factorization
- Thread starter nickname
- Start date
D
Deleted member 4993
Guest
nickname said:
Pair up x^3 and -x.
Just by looking, it is certain that there are no quantites you can factor out of each term as it stands. That strategy will not work in this case. Therefore, you will have to do some "reverse foiling"
Here is an example of this in an equation of order 2...
Factor x^2 + 6x -16
There exists 2 numbers (a) and (b) that, when written in the form
(x-a)(x-b) ,
will be equal to my example question. I have to find a and b. Notice that there is no x^2 term anymore, and only x's of first order remain. This is what we call factoring. Factoring in this case is breaking down higher order equations into lower order ones.
VERY IMPT TO KNOW: You notice that if we started with (x-a)(x-b) and "foiled", meaning that we multiplied out the expression to get rid of the parenthesis, we would get
x^2 -bx - ax + ab
or x^2 -(a+b)x +ab. Each term in the first parenthesis (x-a) must be multiplied by each term in the second parenthesis (x-b) to get this answer, and you must make sure to get the +'s and -'s right between the terms. Do this simply by following the rules of multipication for signs (like x multiplied by a negative b is -bx, or two positives multiplied equals a positive...if you have trouble, remember that (x-a)(x-b) = (x + (-a))(x + (-b))
So in my case I need to find 2 numbers a and b that I can stick into (x-a) and (x-b) that, when foiled, will equal x^2 + 6x -16.
At this point, most people will try to guess and check, and once you get used to it, most people can see what the numbers need to be. However, you do not have to do it this way. Since we know from above that..
x^2 -(a+b)x +ab = (x-a)(x-b), and our original equation was x^2 + 6x -16, we see in this specific case that:
-(a+b) = 6,
ab=(-16)
The coefficients of the squared terms must be equal to each other, the first order terms must be equal, and so on.
Two equations, two unknowns...solve for a or b in one of the equations, and sub accordingly into the other equation to isolate a or b. Once you know one of em, plug it into one of the formulas to get the other!
***You have a cubic equation, which will be factorable into the form (x-a)(x-b)(x-c). Doing it as shown above will get you at least three equations relating a, b, and c which is all you need to find them. The x^3, x^2 and x coefficients of your problem and the one you could derive from (x-a)(x-b)(x-c) will be equal! To do this, "foil" 2 of the three terms first, than foil the newly combined term with the third term.
I might have mentioned this earlier, but you can also use polynomial division, but it doesn't sound like you are that far yet...you could look up how to do this as well, as I am sure that there are a lot of great examples out there online. My way doesn't require guessing, although it is a little more involved, but it is great practice for people that are interested in deriving their own equations to figure out problems.
Here is an example of this in an equation of order 2...
Factor x^2 + 6x -16
There exists 2 numbers (a) and (b) that, when written in the form
(x-a)(x-b) ,
will be equal to my example question. I have to find a and b. Notice that there is no x^2 term anymore, and only x's of first order remain. This is what we call factoring. Factoring in this case is breaking down higher order equations into lower order ones.
VERY IMPT TO KNOW: You notice that if we started with (x-a)(x-b) and "foiled", meaning that we multiplied out the expression to get rid of the parenthesis, we would get
x^2 -bx - ax + ab
or x^2 -(a+b)x +ab. Each term in the first parenthesis (x-a) must be multiplied by each term in the second parenthesis (x-b) to get this answer, and you must make sure to get the +'s and -'s right between the terms. Do this simply by following the rules of multipication for signs (like x multiplied by a negative b is -bx, or two positives multiplied equals a positive...if you have trouble, remember that (x-a)(x-b) = (x + (-a))(x + (-b))
So in my case I need to find 2 numbers a and b that I can stick into (x-a) and (x-b) that, when foiled, will equal x^2 + 6x -16.
At this point, most people will try to guess and check, and once you get used to it, most people can see what the numbers need to be. However, you do not have to do it this way. Since we know from above that..
x^2 -(a+b)x +ab = (x-a)(x-b), and our original equation was x^2 + 6x -16, we see in this specific case that:
-(a+b) = 6,
ab=(-16)
The coefficients of the squared terms must be equal to each other, the first order terms must be equal, and so on.
Two equations, two unknowns...solve for a or b in one of the equations, and sub accordingly into the other equation to isolate a or b. Once you know one of em, plug it into one of the formulas to get the other!
***You have a cubic equation, which will be factorable into the form (x-a)(x-b)(x-c). Doing it as shown above will get you at least three equations relating a, b, and c which is all you need to find them. The x^3, x^2 and x coefficients of your problem and the one you could derive from (x-a)(x-b)(x-c) will be equal! To do this, "foil" 2 of the three terms first, than foil the newly combined term with the third term.
I might have mentioned this earlier, but you can also use polynomial division, but it doesn't sound like you are that far yet...you could look up how to do this as well, as I am sure that there are a lot of great examples out there online. My way doesn't require guessing, although it is a little more involved, but it is great practice for people that are interested in deriving their own equations to figure out problems.
x+1)(x^2-1)\)