can someone please give me tips on solving quadratic equations by factoring :) thanks!
i found this website, but its the best i can give you, i have the same problem http://www.wolframalpha.com/input/?i=x^2%2By^2-18y%2B56%3D0%2C+x^2%2By^2%2B9x-15y%2B74%3D0
Honestly, I don't know any tips, it just takes practice. Practice a lot and it'll be really easy and fun :)
how do you do them tho?
ok let me say somethimg here guys,
first of all it's not all quadratix experssions are factored directly,
becasue some are done only using the quadratic formula
second. in the direct factorization of quadratic experssions we follow trail and error method
and such method need some practising and showing some patcience at the begining.
it would'nt take time then to find your self an expert in that.
my strategy is to have explanation with example approach, so let's have some:
Ex#1) x^2+2x+4
have 2 brackets opened : ( ) ( )
take the x^2 term and distribut it as x * x , put each x in one of the bracket
(x ) (x )
for example how do i get positive at the 2nd term and negative at the third term... something like that
then we come to the very sinsitive step, that factorizing the last term
I'll show that now
we look at the third terim , that 4 and try to check 4 comes from multiplying 2 numers , what are the possibilites here
as we know 4=2*2 or 4=4*1
so what will be our choice here?
here, I have to change my example because it's not valid in the 2nd term to explain the idea, I'll do little change
Ex. 1) x^2 +4x+4
\[(x^2+2x-15)\] (x ) (x ) Ask yourself: What times what equal 15? Answer: 5 and 3 (x 5)(x 3) Ask yourself: What signs corresponds with which number? Answer: 5 and -3 because 5*(-3)=15 and -3+5=2 <-The middle term. Therefore, the answer is (x+5)(x-3) Of course, this process, like Raheen said, is trial and error, but after enough practice, you should be able to quickly recognize how to "mix and match."
now we can see 4=2*2 we can reach to the middle term by summing of the factors of 4 , that is 2+2
so x^2+4x+4= (x+2)(x+2)
regarding the signs we usualy look at the sing of the last term, if it is + , then we know that our two signs in the brackets would be identical, that is either +,+ or -,- , depending on the sign of the middle term
but if the sign of the third term was negative, then our 2 brackets signs would be +,- or -,+
Ex2) X^2 +5x+6 ,X^2 -5x+6 , X^2-5x-6 , what is the differnce between factoring these three quadratic experession ?
let's try the first one X^2 +5x+6 = ( x ) (x )
we distribute x^2 into x*x
then it's -,-
or +,+
well, it's +,+ or -,- , depending on the sign of the middle term
how do I solve it if the last term is a prime number?
the prin is 1* itself
X^2 -5x+6 , X^2-5x-6 , can you do the first one here
they are to let you know the differnce
(x-6)(x-1) x^2-x-6x+6 x^2-5x+6 ???
your posting looks to me interfering
X^2 -5x+6 = (x-3)(x-2)
why 3and 2? and why -,- ?
because 3*2 is 6
right
and - * - is +
exactly
I tel a point that you can test your facors in a fast way
anyway how did my post look interfering?
just in this (x-6)(x-1) x^2-x-6x+6 x^2-5x+6 ???
not all posts
ok try this x^2-5x-6
|dw:1315385954278:dw|
(x+1)(x-6) x^2-6x+x-6
then I'd get x^2-5x-6
great
ok let's have this x^2-4x-5
(x-5)(x+1) x^2+x-5x-5
then I'd get x^2-4x-5
great too
yepee :)
ok let me draw a fast method to check any time you carry this type
ok :)
|dw:1315386503044:dw|
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