For anyone who is a genius at math?

Question

How do you factor out equations in the ax2+bx+c form that have fractions? For example, how do you factor x2 - 5/2x + 25/16 ??

That x2 is suppose to mean x squared. Whoever gives the best step by step outline gets 10 points :)

Answer 1

Hey, if ur talking only about this particular example, then have a look at c, it can be rewritten as: (5/4) ^ 2.
if u have a look at b, you can see that it is (5/2) * 2, so it becomes simple factorization rule: (a-b)^2 = a^2 - 2*a*b + b^2, which can be applied to this example to get the following answer:
(x - 5/4)^2
if there is another example which is not as straight forward as this one, u can just use the Delta formula to find the roots ;)

Answer 2

It does not matter whether it has got fractions or not the method is same for fractions it is slight different. Either it has got rational factor in the sense (x-m) where m is rational or it does not have.

look at c that is 25/16
it m is raional then
m has to be of the from p/q where p is 1,5, or 25 or -1 or -5 or -25
q has to be 1,2,4,8,16

by tring out all cases we see m = -5/4

so we get (x-5/4) and dividing we get it is (x-5/4)^2

but this partiular problem is even simpler

as b^2 - 4ac = (5/2)^2 - 4(5/16) = 0 it is a perfect square

so it if (x-5/4)^2 as (5/4)^2 = 25/16 and x coffecient is -ve

Answer 3

first, u have to clear the fractions, meaning u have to get rid of them. this is not that difficult. it is like this:
x2-5/2+25/16 would turn to
x2(16)-5/2(16)+25/16(16)=
16x2-40+25. Im assuming u know how to factor
basically, all i did was multiply the denominator by the least common denominator of every fraction. Then u would multiply them to get 2 whole numbers which u then multiply to get whole numbers only.
Hope this helps!!

Answer 4

I'd be doing the factoring the way fatcat describes it, i.e. first I'd get rid of the fractions.

So multiply everything by 16. Then factor it like you would any quadratic, i.e. try some numbers that at least make the outside coefficients come out correctly, and, if that doesn't work, use the quadratic formula.

Answer 5

Ok, let me assist you with this one

the trick is to figure out the linear combination of terms that will define the middle and constant terms.

-5/4 - 5/4 = -10/4 = -5/2

and (-5/4)^2 = 25/16

so the equation can be factor as

(x-5/4)(x-5/4)

good luck.

Answer 6

y=x^2-5/2*x+25/16 rearange this to:
y=x^2-5/4*2+(5/4)^2
This has the form y=x^2+2*c+c^2 where c=-5/4
The factors are:
y=(x+c)^2 or y=(x-5/4)^2

Answer 7

x^2-(5/2)x+25/16 =
(16x^2-40+25)/16 =
(4x-5)(4x-5)/(4*4) =
(x-5/4)^2
Some people can solve this by inspection. I ain't one of them.

Answer 8

(x - 5/4)(x - 5/4)

Answer 9

-5/2x^3+23/16