VCE Stuff > VCE Mathematical Methods CAS
how do I factor a quadratic equation?
annaoh_2003:
--- Quote from: Bri MT on January 23, 2022, 06:17:30 pm ---I'm adding to Tapioca's explanation since it seems like you may not have covered the polynomial theorem in school yet which teaches that if f(a) = 0 then (x -a) is a factor of f(x). It's common to test easy numbers first (like 1, -1, 2, -2) when trying to find the first factor.
It might be easiest to first see this looking from the other perspective.
if you have (x-a)(x-c)(x-d) then we know we can make the whole thing = 0 by making one of the brackets equal zero. (null factor law).
I.e. if (x-a) is a factor, then making x = a means that the equation equals zero.
--- End quote ---
--- Quote from: Tapioca on January 23, 2022, 05:10:17 pm ---Let f(x) = 2x^3 - 3x^2 -3x + 2
Because f(-1)= 0, then x+1 is a factor:
Since you wanted to find the X-intercepts, we know that y=0:
0= (x-1) (2x-1) (x-2)
So,
(x-1)=0
(2x-1)=0
(x-2)=0
From algebraically solving, the X-intercepts: x=1 and x=1/2 and x=2
--- End quote ---
ohhh wait so my brain just clicked- you're actually using the -1 in place of the x - okay got it now :)
so when you factorise how did you find = 2x^2(x+1) - 5x(x+1) +2(x+1) ? I can see from working it out we get our original equation - but how did you work backwards to find those values in red?
Tapioca:
.
Billuminati:
--- Quote from: annaoh_2003 on January 23, 2022, 06:55:12 pm ---ohhh wait so my brain just clicked- you're actually using the -1 in place of the x - okay got it now :)
so when you factorise how did you find = 2x^2(x+1) - 5x(x+1) +2(x+1) ? I can see from working it out we get our original equation - but how did you work backwards to find those values in red?
--- End quote ---
I failed methods myself but this is my way of going about the factorisation to get the quadratic component: https://drive.google.com/file/d/106qGlePQrAL0OAmtfH_RgROkOKwYN45q/view?usp=drivesdk
annaoh_2003:
--- Quote from: Tapioca on January 23, 2022, 07:27:11 pm ---
About the factorising step- you are basically taking (x+1) out of the equation. Take this out of each term, then think about how to get to the original equation each time you factorise a term, then work your way to the end. After this collect the terms outside brackets with positives and negatives, then factorise to simplest form.
--- End quote ---
--- Quote from: Bri MT on January 23, 2022, 06:17:30 pm ---I'm adding to Tapioca's explanation since it seems like you may not have covered the polynomial theorem in school yet which teaches that if f(a) = 0 then (x -a) is a factor of f(x). It's common to test easy numbers first (like 1, -1, 2, -2) when trying to find the first factor.
It might be easiest to first see this looking from the other perspective.
if you have (x-a)(x-c)(x-d) then we know we can make the whole thing = 0 by making one of the brackets equal zero. (null factor law).
I.e. if (x-a) is a factor, then making x = a means that the equation equals zero.
So to restate, Tapioca was using the similar idea that if have x = a makes the whole thing equal zero, then that means (x-a) is a factor.
That's the first step. The 2nd step may also be a bit tricky but before we move on I think it's important to check that you get the first one :)
Kudos to you for reaching out for help and to Tapioca for the explanations :)
--- End quote ---
thank you guys ! im doing an intensive six week method course at unimelb and its extremely difficult + I didn't have any methods background knowledge. im just relying on the atarnotes community at the moment (you can see ive been asking quite a few questions on the methods thread) and im extremely grateful for all your help :)
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