The first thing you should always do when factoring is to take out a common factor. This is the simplest technique of factoring, but it is important even when you learn fancier techniques, because you will make your later work much easier if you always look for common factors first. Taking out common factors is using the distributive property backwards. The distributive property says

a(b+c)=ab+ac.

The idea behind taking out a common factor is to look for something like the right side here where there is a common factor, here it would be a, and turn it into something like the left side to factor it.

Example:

4x⁵+12x⁴-8x³=4x³(x²+3x-8)

A good trick for finding the largest common factor when you are using this method to factor polynomials is to find the greatest common factor of the numbers and the smaller power of the variable, so here the greatest common factor of the numbers is 4 and the smallest power of x is 3, so we can take out 4x³ as a common factor.

Grouping

Grouping is a fancier technique that is based on taking out common factors. For grouping we split the polynomial in two pieces and take out common factors in each of them. If we get the same thing left over in each piece, then we can take that big thing out as a common factor, and this will factor the polynomial. It is, however, important to note that this only works if we do get the same thing. Simply taking out common factors of pieces of a polynomial is not factoring it. To factor a polynomial we must write the whole polynomial as a product of two polynomials.

Example:

5x³+10x²+3x+6=5x²(x+2)+3(x+2)=(5x²+3)(x+2)

Here we take out a common factor of 5x² from the first two terms and a common factor of 3 from the second two terms. This would not be a good method of factoring except that this polynomial is kind of special in that what is left when taking out the common factors in both cases is x+2. Then what we do is take out x+2 as a common factor of the two big chunks. The idea is that anything we can do with x we can do with x+2, because it represents a number too. If it helps, blur the (x+2) or think of it as one big strange looking letter.

General Quadratics

To learn to factor second degree polynomials, also called quadratics we first make it simpler by looking at ones where the coefficient on x² is 1. To figure out how to factor these polynomials we need to look at what happens when you multiply things of the form (x+A)(x+B), like for example (x+3)(x+4). Applying FOIL to this multiplication we get

x²+4x+3x+12=x²+7x+12.

Notice that the coefficient on x comes from adding the inners and outers, so it will always be the sum of the two constants. The constant term will always come from the lasts, so it will be the product of the two constants. So really we could use this as a special trick for multiplying anything of this form. We just add the constants for the middle coefficient and multiply them for the last term. So to multiply

(x+5)(x-2)

we can use this shortcut and get

x²+3x-10

by just adding the 5 and the -2 to get the 3 and multiplying the 5 and -2 to get the -10. This means that in order to go the other way around, that is to factor into this kind of form, what you need to find is two number that multiply to the last term and add to the the middle coefficient. So that means that to factor the above polynomial all you need to do is find two numbers that multiply to -10 and add to 3, and that would be 5 and -2, and that tells us that it will factor into

(x+5)(x-2).

Of course we could cheat this time, because we already knew the answer, but for many polynomials it wouldn't be too hard even if we didn't, but there is a little bit of guess work here, so it may take a little practice to get used to it. One thing you can do if you get stuck is to list all of the ways to factor the last term and then play around with them until you find one that adds up to the right thing. You can even simply try all of them, which might seem like a lot of work, but by doing it you will get a feel for how it works and before too long you won't really have to try all of them. Let's say you wanted to factor

x²+17x+60.

This one is a little bit tricky, because there are a lot of ways to factor 60. To get an idea about what happens with these, lets list them all.

1x60
2x30
3x20
4x15
5x12
6x10

1+60=61, no good.
2+30=32, no good.
3+20=23, no good.
4+15=19, closer, but still no good.
5+12=17, eureka, we've found one.

This means that we can factor the polynomial as

(x+5)(x+12).

One way to avoid simply trying them all is to do some estimating and also notice some patterns. One pattern that is helpful is that when the numbers are farther apart, like 60 and 1, they add to bigger numbers and when they are closer together, like 6 and 10, they add to smaller numbers. There are some other tricks you can play as well. One I find handy sometimes is swapping prime factors. Suppose in this one you guessed 6x10 and say that it was 16, which is just a little too small. Since it is too small, what you want is to factor it so that the factors are a little farther apart. Sometimes you can come up with this by prime factoring the two numbers and then switching some factors.

6=2x3
10=2x5

One way to make these farther apart would be to switch the 3 and the 2 and get

2x2=4
3x5=15,

but that gives us a sum that is too big. The way to do it that does work is to give the 10's 2 to the 6 and get

2x2x3=12
5=5.

It is also important to consider the signs of the coefficients and what happens when you multiply and add plus and minus numbers. The simple case is when both coefficients are plus like the one above where you only need to use positive numbers. The case where the middle coefficient is negative and the last terms is positive is almost as easy. What you need then is two numbers that multiply to a positive number and add to a negative number. The only way to get that is if they are both negative. Then because when you add two negative numbers you add and your answer is negative, you can think of it pretty much the same way you do when they are both positive. The only thing you have to do different is make them both negative instead of both positive. So for example, suppose you want to factor

x²-5x+6.

Then you still just need to find two numbers to multiply to 6 and add to 5, which would be 2 and 3, but since we really want -5, we would use -2 and -3 and factor it

(x-2)(x-3).

If the last term is negative it is a little bit different. Then regardless of what the sign of the middle coefficient is, one of your numbers has to be positive an the other negative to multiply to a negative number. That means that when you add those numbers, in terms of positive number arithmetic, it is a subtraction. When you add a positive and a negative number you subtract and use the sign of the larger one, so you need to think in terms of subtracting to get the middle term, and if the middle coefficient is positive you want the positive one to be bigger, but if it is negative, you want the negative one to be bigger. For example if we wanted to factor

x²+x-12,

we could do it by finding two numbers to multiply to 12 and subtract to 1. That would be 4 and 3. Then since the middle coefficient is 1 and not -1, we would use 4 and -3 and factor the polynomial as

(x+4)(x-3).

On the other hand, if we wanted to factor

x²-x-12,

we could think about it the same way, what two numbers multiply to 12 and subtract to 1 and get 4 and 3, but since this time we want a -1, we would use -4 and 3 and get

(x-4)(x+3).

Next we turn to the more difficult problem of when the first coefficient is not 1. Here the guessing is a little more complicated, because the coefficient you get from the inners and outers will not be just the sum of the two constants, because they will get multiplied by the x coefficients. For example, lets say we want to factor

3x²+11x+10.

We need to make the product of the firsts give us 3x². In this problem that isn't too difficult, because there is only one way to do it, and that is 3x and x, so we want a factorization of the form

(3x )(x ).

Now, the product of the lasts still has to be 10, but it is not their sum that has to be 11, it is something a bit more complicated, because in the sum of the inners and outers one of the numbers will be multiplied by 3. For example, lets suppose we tried 2 for the first one and 5 for the second one. Then it would be

(3x+2)(x+5).

That wouldn't work, because the sum of the inners and outers wouldn't be 11x. Instead it would be 15x+2x=17x. The first and last terms would be right, but that is not good enough. But if we put the 5 in the first factor and the 2 in the second factor, it would work, since then it would be

(3x+5)(x+2)

and the sum of the inners and outers would be 6x+5x, which is indeed 11x. But how are you to come up with that arrangement other than trial and error? To some extent you can't, you just get better at the trial and error with practice. Again, you could if worse came to worse try all possible combinations. For most of the problems you will be expected to do in algebra classes there won't be that many possible ways to factor the last term anyway. For 10 it is just 1x10 and 2x5, but here this makes a few more possibilities than last time, because the order will matter. But still for this problem there are only 4 possibilities,

(3x+10)(x+1),
(3x+1)(x+10),
(3x+2)(x+5),
(3x+5)(x+2).

In the beginning it won't hurt you to just try them all and see what happens. Again, this can be quite helpful in getting a feel for what happens with this. It sounds like a long process, but after you do it a few times, you will find that you can eliminate a lot of possibilities without fully trying them. There are again some techniques you can use involving estimating sizes. In general you will get larger numbers by using factors that are farther apart and by placing the numbers so that the large numbers multiply large numbers and the small numbers multiply small numbers and you will get smaller numbers by doing the reverse. So for example, here (3x+1)(x+10), where the 10 is multiplied by the 3 and the 1 is multiplied by the 1, gives us the largest possible middle term and (3x+5)(x+2), where the 5 is multiplied by the 1 and the 2 is multiplied by the 3, gives us the smallest possible middle term. Also in these problems you can eliminate some possibilities right away by estimating. If I was approaching this problem I would see right away that I couldn't put the 10 so that it multiplied the 3, because the outers alone would be too big. I would even be pretty suspicious of using the 10 at all, because even 10 times 1 is almost 11, so there wouldn't be too much left for the other term. So I would guess it would be 5 times 2, and putting the 5 where it gets multiplied by 3 gives 15 which is already too big on its own, so there is really only one possibility that has a chance. With a little practice you may find it surprising that you will be able to guess right the first time a good amount of the time, and when your first guess isn't right, usually your second one is. I find when doing these that sometimes the right solution just sort of appears before me magically, but I have been doing these in front of classes for a lot of years. Just the same, with practice you may find that this happens to you more and more often.

More Examples

For these examples the instruction is to factor the polynomial.

Example 1:

2x²+9x-5

Solution:

(2x )(x )

There is only one way to factor 5, 5 and 1, so those are the numbers we are going to use. The only question is where to put them. Here since the last term is minus, one of the numbers must be minus and the other plus. That means that the inners and outers will subtract instead of add. Since 9 is large, particularly since it will be a difference we want to split things unevenly, so the best guess is to place the 5 so that it gets multiplied by the 2 and the 1 so it gets multiplied by 1.

(2x 1)(x 5)

Since we want to get a positive 9, we want the one that gives us the bigger product to be positive and the one that gives us a smaller product to be negative. That gives us

(2x-1)(x+5).

And that should work, because we get 10-1=9. The firsts and lasts will work without problems. Check again to make sure the inners and outers do in fact work. We get

10x-1x=9x,

so it works.

Example 2:

6x²-31x+35

Solution:

This one is a bit harder, because we have a choice for the firsts. We can use 6x and x, or 2x and 3x, and we don't really know which one will work. Sometimes it is possible to use the relative sizes of the middle and last terms to help us guess, but there isn't really any way to know for sure. Mainly we just have to try one, and see if we can find anything for it, and if we can't try the other. For this one, I'm not really sure at all which one it is going to be, so let's try 6x and x first and see what happens.

(6x )(x )

We need to multiply to get 35, so it must be 1 and 35 or 5 and 7. There are some possibilities that we can eliminate right away. Since the last term is plus, the signs of the two numbers will be the same, so for the middle term we will be adding. The only difference the minus on the middle term makes is that both of the numbers will be minus. We can't put the 7 or the 35 so they get multiplied by the 6, since that would be already too big. Even putting the 5 so that it gets multiplied by the 6 doesn't look like a good idea, because that only leaves 1 for the inners, but that is really the only possibilities with this set up, since actually we can't use the 35 at all, since it alone is bigger than 31. If we put the 5 with the x, so that it would get multiplied by the 6, then we would have to put the 7 with the 6x to be multiplied by the x. In case you have gotten lost, that would be

(6x+7)(x+5).

But without even precisely working that out, we can tell that this would be too big, since 7 is bigger than 1. So we have come to the end of the line with 6x and x and must see what we can get with 2x and 3x.

(2x )(3x )

Okay, we have already figured out that there is no way that we can use 35 and 1, because the 35 alone would be too big, so we have to use 5 and 7. Let's just try both ways to see what happens.

(2x-5)(3x-7)

This gives us 14+15=29, close, but that's not good enough.

(2x-7)(3x-5)

This give us 21+10=31, which works!!

Notice that without too much difficulty we eliminated all but two possibilities, by some simple estimating, and even if you had to try the two remaining possibilities simply by trial and error, that wouldn't be that bad. With some practice, maybe you wouldn't even have to do that. I might guess the right answer on this one right away, because of seeing that if I put the 7 to be multiplied by the 3, I will get 21, which is 10 off from what I want to get, and I know at least that 10 is divisible by the other numbers, so there is a chance that they might multiply to 10. After you do these for a while you sort of pick up on patterns like that. Another thing you can do is to save time when you are trying the numbers out, don't bother with the signs, and sort those out after you see that the numbers work, and of course don't bother with the variable either, because you know that will be there. So in this correct arrangement you would simply figure 3x7 for the inners and 2x5 for the outers, and check to see if that adds up to your middle coefficient.

Example 3:

4x²-5x-6

Solution:

For this one again we have to be careful, because we get a choice between 4x and x or 2x and 2x. This time I am going to deliberately choose the wrong one, because I know the answer to this problem, because I cheated by multiplying the factors to get the problem. Let's try 2x and 2x, since I know it is really 4x and x.

(2x )(2x )

Since the last term is minus, we will be looking for the middle term to be the difference of the outers and inners. 6 can be factored 6x1 and 2x3 and we need a difference of 5, but remember, that includes multiplying by the 2's. This time since the coefficients on the x's are the same, it won't matter which order we put them in. 6 and 1 will give us 12-2=10 and 3 and 2 will give us 6-4=2, so neither will work. If you are clever about it you don't have to go through all that to see it, because since both numbers are getting multiplied by 2, you will be getting the difference of two even numbers, which must be even, and 5 is odd. You may not be able to precisely explain it like that, but you still might be able to intuitively look at it and see that the inner outer difference just doesn't look like it could come out to 5. So any way we do it, we can see that 2x and 2x, won't work, so we will have to try the other possibility 4x and x.

(4x )(x )

6 and 1 just looks like it won't work. One way the difference looks too big and the other way it looks too small, so lets try 2 and 3. We are looking for a difference of 5. If we put the 3 so it gets multiplied by the 4, that looks like it will make the difference too large, since the 3x4 part is 12 and the other part is small, so it looks like the other way around is the best choice. Let's try it.

(4x 3)(x 2)

Write the numbers in first without signs, because that is quicker and we can adjust the signs later. This gives us 8-3=5, so that looks good. Now adjust the signs. We want it to be a -5, so we want the negative one to be bigger. Now here we have to be careful, because it is not the bigger of the two numbers that should be negative it is the one that is bigger after multiplying by the x coefficients. The 2 is smaller than the 3, but it is getting multiplied by the 4 for make 8 and the 3 is multiplying by 1 to get 3, and it is these products that we are adding together. So we want to get -8 and +3, so that they add up to -5, so that means that even though the 2 is smaller, it is the 2 that will get the minus sign, so the final answer is

(4x+3)(x-2).

Let's check that to make sure. For the firsts we get 4x² as planned. For the outers we get -8x. For the inners we get 3x. And for the lasts we get -6. Putting that together we get

4x²-8x+3x-6=4x²-5x-6,

so it works!!

With some practice, even though this method may seem a bit fiddling at first, you should be able to get good enough at it so that you can manage most problems. But in case you have one where both the first and last terms can be factored a lot of different ways and you get stuck, there is another method based on grouping you can use. It is longer, but involves less guesswork. Because it involves less guesswork it can be tempting to give up on the guesswork and only use this method, but I don't think this is a good idea. It is better to save this method for problems where you get really stuck doing it the other way, because otherwise you won't get the practice that it takes to get good at doing it the other way. In other words, if I teach you this other way, you must promise not to abuse it. So read this part only if you can make that promise.

Okay, now that I know I am only talking to those who are ready for this, here is the trick. We multiply the first and last coefficients and then play the game we played for ones with a first coefficient of 1, but with this number instead of just the last term. We take that number and look for two numbers that will multiply to it and add to the middle coefficient. Then we use those numbers to break up the middle term, and if we do that it turns out that the 4 term polynomial we get will always factor by grouping. This might be a bit hard to grasp without examples, so I will show you how to use this method for the above three examples.

Example 1:

2x²+9x-5

Alternate Solution:

Multiply the 2 and the -5 to get -10

Now find two numbers that multiply to -10 and add to 9.

That would be 10 and -1

Now use these numbers to break up the middle term and the polynomial becomes

2x²+10x-x-5.

It doesn't matter what order we put them in. We could use -x+10x as well, and in the end we would come out with the same answer.

Now I claim we can factor this by the grouping method. For the first two terms take out a common factor of 2x and for the second two terms take out a common factor of -1, and we get

2x(x+5)+-1(x+5)

and look, isn't that amazing, we get a common factor of x+5 that we can take out and we get

(2x-1)(x+5),

the same thing we got before.

Example 2:

6x²-31x+35

Alternate Solution:

Multiply 6 times 35 to get 210.

Find two numbers to multiply to get 210 and add to get 31.

Not quite as easy as last time, but -21 and -10 will do.

Now use these numbers to break up the middle term and we get

6x²-21x-10x+35.

Now factor this by grouping. Take out a common factor of 3x from the first two terms and -5 from the second two terms and this becomes

3x(2x-7)-5(2x-7)

Again the leftover parts are both the same, this time 2x-7, so we can take that out as a big common factor and get

(3x-5)(2x-7).

Example 3:

4x²-5x-6

Alternate Solution:

Multiply the 4 and the -6 to get -24.

Then find two numbers that multiply to -24 and add to -5.

-8 and 3 will work.

Use these numbers to break up the middle term and we get

4x²-8x+3x-6.

Now factor this by grouping. Take out a common factor of 4x from the first two terms and a common factor of 3 from the second two and we get

4x(x-2)+3(x-2)

Then take out x-2 as a factor to get the answer of

(4x+3)(x-2).

Just make sure to keep an open mind. It is never a good idea in mathematics class to get stuck on one way of doing something.

Difference of Squares

There is a special formula for factor a difference of squares that comes from applying a shortcut multiplication formula backwards. It turns out that whenever you multiply the sum and difference of the same things, the inners and outers cancel out and you only get two terms, and this gives you the formula

(A+B)(A-B)=A²-B² .

To use the formula for factoring we need to write it backwards as

A²-B² =(A+B)(A-B).

So if we see something that is a difference of squares, we can factor it as the sum and difference.

Example:

x²-9=(x+3)(x-3)

Sometimes we can use the formula more than once.

x⁸-1=(x⁴+1)(x⁴-1)=(x⁴+1)(x²+1)(x²-1)=(x⁴+1)(x²+1)(x-1)(x+1)

This brings up another point. Factoring is like prime factoring of numbers. Always factor completely, that is keep factoring the factors until nothing more will factor. The ones that can't factor come along for the ride. Sums of squares can't be factored, so they come along for the ride.

Perfect Squares

Sometimes we can also use the squaring formulas,

(A+B)²=A²+2AB+B²
(A-B)²=A²-2AB+B²

in reverse for factoring. If you see a polynomial with three terms and the first and last terms are perfect squares, you can check and see if the middle term is right for this kind of factorization.

Example:

x²+14x+49

The first and last terms are perfect squares. If this were to fit into the pattern of the first of these formulas, it would have to be

(x+7)²

The first and last terms are right for that. Is the middle term right. The middle term should be twice the product of 7 and x, which is 14x, and yes, that is what we have, so this factorization works. For this problem we could have also used the method above for general quadratics, but for some, particularly like ones with fractions, we can get it easier this way.

Sum and Difference of Cubes

There are special formulas that need to be memorized for factoring the sum and difference of cubes.

A³+B³=(A+B)(A²-AB+B²)
A³-B³=(A-B)(A²+AB+B²)

Notice that although it isn't possible to factor a sum of squares, it is possible to factor a sum of cubes. It turns out in general that it is always possible to factor sums of odd powers, but not even powers. The way I always remember these formulas is that the first factor is the original without the cubes and the second factor is almost like the squaring formulas, but without the 2 and the sign on the second term of it is always the opposite of the sign in the original and in the first factor. In other words these formulas almost look like

A³+B³=(A+B)(A-B)²
A³-B³=(A-B)(A+B)²

except that there is no 2 on the middle term. Also when you are factoring, once you have written down the first factor, you have A and B handy, so you can easily read them off from it for writing down the second factor.

Example 1:

x³+8

Solution:

Since the first term is x cubed and the second term is 2 cubed, this is a sum of cubes. For the first factor we just remove the cubes and write down

(x+2).

Then for the second factor, just take these two terms that are written down in the first factor and plug them into the formula for the second factor, which is just like the formula for the square of x-2, except that you don't double for the middle term, the middle term is just the product of the two terms, so we square the x and then take the product of 2 and x and then square the 2 and get

(x+2)(x²-2x+4).

Example 2:

x³-27

Solution:

This one is a difference of cubes, where A is x and B is 3, so for the first factor we get

(x-3).

Now for the second factor we just plug these two numbers into the formula for the second factor. Square the x for the first term, multiply 3 times x for the second term and square the 3 for the last term and we get

(x-3)(x²+3x+9).

Sum and Difference of Higher Powers

There are formulas for sums and differences of higher powers, but mostly people don't bother to learn them. You can either look them up, or work them out by long division. But some higher powers can be done by using the difference of square and sum and difference of cubes formulas, namely any power that is a multiple of 2 or 3. Any power that is a multiple of 2 is a perfect square and any power that is a multiple of 3 is a perfect cubes. This is true because powers to powers multiply. So for example x¹⁶ is a perfect square, because it is (x⁸)² and x²⁴ is a perfect square and a perfect cube, since it is (x¹²)² and (x⁸)³. So you can factor differences of even powers by using the difference of squares formula and sums and differences of powers that are multiples of 3 by using the sum and difference of cubes formulas. And sometimes you can factor more than once this way. Sometimes when you approach it two different ways it can look like you get different answers. An interesting example of that is when you have a difference of 6th powers where you can treat it either as a difference of squares or a difference of cubes.

Example:

x⁶-1

Method 1:

Look at it as a difference of squares. (x³)²=x⁶, 1²=1.

x⁶-1=(x³-1)(x³+1),

but we can go further, because each of these factor can be factored using the sum and difference of cubes formulas.

x³-1=(x-1)(x²+x+1)

and

x³+1=(x+1)(x²-x+1),

so putting this all together, we get

x⁶-1=(x³-1)(x³+1)=(x-1)(x²+x+1)(x+1)(x²-x+1).

Method 2:

Look at it as a difference of cubes. (x²)³=x⁶, 1³=1.

x⁶-1=(x²-1)(x⁴+x²+1)

The first factor can be factor further as a difference of squares, and we get

(x-1)(x+1)(x⁴+x²+1),

but it looks like that is all we can do, because it is hard to see how to factor the other factor.

But we should get the same answers with both methods, because after all, they are equal, so it must be true that

x⁴+x²+1=(x²+x+1)(x²-x+1),

but it is not at all clear how you would figure that out if you did it that way. For this reason it is generally better when you have problems like this where there is a choice to look at it first as a difference of squares.

But there is a way you can get the complete factorization the other way, and it involves a fancier factorization that you can use for some other problems as well. I don't know any official name for it, but I call it completing the square from the inside, because it involves adding and subtracting something that will change the middle term so that the first chunk is a perfect square, and then factoring it as a big difference of squares.

Completing the Square from the Inside (Extra for Experts)

This is a method of factoring that we can use when we have a 4th degree polynomial that has no odd terms, and we can't factor simply by treating x² as the variable. For some polynomials of this sort we can factor them much easier by treating x² as the variable. For example if we want to factor x⁴-5x²+6, we can do that. We just ask ourselves what two numbers will multiply to 6 and add to -5, and since that is -2 and -3, we can factor it to (x²-2)(x²-3), just like if it were x²-5x+6. But it turns out that if we have one like that where we can't find numbers to multiply to the last term and add to the middle coefficient, we can often still factor them by using this method of completing the square from the inside.

Example:

x⁴+x²+1

We can't factor this one like the last one, because we can't find two numbers to multiply to 1 and add to 1. What we do with a problem like this is try to find something to add to the middle term so that the polynomial becomes a perfect square, and we add and subtract it. By doing this we can turn it into a difference of squares. In this case if there was a 2 coefficient on the x², we would have x⁴+2x²+1, which would be a perfect square, so what we do is add x² and subtract it, to get

x⁴+2x²+1-x²=(x²+1)²-x²

If we couldn't see by guessing what the coefficient would need to be for this we could get it by doing the opposite of what we normally do for completing the square, take the square root of the constant term, and then double that, so here this would be

sqrt(1)=1
2·1=2.

Then to get that coefficient we figure out what multiple of x² we need to add to the existing term to get it. In this case since we already have 1, we need to add another 1 to get 2.

After we do this we get a big difference of squares, so we can use the difference of squares formula to factor it and get

[(x²+1)+x][(x²+1)-x].

Then getting rid of the extra parentheses and writing things in standard order this becomes

(x²+x+1)(x²-x+1),

just like we got from the other method.

Here are a couple more examples of this method.

Example 1:

x⁴+2x²+9

Solution:

sqrt(9)=3
2·3=6

So here the middle term needs to be 6. Since it is 2 now, we need to add and subtract 4x². Then we get

x⁴+2x²+9=x⁴+6x²+9-4x²=(x²+3)²-4x²=[(x²+3)+2x][(x²+3)-2x]=(x²+2x+3)(x²-2x+3).

Example 2:

x⁴-19x²+25

Solution:

Here if the middle coefficient were a -10, then it would be a perfect square, so add and subtract 9x².

sqrt(25)=5
2·5=10

x⁴-19x²+25=x⁴-10x²+25-9x²=(x²-5)²-9x²=[(x²-5)+3x][(x²-5)-3x]=(x²+3x-5)(x²-3x-5).

Tips

1. First look for a common factor.
2. Always factor completely. This is like prime factoring.