How do you solve the following:
Directions:Factor
Can you show me!!!
(1) x^2-x-12
(2)x^3+2x^2-3x
Here is how I do it.
First make two sets of parentheses.
( )( )
The first in each set of parentheses must give x2. Obviously that is x and x. Looks like this.
(x )(x )
The sign in each must be + for one and - for one. That is the only way to get a - sign for the last number.
(x+ )(x- ).
The last number must be factors of 12 that multiply to give 12 but when subtracted gives 1 (for -1x). The factors of 12 are 12*1; 6*2; and 4*3. the 4*3 combination is the only one that has a difference of 1 in the two factors, so place those in. You must get them in the right slots to give it the right sign.
(x+3)(x-4). Now we check it.
First: x*x = x2
Last: +3*-4=-12.
The middle term comes from the sum of the outer set + the inner set like this.
outer: x*-4=-4x
inner: +3*x=3x
sum of outer and inner: -4x+3x=-1x
I always check each time I do it to see if these are the real factors. Keeps me from making too manyh mistakes.
both of the problems are separate equations. how do u solve them i am lost
I know they are separate. But I showed you how, in detail, to do the first one.
The idea is that if you learn to do that one you can do any of them. It does take some practice, of course. I shall be happy to answer any questions you have about the first one. You try the second one.
Oh because it seem to me like your answer was involving two of them. How did you get the three and four in there. I want to try the second one but i need tolearn to do the first one so i can see how to actually do them.
I have seen this called FOIL on the internet. F stands for first and is the first letter in each set of parentheses, O stands for outer, I for inner, and L for last. The First one is just the product of the two first letters. The Last one is the product of the last two numbers. The middle term of the orginal equation is the sum of the outer and inner, multiplied and added as I did in the example. So how did I know to pick 4 and 3? In this case, the number of -12 can be made up of 12*1 or 6*2 or 4*3. Assuming the equation is factorable, you want to pick two numbers whose product gives 12 but the difference gives 1. 12*1 gives 12 but the difference is 11. 6*2 gives 12 but the difference is 4. 4*3 gives 12 AND the difference is 1; therefore, 4 and 3 are the two numbers to use and you know it will work. The only thing you need to worry about is how they are placed in the parentheses. If you put them like this (x+4)(x-3) it gives you +1x for the middle term (x2+x-12 but if you place it like this
(x+3)(x-4) it gives you -1x for the middle term and that is what you want. That is x2-x-12.
okay so i'll try the second one.
Great! Let's work on the second equation: x^3 + 2x^2 - 3x.
To factor this equation, we first look for common factors. In this case, we see that each term has an x. Therefore, we can factor out an x:
x(x^2 + 2x - 3).
Now we have x multiplied by a trinomial, which we can further factor. We use the same method as before:
First, make two sets of parentheses: ( )( ). The first in each set of parentheses must give x^2. Let's put x and x in the first positions in each set of parentheses: (x )(x ).
Next, we look for two numbers that, when multiplied together, give -3 (the last term) and when added together, give 2 (the coefficient of the middle term). The only combination that works is 3 and -1. So, we can put 3 in the first set of parentheses and -1 in the second set of parentheses: (x + 3)(x - 1).
Now, we have factored the equation completely: x(x + 3)(x - 1).
To check if our factoring is correct, we can expand the equation. If it matches the original equation, then our factoring is correct:
(x + 3)(x - 1) = x^2 + 2x - 3.
And this matches the original equation x^3 + 2x^2 - 3x.
So, the factored form of x^3 + 2x^2 - 3x is x(x + 3)(x - 1).