# Can't figure out a simple factorization

1. Jan 2, 2008

### nanoWatt

I'm doing some basic factoring now to refresh my skills.

On one problem, it is asking to factor: $$(y^2 + 8)^2 - 36 y^2$$

I was able to get to where I factor the trinomials: (y-4)(y-2)(y+4)(y+2)

When it says to give the summary of factorization, it gives: $$(y^2 + 8)^2 - (6y)^2$$

I'm going through a tutorial and this last step is the one I can't figure out.

For some reason, I thought that (y-4)(y-2)(y+4)(y+2) = $$(y^2 + 8)^2$$

but instead: (y-4)(y-2)(y+4)(y+2) = $$(y^2 + 8)^2 - (6y)^2$$

My question is how the extra $$(6y)^2$$ comes about.

Last edited: Jan 2, 2008
2. Jan 2, 2008

### arildno

"(y-4)(y-2)(y+4)(y+2) = (y^2 +8)^2

Why on earth should this be correct??

Whereas the left hand side has 4 zeroes, the righthand side has no zeroes.
Therefore, they cannot be identical.
Do you understand that?

3. Jan 2, 2008

### nanoWatt

What do you mean by 4 zeros? (I've been out of college for 7 years so I've forgotten quite a bit).

Does it mean that constants (like -4, -2) are zeros?

I guess I'm having trouble with reverse factoring. I just see this as a pair of squares.

4. Jan 2, 2008

### dst

From the start - compare:

(y^2 +8)^2 - 36y^2 --- (1)

And the book also gives:

(y^2 +8)^2 - (6y)^2 --- (2)

Since you know these have the same factors multiplied out then you know (1) = (2).

In any case, note that (6y)^2 = (6y)(6y) = 6*6*y*y = 6^2y^2 = 36y^2 as seen at the start.

The 6y comes straightaway when you multiply the first 2 brackets (y-4)(y-2), and it should eventually become (6y)^2.

5. Jan 2, 2008

### dodo

I ocassionally have problems with factorizations. Maybe I could tell you how would I do that one (it is easy), if it is of any help.

I would try to 'unwrap', or 'flatten', the whole expression, to get it into the form of a polynomial, ax^4 + bx^3 + cx^2 + dx + e; that is, get rid of the parenthesis, so that you can group similar exponents together. (This way at least you would get a different feeling of the expression, what its degree is, or whatever.) In your case, once you group similar exponents the answer should be pretty straightforward.

• Checking for two numbers that multiplied give C and added give B, when you have an expression like x^2 + Bx + C. If those two numbers are p and q, then you know the factorization is (x + p)(x + q). The numbers p and q can be positive or negative.
• "Completing the square"; not always leading to a complete factorization, but very handy. (You can google for that term.)
• Multiplying/dividing the whole expression by some constant, for example if you need to get rid of one of the coefficients; then you may apply other techniques, and then restore back the factor that you removed (if you divided by 3 and the did some trick, you must multiply by 3 at the end to restore the original problem).

What Arildno was saying above about some zeroes, is that (y-4)(y-2)(y+4)(y+2) becomes 0 when y = 4, 2, -4 or -2; while the right side (y^2 +8)^2 is always positive no matter the value of y; so they cannot be equal.

Hope this helps.

Last edited: Jan 2, 2008
6. Jan 2, 2008

### HallsofIvy

Staff Emeritus
He means that y= 4, y= -4, y= 2 and y= -2 all make one of the factors of (y-4)(y-2)(y+4)(y+2) equal to 0 and so make the entire polynomial 0. On the other hand y^2+ 8 cannot be smaller than 9 and so (y^2+ 8)^2 is never 0 for any value of y. They can't possibly be the same!

You said you see $(y^2+ 8)^2- 36y^2$ as "pair of squares". I assume you are talking about the "difference of square" formula: $a^2- b^2= (a+ b)(a- b)$
Okay, $(y^2+ 8)^2- 36y^2$ is of that form with $a= x^2+ 8$ and $b= 36y^2$. That means $a= y^2+ 8$ and $b= 6y$
$(y^2+ 8)^2- 36y^2= (y^2+ 8+ 6y)(y^2+ 8- 6y)$. You can then factor $y^2+ 6y+ 8= (y+4)(y+2)$ and $y^2- 6y+ 8= (y- 4)(y- 2)$.

7. Jan 2, 2008

### symbolipoint

You want to understand factoring trinomials AND recognize the difference of squares.

$$$\begin{array}{l} ((y^2 + 8) + 6y)((y^2 + 8) - 6y) \\ (y^2 + 6y + 8)(y^2 - 6y + 8) \\ (y + 2)(y + 4)(y^2 - 6y + 8) \\ \end{array}$$$