10 is ....................
since 2*5=10 and 2+5=7

Answer 46

another example
the factored form for x^2+x-6 is (x+3)(x-2)
since 3(-2)=6 and 3+(-2)=1

another example
the factored form for x^2+6x+8 is (x+2)(x+4)
since 2(4)=8 and 2+4=6

Answer 47

so 8 and 6

Answer 48

wait no 4

Answer 49

what are you saying about my example?

Answer 50

7x and 10 is the replacement

Answer 51

I don't understand why you are having trouble with this...
the factored form for x^2+7x+10 is (x+2)(x+5)
since 2*5=10 and 2+5=7

Answer 52

so 2and 5 is the replacement for m and n right

Answer 53

yes this has been mentioned before :p

Answer 54

x^2+bx+c has formed form (x+m)(x+n)
if m*n=c and m+n=b

since we had 2*5=10 and 2+5=7 then m=2 and n=5
or vice versa m=5 and n=2
doesn't matter

Answer 55

anyways I have to go for now
I have mentioned several examples above of using the general thing I mentioned which was
"
The factored form for x^2+bx+c is (x+m)(x+n)
if m*n=c and m+n=b
"