#Need help with factorising and cadric equation
43 messages · Page 1 of 1 (latest)
note that (3x-2)(4x+7) is not in the form of a(x-x1)(x-x2)
But 12(3x-2)(4x+7) is
Its because i hade to switch the fraction into numbers
I wrote it using the fractions next to it too
yes so you have $12x^2+13x-14=12(x-\frac{2}{3})(x+\frac{7}{4})$
Smiley ッ
Yes
But to get ride of the fraction you need to multiply the x of each weird box but the denominator of the fraction
To get ride of the fraction
what for?
Why not xD and also because i dont know if my teacher will put me a error if i dont do it
both forms would be valid answers but ig it's up to your teacher
note that the lcd in (x-2/3) is 3, while the lcd in the other is 4
note 12 = 3*4
multiply the first factor by 3 and the 2nd by 4
and you end up with (3x-2)(4x+7)
Smiley ッ
$12(x-\frac{2}{3})(x+\frac{7}{4})$
PAWZ
yes, because it distributes
Ohhh
One last question
For the factorisation method for (ax^2+bx+c) there is a method where you need to find the 2 term that is equal to a.c when multipied and b when addition
Well my teacher said that there was a speacial mode on some calculator that lets us find the 2 term quickly but that she wont teach us how to do it since its not fair to the other who dont have it fo you know what that is ?
Do you have a video or something to help me do that cuz i dont have any idea what is a graph
Is English your native language?
no french
You just left the 12 there on accident on the work to the right of the standard form