Do you have a question you would like to ask Professor Puzzler? Click here to ask your question!

*"Good day sir, pls sir I need explanation on how to solve an equation like this x ^{4 }- 3x^{2} + 2. although I don't have problem on factorization but I don't know what to do when the power of "X" is greater than 2. looking for your reply Sir"*

I can give you several answers to this question. One is a short quick answer that works for this problem. The other answers are more detailed, and give you more information than you need to solve this specific problem. Let's start with the short answer.

## Short Answer

Notice that all the exponents in this expression are even. This means that we could do a substitution like this: y = x^{2}. Doing this substitution changes the expression to:

y^{2} - 3y +2

As you said, you know what to do when it's a quadratic (the highest exponent is 2). This factors as follows:

(y - 1)(y - 2)

But wait - you're not done, because the problem had x in it, not y, so you have to put the x's back in:

(x^{2} - 1)(x^{2} - 2)

Now that you've substituted x back in, you should realize that one of those can be factored:

(x - 1)(x + 1)(x^{2 }- 2). And that's your factorization. Technically, that's your factorization over rationals; if you're including irrationals, the second binomial can be factored as well:

(x - 1)(x + 1)(x - √2)(x + √2)

## Longer Answer

The method shown above doesn't just work if the exponents are 4 and 2; they also work if the exponents are 6 and 3, or 10 and 5, or any other such combination. If you wanted to get really ugly, I suppose it would also work if you had fractional exponents like 5 and 5/2. The method works if it's a trinomial (three terms) and the exponent of the highest degree term is twice the exponent of the middle term, and the lowest degree term is a constant. For example, consider this one:

x^{6} + 8x^{3} + 12

We'll substitute y = x^{3}.

y^{2} + 8y + 12

(y + 6)(y + 2)

(x^{3} + 6)(x^{3} + 2)

Of course, if you were factoring over reals, and you happen to know the rule for factoring a sum of cubes, you could factor this further. But since you mentioned not knowing what to do for exponents larger than two, we can save that for another day!

Similarly, if it was x^{10} +8x^{5} + 12, you would do the substitution y = x^{5}.

## Longest Answer

Again, this only works for trinomials under certain circumstances. But what if the expression doesn't match those criteria? Let's take an example:

x^{3} + 5x^{2} - 2x - 10

In this case, we can't do a substitution to simplify it, so we'll try doing some *grouping*. Since there are four terms, I decide to try breaking it into two groups of two terms each:

(x^{3} + 5x^{2}) - (2x + 10)

Now we can factor each group:

x^{2}(x + 5) - 2(x + 5)

Oh! Since both binomials in parentheses are the same, we can rewrite this as:

(x^{2} - 2)(x + 5).

This technique *sometimes* works. Sometimes grouping works only if you get creative in your groupings. Sometimes you need to rearrange the terms. Sometimes you might need to group them into groups of three terms, or groups of four terms.

Here's an example of rearranging terms:

x^{3} + 3x^{2} + 3x + 1

(x^{3} + 1) + (3x^{2} + 3x)

(x + 1)(x^{2} - x + 1) + 3x(x + 1)

(x + 1)(x^{2} + 2x + 1)

(x + 1)^{3}

## Confession Time

When I first saw your problem I did *not* come up with the simple solution. I wasn't paying attention, I guess, and my first thought was: I'll do this by grouping. I know. I made it harder than it really is. But even when you make a problem harder than it really is, sometimes you can come up with something interesting. In this case, I looked at your expression and rewrote it splitting the middle term into two pieces* like this:

x^{4 }- 3x^{2} + 2

x^{4 }- x^{2} - 2x^{2 }+ 2

(x^{4}^{ }- x^{2}) - (2x^{2 }- 2)

x^{2}(x^{2 }- 1) - 2(x^{2} - 1)

(x^{2} - 2)(x^{2} - 1)

(x^{2} - 2)(x - 1)(x + 1)

And I got the same answer, even though I took a silly, round-about method.

Hope all of this was helpful - or at least interesting!

* And if you're wondering, "Why in the world did you split it in that particular way?" The answer is, "I don't really know. I just had a gut feeling it would be helpful."

# Ask Professor Puzzler

Do you have a question you would like to ask Professor Puzzler? Click here to ask your question!**Over 3,000 Pages of Free Content**

We've been providing free educational games and resources since 2002.

Would you consider a donation of any size to help us continue providing great content for students of all ages?