help me simplify this

[mystic brook]

idk where to start

[twilit lake]

Factorise the numerator

[mystic brook]

how tho

i tried

idk how to start

[twilit lake]

There are two ways

To solve this, either simply using the factoring method which is a very straightforward method or you can use the quadratic formula to find the zeros

Let's go with the factoring method because it's less tedious

[mystic brook]

could u show me this method

[twilit lake]

Yes

I'll show the quadratic formula too, so you know the difference

[mystic brook]

ok

[twilit lake]

let me just test the latex

$asad$

[meager scaffold]

せんく(It's Quantum)

[twilit lake]

great it works

factoring method:\
$x^2+2x-3$, find two numbers when multiplied with each other gives us -3 and and when added gives us 2. The best candidat for that is, 3 and -1. \
so our zero's are at point, -3 and 1. Which imples that we can rewrite our function to: (x+3)(x-1)

[meager scaffold]

せんく(It's Quantum)

[mystic brook]

so u did 2 -3

then added 3

?

[twilit lake]

quadratic formula method: \
This method goes to find the zero's of the functions i.e x values for which the function gives us zero. for that we apply the quadratic formula:\
for functions with form ax^2+bx+c \
gives the roots of the function \
substituting the coefficients we get similar solutions as one above.

[twilit lake]

I'm coming

$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

[meager scaffold]

せんく(It's Quantum)

[twilit lake]

quadratic formula method: \
This method goes to find the zero's of the functions i.e x values for which the function gives us zero. for that we apply the quadratic formula:\
for functions with form ax^2+bx+c \

[mystic brook]

yea i know that one

[mystic brook]

it gave me different answers

i got 1 and -3

from it

when i used that

[twilit lake]

but those are the zero's, x values when we substitute in we get zero
so (x-1) would be one zero, because we substitute in 1 we get 0

and (x+3), because when we substitute in -3 we get 0

so in the end we have (x-1)(x+3)

as our factored form

[mystic brook]

did i do something worng?

[twilit lake]

well you didn't get the wrong zero's or roots. U just used it wrong when changing it into factored form

[twilit lake]

$\frac{x^2+2x-3}{x-1}$ would now be $\frac{(x-1)(x+3)}{x-1}$

[meager scaffold]

せんく(It's Quantum)

[mystic brook]

how was i supposed to use it?

[twilit lake]

the solutions you get from the quadratic formula will should be zero when substituted in into the factors. for example
if your solution is 3 and 2, then it would be (x-3) and (x-2) in factored form for that function because when we replace x with 3 and 2 respectively we get 0, which is the point

so basically change the sign, i.e negative or positive for solutions, which is a way of looking at it.

[mystic brook]

omg

im not getting it

sorry

[twilit lake]

Tell where it's confusing and I'll try to clear it up

[mystic brook]

how 1 and -3

turned into

(x+3)(x-1)

[twilit lake]

1 and -3 are x values for when our function is 0.
So that implies that, that it should be true in it's factored form as well

(x+3) becomes zero when x = -3

and likewise, (x-1) becomes 0 when x = 1

hence our factored form becomes, (x+3)(x-1)

[mystic brook]

oh

so im allowed to make them positive/negative no matter what

?

[twilit lake]

well that is how it seems, remember that we should be able to get their factored form to 0 when replacing the x with the solutions from the quadratic formula

[mystic brook]

this goes for other calculations related to this?

[twilit lake]

Verily

[mystic brook]

so so here im just supposed to turn it into -1 and 3 and continue it?

[twilit lake]

You are missing the point, you are not changing the solutions you are just writing it in another form, this case being factored form, where the factors when we replace the solutions we get 0.

[mystic brook]

im using this form

wasnt i supposed to use it?

[twilit lake]

that is a formula not a form. I could as well have said that our zero's or roots of the functions are 1 and -3, but we would have said the same thing if we wrote it like this (x-1)(x+3)

[mystic brook]

oooh

so im changing the values since i would get 0 if i used those

right?

[twilit lake]

Yeah

[dreamy saffron]

There's an easier way.

What's (x - 1)^2?

[young frigate]

u have to use long diviosn

[young frigate]

or if u know how to factor out x-1

you already know -1 has a digit so find two numbers that multiply to -3 and add to 2. you know alread -1 has to be one so the other number can only be 3

Test:
-1 x 3 = -3
-1 + 3 = 2

So it will be simplified to (x-1) (x+3)/ (x-1)

Now divide both sides by x-1 and you will get x+3

Hence x = -3

Enjoy

x= -3, enjoy

[dreamy saffron]

Divide both sides of what?

[twilit lake]

@mystic brook factor

[cosmic birch]

how is this a university question?