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Have you learned factoring?
This is an example of factoring trinomials :)
can you please help
im stuck and need to be done within the hour
this one factors
do you know how it factors?
ehh not really :(
me neither i cheat
factors as \[(2x-1)(2x-5)=0\] would you like to know how i know it?
i will wait and let @pooja195 explain maybe i will learn something
i have never figured out a way to factor other than to grind it til i find it
thanks :) Ok well fist you want to multiply the a and c value what do you get?
this is algebra 1 incase you where wondering!
Whats \[\huge 5 \times 4 \]
good and our B value is -12 right?
Now ask yourself what multiplies into -20 BUT also adds up to -12?
You will have 2 numbers...
2 and 10?
multiplies into 20...
or 4 and 5
oh right multiplies into 20 but adds to -12 you are close with 10 and 2
you need a sign on both numbers...
-2 and -10?
or -2 and 10
yes! :) Now... \[\huge~(4x^2-2)+(10x+5)\]
Now find the GCF
After @pooja195 is done, I can summarize it all into one huge message of steps so it won't seem all over the place :)
-2 for the first and 5 for the second?
thank you aureyliant!
Are you trying to find the zeros?
You have the signs messed up but 2x and 5 is right :) \[\huge 2x(2x-1)+5(2x-1)\]
so 2x(2x-1) - 5 (2x-1)?
i know but im making sure that was correct XD
(2x - 5)(2x - 1)
now you just combine the parenthese and the outter numbers (2x-5)(2x-1) and you end up with Sat's grinding :P
You need to combine the first loner and the second loner and bring them together
ok is that all to it?
Then you set both equations to 0
Trying to factor by splitting the middle term 2.1 Factoring 4x2-12x+5 The first term is, 4x2 its coefficient is 4 . The middle term is, -12x its coefficient is -12 . The last term, "the constant", is +5 Step-1 : Multiply the coefficient of the first term by the constant 4 • 5 = 20 Step-2 : Find two factors of 20 whose sum equals the coefficient of the middle term, which is -12 . -20 + -1 = -21 -10 + -2 = -12 That's it Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -10 and -2 4x2 - 10x - 2x - 5 Step-4 : Add up the first 2 terms, pulling out like factors : 2x • (2x-5) Add up the last 2 terms, pulling out common factors : 1 • (2x-5) Step-5 : Add up the four terms of step 4 : (2x-1) • (2x-5) Which is the desired factorization Equation at the end of step 2 : (2x - 5) • (2x - 1) = 0 Step 3 : Solve (2x-5)•(2x-1) = 0 (This goes towards solving for the zeros but that's only if the Q is asking that, too)
Solve it .-.
Now you could either do that or use the quadratic formula
But thats a story for another day
Just set it to 0 its sooooo much more easier
Solving a Single Variable Equation : 3.2 Solve : 2x-5 = 0 Add 5 to both sides of the equation : 2x = 5 Divide both sides of the equation by 2: x = 5/2 = 2.500 Solving a Single Variable Equation : 3.3 Solve : 2x-1 = 0 Add 1 to both sides of the equation : 2x = 1 Divide both sides of the equation by 2: x = 1/2 = 0.500
can you guide me through setting through 0?
It depends on the type of factoring Q Pooja
2x-5=0 we want to isolate X
It may not be as easy to split the middle term
so start by adding 5
that way on the left side the 5 will cancel what does the equation look like now?
ok so 2x=5?
Yes now just divde both sides by 2
what do you get?
keep the 5
Now we know the two values for x... also I summarized solving for the zeros above @cacaman
you get x=5/2
-_- i gave u the answer :P
I think you answered this one earlier...
YES! now what are the 2 answers?
please get this right -.-
5/2 and 1/2
lol im nervous!
why so? :P and why did we choose this do you know?
You tell me because i dont know either.
We chose this because it was an easier method in my books?
Find a better reason -_-
I guess you can write because it is a easier way to do it :/ and since its a trinomial grouping is involved...
there is another method to factor quadratic equation cross multiply instead group method |dw:1433299203705:dw| factor first and last term then cross multiply after that combine them if you get middle term then (2x-5) and (2x-1) are factor of that equation