# Thread: I need help with algebra. Factoring.

1. Hello!
I'm homeschooled and need help with algebra.
I understand most of what to do in the problems but I can't figure out this one.
Factor: 2x^2+5x-3
I tried to use
Factoring Calculator here:
And I've got:
(x+3)*(2*x-1)
Seems like it is the right answer, because textbook shows me the same.

But I really need help on this and I would really appreciate a step by step way to do this so I can understand what to do in these situations in the future.  2.

3. Well, first off, if you want to learn how to factor on your own, don't use the factoring calculator. Unlike some tools, it won't help, since the whole battle is getting some candidates to check out.

Now, a question: how would you determine for yourself whether (x+3) and (2*x+1) were the right factors? [I'll have more questions later].

Edit: Meaning after you answer this one.  4. Some math teachers are annoyed by this, but you can always use FOIL to check. That means First Outer Inner Last. You multiply the first two terms in each part by one another, then the outer, then inner, then last. Then, you combine like terms and check to see if the answer is the same as the equation in the original question.

I know MVB wants you to answer that part yourself, but if you don't know what process to apply, I doubt you're going to figure it out on your own (sans Google).

Also, it might help to know what you're comfortable with up to this point. Is this the first time you've factored?  5. mathsudent150 - what you wnat to factorize is called a "polynomial". And, if I may be so bold, both mvb and Flick have entirely missed the point. Which was not, as far as I can see, whether or not the OP polynomial factors as it clearly does, but rather......

Given a polynomial, how does one factor it - if at all?. The answer to that is extremely non-trivial. but in the OP case is rather straightforward.

Note first the fact (which is irrelevant in the present case, but may not be in others) that in we may more exactly write . Can you see why (check the law of exponents).

Now the numbers in front of the "x's" are called coefficients for this polynomial, and you will note they are all prime numbers. Notice also that So you have very little choice in this case - you have only a few possibilities to check.

Say what they are!

But note that for other polynomials this "trial and error" procedure is not likely to be productive, and the methods available (short of computer programmes) are a bit tricky, not say daunting  6. Mathstudent150, I think Guitarist rightly perceived the situation. Ultimately, you are looking to find two factors, ax+b and cx+d, when multiplied together produce, in this case, 2x² + 5x  3. Montana gave you the FOIL mnemonic to help you with the steps to perform. So, go ahead and multiply ax+b and cx+d as they are, and then simplify by combining like terms. Then compare your results with the polynomial you intend to factor, focusing on the coefficients. This will give you possible coefficients, and then it's a matter of trying out the different combinations of coefficients until you find the ones that produce the polynomial that you began with.

I'll give you a cool calculator shortcut to check your combinations of coefficients. Let's say you think that x+3 and 2x1 might be the factors. To see if they are, extract the coefficients from them: +1 and +3, and +2 and 1. Key them into your calculator (even the simplest calculator) as follows. The calculator's rightmost three digits represent the coefficient, the next three digits to the left represent the coefficient, the next three the coefficient, the next three the coefficient, and so on. Some calculators are nice enough to separate these groups with commas.

Even if I've confused you at this point, perhaps it's easier to see here how x+3 is represented by 1,003 (because 1,000 plus 3 equals 1,003) and 2x1 is represented by 1,999 (because 2,000 plus 1 equals 1,999). Now all you need to do is multiply these two representative numbers: 1,003 × 1999 = 2,004,997. Just as I said above, the 2 is the coefficient. The 4,997 is really 5,000 plus 3, so the 5 is the coefficient, and the 3 is the coefficient. Insert these coefficients into the polynomial and obtain 2x² + 5x  3, which is what you started with. So you have found the factors of this polynomial.

Let's say that, while checking different combinations of coefficients, you want to consider x1 and 2x+3. These two factors are represented by 999 (that is, 1,000 plus 1) and 2,003. As I said, simply multiply these two representative numbers: 999 × 2003 = 2,000,997, which converts into 2x² + x  3 (remember that 0,997 is 1,000 plus 3), which is not what you began with, so these factors are wrong.

Now, knowing this kind of "representative" polynomial math, you can now use division to try to find the answer. So, beginning with 2x² + 5x  3, you think that x+3 might be a factor, so you simply divide 2,004,997 by 1,003 and you obtain 1,997 as the other factor. Because it's a whole number (and you know you're dealing with only whole numbers here), you realize that x+3 is a factor and so is 2x3 as indicated by the 1,997 that you just obtained.

Using the wrong factors example in the fourth paragraph above, you think that 2x+3 might be a factor, so you divide 2,004,997 by 2,003 and obtain 1,000.997004493... . Wow, this might represent something such as x+13/x+4/x²+493/x³... . Definitely not a simple polynomial, so 2x+3 is not a factor.

If you really want to put this method to the test, try squaring 2x² + 5x  3. You go ahead and struggle doing it longhand with pencil and paper, and I'll merely type the representative number 2,004,997 into my computer's calculator and square it to obtain: 4,020,012,970,009. This answer converts into 4x4 +20x³ + 13x² 30x + 9. Remember that ...,012,970,... is really 13,000,000 plus 30,000. (When you see ...,9... in a representative number, it usually means that the number in that set of three digits is probably negative, so you'll need to find the complement (for example 1,000  970 = 30) and throw a negative sign in front of it, and remember to add 1 to the coefficient to the left of the negative coefficient (in this case, 12 + 1 = 13).

I think that, once you learn polynomial multiplication and understand what you're doing and why, then it's perfectly okay to use this "representative" polynomial math, much the same as using a calculator after learning how to multiply and divide longhand.  7. More easier way is just to guess the factor. Of course an algorithm for computers are more specific than guesses, but in school they do not teach such tedius algoritm. In school they just taught you to guess the factor and then methodically check them if they fit the equation.

p/s: Also, (in school) they won't just give you random equation to solve, they will only give you sensible equation. And they never taught anyone to solve a polynomial equation! (only quadratic, to power of 2, or x^2 is important)  8. Since our young friend shows no sign of acknowledging the help (s)he has been given, let alone giving any thanks, let me go on a bit of a ramble......

The degree of a polynomial is the number that appears as the highest exponent on the indeterminate with non-zero coefficients, say . So for example, in the simple case , this is a polynomial of degree 2.

Note from this construction, a madman might justifiably write .

Now a polynomial is said to be reducible if it can be factored as 2 or more polynomials of lesser degree. But here is the important point....

A polynomial may be reducible over one field, but not over another. Two simple examples will suffice. is clearly not reducible over , the rational number field but factors as over the real number field.

Conversely, is not reducible over but factors over the complex number field as .

You might also like to know about Einsentein's Irreducibly Criteria (no, there is no typo here!). At first sight it looks like magic, but it;s not.....  9. Hate to do this but its actually Eisenstein, not Einsenteins lol  10. Originally Posted by Guitarist Since our young friend shows no sign of acknowledging the help (s)he has been given, let alone giving any thanks,
Our young friend was probably just posting a spam link. I deleted it.  11. Originally Posted by TheObserver Hate to do this but its actually Eisenstein, not Einsenteins lol
Actually, Guitarist is referring to the somewhat obscure mathematician, Earl "Goober" Einsentein, who developed a criterion which fully subsumes Eisenstein's.  12. Ha! And who was the pompous ass who asserted there was no typo?

Why are you looking at me like that.........?  13. Anyway, back to Eisenstein.

There are 2 equivalent ways to formulate his criteria for irreducibly.

Easiest first....... Suppose a polynomial then is irreducible over if and only is, for some prime number that

1. exactly divides every except 2. does not exactly divide There is a famous theorem of Gauss that states that if a polynomial is irreducible over then it is of necessity irreducible over . Anyone foolish enough to be reading this will note that from an earlier post of mine, the converse is false

The alternative uses modulo arithmetic over , and is ultimately more rewarding in that it leads to a nice "shortcut", but it requires a knowledge of quotient rings and the like.  14. As a general rule, you do not need to guess the factors of a quadratic equation - just complete the square, solve, rinse, repeat  15. They don't just go directly to quadratic formula in school, first they teach you to guess, then the formula. IMO its for demonstrating the concept of factoring (ie: (x+?)(x-?)=0).  16. Well the OP seems to be gone, so: Factoring is normally easy and you do it mentally. Given Ax^2 + Bx +C you are supposed to find (ax+b) and (cx +d) so that [x^2 means x-squared; I haven't mastered algebra in the editor yet]

Ax^2+Bx+C = (ax+b)(cx+d)
= ac x^2 + (ad+bc) x +bd

Clearly a and c are factors of A, while b and d are factors of C. Normally there won't be many choices, and you just figure out how to shuffle the factors so that you get the right coefficient of x in the polynomial.

There is no point in doing anything fancier unless there are too many possible factors to keep track of in your head. Hence being able to check the result is the most important skill.

Of course if you are in the habit of doing all your arithmetic on a calculator, this method doesn't work very well.  17. Nobody offering the quadratic formula?  18. Suppose we want to factor , let and so We can rewrite this last equation as which is a difference of two squares and so can be easily factored.

Back substituting gives The trick of substituting in for a degree n polynomial is quite useful as it generally kills the n-1 degree term term in the polynomial   19. Very nice ratty.

But here's something I still find astonishing after all these years (as we are talking about polynomials)....

First I assume we all recall from high-school the Prime Factorization Theorem - any integer can be factored as the product of a unit (i.e. ) and a set of primes and that this factorization is unique if and only if we disregard the order in which the factors entering into the product are written.

It seems there is a somewhat similar theorem for polynomials....

Any polynomial can be factored as the product of a constant and a set of irreducible monic polynomials, and this factorization is unique, again up to order, if and only our polynomial is defined over a field (and not a ring, say). BTW a polynomial is said to be monic if and only if the coefficient Usually it is considered hygienic to prove uniqueness - left as a (fairly) easy exercise for the reader.

The implication is clear - the irreducible monic polynomials that enter into the unique factorization of any arbitrary polynomial play the role of the prime numbers in the unique factorization of the integers.

So. Does this mean that the space of all polynomials over a given field (this space being a ring) is somehow equivalent to the space of all integers (also a ring)?

Something for you all to think about....  20. Let's keep it simple:

a*x^2+b*x+c = 0 -> x^2 + (b/a)*x +c/a = 0 -> define b/a := b' & c/a := c' then: x^2 + b'*x+c' = 0 = (x + d)*(x+e) = x^2 + (d+e)*x + d*e
From that we deduce:
d+e = b' =b/a & c' = d*e = c/a

In most cases it is then a piece of cake.  Bookmarks
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