Factoring... Yay... :)
First things first, do you want to.... explain this format? I don't do factoring like.....-that- lol
oh
its tic tac toe method
wat is that-
the fromula is ax^2 + bx + c
top left is a top middle is c top right is a*c
middle right is a factor of a*c that adds up with the other factor to equal b
@AZ @angle i've never done something called the 'tic-tac-toe method'
bottom right is the other factor
do you have to factor this way-
thats how i was tought
damn, they dipped
im open to more methods
sorry, had to go delete a spam post I've never learned this before but it seems similar to factoring by grouping
yes
ill make a template
so the arrow shows which way the rows multiply
a is in square 1-1
c square 1-2
a*c in square 1-3
now square 2-3 and 3-3 are factors of a*c that add up to b
once you put those there, square 2-1 * 2-2 = 2-3
same with row 3
then 3-1 * 2-1 = 1-1
as with column 2
Let me try an example and you tell me if I'm understanding your method correctly: If I had an equation: 2x^2 - x - 3 a = 2 b = -1 c = -3 Then my box would look like: |dw:1616800872750:dw|
so, instead of -6 and 1, you would do -3 and 2 cuz -3 + 2 = -1, b
Are you required to use this method? ;-;
no
good, let's try factoring by grouping instead :D
which problem do you want to try first?
any also, this would be ur problem
6. Factor 2x4−13x2−45 2x^4−13x^2−45 =(2x^2+5)(x+3)(x−3)
ohh I wasn't too far off, that would make sense
mhm, u did good
better than i did my first
kynxkatz, howd u get that?
Solved it?
yea
N just call me Kynx
That’s my first name
ight
Since katz is so eager to try out number 6, let's take a look at how that would look in one of your boxes 2x^4−13x^2−45 |dw:1616801365579:dw|
Imma go through the questions n solve them
Wait wht
correct Angle
When did i give direct answers
\(\color{#0cbb34}{\text{Originally Posted by}}\) @KynxKatz 6. Factor 2x4−13x2−45 2x^4−13x^2−45 =(2x^2+5)(x+3)(x−3) \(\color{#0cbb34}{\text{End of Quote}}\) that's a direct answer, mate
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Angle \(\color{#0cbb34}{\text{Originally Posted by}}\) @KynxKatz 6. Factor 2x4−13x2−45 2x^4−13x^2−45 =(2x^2+5)(x+3)(x−3) \(\color{#0cbb34}{\text{End of Quote}}\) that's a direct answer, mate \(\color{#0cbb34}{\text{End of Quote}}\) Oh it u, thought it was sumone else
so u would make 2x^4 - 13x^2 - 45 2u^2 - 13u - 45 | u=x^2
that allows u to make the table
I dont even have to strength to argue wit u, I’m just here to help ppl
7. Since both terms are perfect cubes, factor using the difference of cubes formula, a^3-b^3= = a^3-b^3=(a-b)(a^2+ab+b^2) where a=4x a=4x and b=1b=1 (4x−1)(16x^2+4x+1)
Srry that took so long i just got out of chemo like 1 hr ago so I’m just kinda tired
ight
Buh i can still continue if u want
yes pls
if u dont bmind
Mk
where a=2x a=2x and b=3 b=3 So (2x+3)(4x^2−6x+9) is the answer Use this formula if ur confused how i got the answer- (a−b)(a^2+ab+b^2) Since both terms are perfect cubes, factor using the difference of cubes formula, a^3−b^3
but they arent perfect cubes though
im confuzzled
8. Explanation: Step 1 : To factor 81x^4 −1 we can use difference of squares formula: I^2−II^2 =(I−II)(I+II) After putting I=9x^2 and II=1 , we have: 81x^4−1=(9x^2−1)(9x^2+1) Step 2 : To factor 9x^2−1 we can use difference of squares formula: After putting I=3x and II=1 , we have: 9x^2−1=(3x−1)(3x+1)
See if wht i wrote just now helps
i gtg now, thx tho for the help, i appreciate it very much
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