#1

Hey there UG. I'm having trouble factoring / factorising trinomials of the type

A

I can do the normal ones, but where there is more than one x in the first part - Ax I get confused and can't seem to work out how to get the answer efficiently. I do stumble upon the correct answer eventually by trial and error, but is there any efficient method to do it? Any help is much appreciated.

A

*x*² + B*x*+ CI can do the normal ones, but where there is more than one x in the first part - Ax I get confused and can't seem to work out how to get the answer efficiently. I do stumble upon the correct answer eventually by trial and error, but is there any efficient method to do it? Any help is much appreciated.

#2

It's pretty much trial and error, I'm afraid.

Just remember, that if it's 2x, then the brackets must be (2x + a)(x + b)

If it's 3x, then (3x + a)(x + b) applies. Same for any prime numbers, but other than those, whenever A > 3, you'll probably take more time. See if you can cancel any of the numbers out by dividing the whole thing by a common factor first, perhaps, to try to get 1x, 2x, or 3x.

Just remember, that if it's 2x, then the brackets must be (2x + a)(x + b)

*Where a and b are numbers within the brackets, I'm hoping you know what i mean*If it's 3x, then (3x + a)(x + b) applies. Same for any prime numbers, but other than those, whenever A > 3, you'll probably take more time. See if you can cancel any of the numbers out by dividing the whole thing by a common factor first, perhaps, to try to get 1x, 2x, or 3x.

#3

x = (-b +/- sqrt(b^2 - 4ac))/2a

#4

x = (-b +/- sqrt(b^2 - 4ac))/2a

The solution to life's problems, right thur.

Um, real answer?

http://www.algebrahelp.com/lessons/factoring/trinomial/

#5

It's pretty much trial and error, I'm afraid.

Just remember, that if it's 2x, then the brackets must be (2x + a)(x + b)Where a and b are numbers within the brackets, I'm hoping you know what i mean

If it's 3x, then (3x + a)(x + b) applies. Same for any prime numbers, but other than those, whenever A > 3, you'll probably take more time. See if you can cancel any of the numbers out by dividing the whole thing by a common factor first, perhaps, to try to get 1x, 2x, or 3x.

Yeah i know what you mean. Its just basically where the number isn't prime, for example 4x. I then am not sure whether to do (2 + )(2 + ) or (4 + )(1 + ). So i guess i'll just have to practise and get into the swing of the trial and error. Thanks anyway.

#6

ax^2 + bx + c

2 numbers that add up to B and their product is C is your answer

and when you get it like (x-3)(x+5)

your answer would be x=3 and x=-5 cause

x - 3 = 0

(add 3 to each side)

x=3

same for the other

wait your doing when A > 1

use the quadratic formula

-B +/- the square root of

________________________________

divided by 2a

2 numbers that add up to B and their product is C is your answer

and when you get it like (x-3)(x+5)

your answer would be x=3 and x=-5 cause

x - 3 = 0

(add 3 to each side)

x=3

same for the other

wait your doing when A > 1

use the quadratic formula

-B +/- the square root of

________________________________

divided by 2a

*Last edited by fenderbassist12 at Feb 15, 2008,*

#7

Hey there UG. I'm having trouble factoring / factorising trinomials of the type

Ax² + Bx+ C

I can do the normal ones, but where there is more than one x in the first part - Ax I get confused and can't seem to work out how to get the answer efficiently. I do stumble upon the correct answer eventually by trial and error, but is there any efficient method to do it? Any help is much appreciated.

It really is just trial and error. After you do enough of them it starts to get easier and easier, till eventually it's almost automatic.

#8

Don't any of yous use the PS method? Find P which equals the product of A and C and S which is equal to B, in the case of TS's equation. Then find 2 numbers for which the product equals P and the sum equals S, hence the name Product-Sum method or PS. Stick the numbers in the equation. Substitute the two numbers you just found, let's call them 'm' and 'n' in the equation in the form Ax^2+mX+nX+C. You can then factorise and simplify ahead from here.

#9

Don't any of yous use the PS method? Find P which equals the product of A and C and S which is equal to B, in the case of TS's equation. Then find 2 numbers for which the product equals P and the sum equals S, hence the name Product-Sum method or PS. Stick the numbers in the equation. Substitute the two numbers you just found, let's call them 'm' and 'n' in the equation in the form Ax^2+mX+nX+C. You can then factorise and simplify ahead from here.

I've never heard of that before in my life.

#10

I've never heard of that before in my life.

It's exactly what FenderBassist explained, look at his if it seems familiar

#11

This is flabergasting stuff, and almost totally useless, I'm afraid.

Sure you can be a math teacher, but look at your math teacher!

Trust me. You can make more pumping gas. Plus you get free coffee and donuts.

Engineers have computers to do this sort of thing. Doctors just add a random integral (say...oh 8!) in front of the three zeroes on your bill:

$8,000

Real world application here? None. Sorry.

Sure you can be a math teacher, but look at your math teacher!

Trust me. You can make more pumping gas. Plus you get free coffee and donuts.

Engineers have computers to do this sort of thing. Doctors just add a random integral (say...oh 8!) in front of the three zeroes on your bill:

$8,000

Real world application here? None. Sorry.

#12

This is flabergasting stuff, and almost totally useless, I'm afraid.

Sure you can be a math teacher, but look at your math teacher!

Trust me. You can make more pumping gas. Plus you get free coffee and donuts.

Engineers have computers to do this sort of thing. Doctors just add a random integral (say...oh 8!) in front of the three zeroes on your bill:

$8,000

Real world application here? None. Sorry.

Exactly, completely pointless. Why do they still teach it in schools? Beats me.

#13

Factorising?

I'm sorry.

I'm sorry.

#14

Quadratic formula.. learn it.. use it.. love it..

Trust me... if you ever had to use MatLab, you'd prefer the ol' paper/pencil/calculator method.

Engineers have computers to do this sort of thing.

Trust me... if you ever had to use MatLab, you'd prefer the ol' paper/pencil/calculator method.

*Last edited by RockThe40oz at Feb 15, 2008,*

#15

x = (-b +/- sqrt(b^2 - 4ac))/2a

Seriously guys. Just use the solution to quadratic equation (above). The guess and check method is okay if it's an easy quadratic, but if you cannot figure it out, just use the solution. You will always get the right answer, even if it turns out to be complex. Once you get your 2 values of x (called your roots: x_1 and x_2 and the _ means a subscript) then you just set it equal to the following:

Ax² + Bx + C = (x - x_1)*(x - x_2)

I am assuming you can add and subtract well enough to use it.

If you need anymore junk on this, just go to Wolram-Mathworld

http://mathworld.wolfram.com/QuadraticEquation.html

By the way, if you want to know how dividing by zero works, just ask L'Hospital.

#16

yeah use the quadratic formula or do you know the box method?

i have 100 in my algebra 2 class

i have 100 in my algebra 2 class

#17

x = (-b +/- sqrt(b^2 - 4ac))/2a

that right there is EXACTLY what we're doing in Algebra 2 right now.

#18

Practical formulae? Try everything in engineering. Everything I've learned so far in my first year of Mech Eng is either differentiation or integration. Which are pretty much the same thing anyway. So yes, it is practical.