Question

factor completely

m^2+7m+10
y^2-13+06
x^2-8x+7
x^2=9x+18
r^2+4r+3
n^2-3n+2
x^2-2x+1
k^2-16k+18
x^2+3x-4
m^2-2m-8
d^2+6d-40
p^2-15p-54
k^2-8k-9
y^2-18y+45

Answer 1

Whoa! That's a lot of questions. Lol.

Answer 2

to factor those you must find their roots, that is for example for what x => x^2+3x-4=0.
So you must solve a lot of quadratic equations...

Answer 3

yeah i know

Answer 4

@abb0t

Answer 5

so you don't know how to solve quadratic equations? or you just someone else to do this for you?

Answer 6

I'm not going to do your homework for you. That won't help you if I do it for you, but i will guide you so you can work on it on your own.

\[ax^2+bx+c\]
What you want to find is basically two numbers when multiplied give you C, but, when you add those two numbers, you get "b".
Does that make sense?
I will do the first one for you:
\[m^2+7m+10\]
notice that it's in the form: \[ax^2+bx+c\]
where b = 7 and c = 10
find those two numbers when multiplied (5 x 2) = 10
but when added (5 + 2) = 7
As you ca see, those two numbers are 5 and 2.
So you factor it out: \[m^2+5m+2m+10 = m(m+5) + 2(m+5) = (m+2)(m+5)\]
You can check to make sure that those two numbers give you the original polynomial by doing the whole FOIL method thing if you are unsure.

Answer 7

k thank you

Answer 8

However, if there does come a problem where you cannot find two numbers when multiplied = c and added give you "b" then you must use the quadratic formula: \[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 9

and you are sure about this ?

Answer 10

Do your homework!

Answer 11

daddy*

Answer 12

Aha Wow