Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

can someone please give me tips on solving quadratic equations by factoring :) thanks!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

http://www.wolframalpha.com/ is the original website

OpenStudy (anonymous):

Honestly, I don't know any tips, it just takes practice. Practice a lot and it'll be really easy and fun :)

OpenStudy (anonymous):

how do you do them tho?

OpenStudy (anonymous):

ok let me say somethimg here guys,

OpenStudy (anonymous):

first of all it's not all quadratix experssions are factored directly,

OpenStudy (anonymous):

becasue some are done only using the quadratic formula

OpenStudy (anonymous):

second. in the direct factorization of quadratic experssions we follow trail and error method

OpenStudy (anonymous):

and such method need some practising and showing some patcience at the begining.

OpenStudy (anonymous):

it would'nt take time then to find your self an expert in that.

OpenStudy (anonymous):

my strategy is to have explanation with example approach, so let's have some:

OpenStudy (anonymous):

Ex#1) x^2+2x+4

OpenStudy (anonymous):

have 2 brackets opened : ( ) ( )

OpenStudy (anonymous):

take the x^2 term and distribut it as x * x , put each x in one of the bracket

OpenStudy (anonymous):

(x ) (x )

OpenStudy (anonymous):

for example how do i get positive at the 2nd term and negative at the third term... something like that

OpenStudy (anonymous):

then we come to the very sinsitive step, that factorizing the last term

OpenStudy (anonymous):

I'll show that now

OpenStudy (anonymous):

we look at the third terim , that 4 and try to check 4 comes from multiplying 2 numers , what are the possibilites here

OpenStudy (anonymous):

as we know 4=2*2 or 4=4*1

OpenStudy (anonymous):

so what will be our choice here?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Ex. 1) x^2 +4x+4

OpenStudy (anonymous):

\[(x^2+2x-15)\] (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."

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so x^2+4x+4= (x+2)(x+2)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

but if the sign of the third term was negative, then our 2 brackets signs would be +,- or -,+

OpenStudy (anonymous):

Ex2) X^2 +5x+6 ,X^2 -5x+6 , X^2-5x-6 , what is the differnce between factoring these three quadratic experession ?

OpenStudy (anonymous):

let's try the first one X^2 +5x+6 = ( x ) (x )

OpenStudy (anonymous):

we distribute x^2 into x*x

OpenStudy (anonymous):

then it's -,-

OpenStudy (anonymous):

or +,+

OpenStudy (anonymous):

well, it's +,+ or -,- , depending on the sign of the middle term

OpenStudy (anonymous):

how do I solve it if the last term is a prime number?

OpenStudy (anonymous):

the prin is 1* itself

OpenStudy (anonymous):

X^2 -5x+6 , X^2-5x-6 , can you do the first one here

OpenStudy (anonymous):

they are to let you know the differnce

OpenStudy (anonymous):

(x-6)(x-1) x^2-x-6x+6 x^2-5x+6 ???

OpenStudy (anonymous):

your posting looks to me interfering

OpenStudy (anonymous):

X^2 -5x+6 = (x-3)(x-2)

OpenStudy (anonymous):

why 3and 2? and why -,- ?

OpenStudy (anonymous):

because 3*2 is 6

OpenStudy (anonymous):

right

OpenStudy (anonymous):

and - * - is +

OpenStudy (anonymous):

exactly

OpenStudy (anonymous):

I tel a point that you can test your facors in a fast way

OpenStudy (anonymous):

anyway how did my post look interfering?

OpenStudy (anonymous):

just in this (x-6)(x-1) x^2-x-6x+6 x^2-5x+6 ???

OpenStudy (anonymous):

not all posts

OpenStudy (anonymous):

ok try this x^2-5x-6

OpenStudy (anonymous):

|dw:1315385954278:dw|

OpenStudy (anonymous):

(x+1)(x-6) x^2-6x+x-6

OpenStudy (anonymous):

then I'd get x^2-5x-6

OpenStudy (anonymous):

great

OpenStudy (anonymous):

ok let's have this x^2-4x-5

OpenStudy (anonymous):

(x-5)(x+1) x^2+x-5x-5

OpenStudy (anonymous):

then I'd get x^2-4x-5

OpenStudy (anonymous):

great too

OpenStudy (anonymous):

yepee :)

OpenStudy (anonymous):

ok let me draw a fast method to check any time you carry this type

OpenStudy (anonymous):

ok :)

OpenStudy (anonymous):

|dw:1315386503044:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!