Question

factor 36-16m+m^2

Answer 1

Hi sammy

Answer 2

hi

Answer 3

How are you doing ?

Answer 4

fine.How about you

Answer 5

I am fine, thanks for asking :)

Answer 6

36 -16m +m^2 is that your question?

Answer 7

m^2-16m+36=0
ax^2+bx+c=0
Let's find the discriminant
D=b^2-4ac
D=256-4*36
D=112
so the roots will be irrational
\[x=(-b \pm \sqrt D)/2a\]
\[m=(16 \pm \sqrt 112)/2\]
\[m=(8 \pm 2 \sqrt7)\]
so the factors are
\[(m-(8+2 \sqrt 7))(m-(8-2 \sqrt7))\]

Answer 8

OK, so sammy, what you want to do here is start by listing the factors of 36. What are they?

Answer 9

yes

Answer 10

so we can't do this question using polynomial equation.Right

Answer 11

do you mean quadratic formula?

Answer 12

There is much easier way
List factors of 36 like Laura was suggesting

Answer 13

the factor is (m-(8 +2*sqrt(7))(m-(8-2*sqrt(7))

Answer 14

\[36=6(6)=2(8)\]
and if you can't find two factors that add up to be -16
then it isn't factorable over the integers

Answer 15

you can use the quadratic formula to factor it over the complex numbers

Answer 16

I know quadratic formula.but my teacher ask me to factor it

Answer 17

yes but you can use that formula to factor it

Answer 18

\[m=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
=>
\[(m-\frac{-b + \sqrt{b^2-4ac}}{2a})(m-\frac{-b - \sqrt{b^2-4ac}}{2a})\]

Answer 19

those are the factors of the polynomial given

Answer 20

we didnt learn the quadratic formula in class.

Answer 21

ok then I will assume that you are suppose to say this not factorable over the integers

Answer 22

but i know how do it.yes that's what i want to do

Answer 23

Thank you so much.

Answer 24

i will write not factorable

Answer 25

You can find that:
\[36=18*2\]
and
\[2-18=-16\]
then:
\[36-16m+m^{2}=(2-m)(18+m)\]

Answer 26

so you are saying -m*m =m^2?

Answer 27

-m*m does not equal m^2

Answer 28

-m^2

Answer 29

you're right, I'm wrong

Answer 30

lol

Answer 31

-3(-13)=39

Answer 32

m^2-16m+36 is not factorable over the integers

Answer 33

There are not two factors of 36 that have product 36 and sum -16.

Answer 34

This implies the polynomial is not factorable over the integers.

Answer 35

However we can factor any quadratic over the complex which was done above .

Answer 36

exactly lol

Answer 37

nasty square roots and fractions

Answer 38

cofeee?

Answer 39

coffee?

Answer 40

Good discussion! :)
Goodnight guys!

Answer 41

this has to be the longest thread on factoring a quadratic. :)

Answer 42

lol

Answer 43

i think so

Answer 44

unless you consider the proof i made to factor any quadratic over the complex without using completing the square, or quadratic formula

Answer 45

I wouldn't consider that...that was for a general quadratic...this one is specific

Answer 46

zarkon i missed the human you

Answer 47

I can only be in human form for short periods of time

Answer 48

i understand

Answer 49

We can take it as:
\[m^{2}-16m+36 = m^{2}-16m+64-28\]
then you got a perfect square:
\[(m^{2}-16m+64)-28 = (m-8)^{2}-28\]
and you can work this as square difference:
\[(m-8+\sqrt{28})(m-8-\sqrt{28})\]

Answer 50

that is beautiful

Answer 51

thank you!

Answer 52

Ok great work and nice difference of square ruizgeorge :)
night!