# Write each polynomial as the product of it's greatest common

1. Jul 20, 2013

### MathJakob

1. The problem statement, all variables and given/known data
Write each polynomial as the product of it's greatest common monomial factor and a polynomial

2. Relevant equations
$$8x^2+12x$$
$$6a^4-3a^3+9a^2$$

3. The attempt at a solution
$$4x(2x+3)$$
$$a^2(6a^2-3a+9)$$

I really don't understand what it's asking me for, how can I factor this more?

2. Jul 20, 2013

### QuantumCurt

The second equation can be factored a little further. The three terms still have one more common factor.

Are these two separate problems? It sounds like it just wants you to factor out the greatest common factor.

3. Jul 21, 2013

### MathJakob

$$4x(2x+3)$$ I can't see anymore common factors...

$$a^2(6a^2-3a+9$$ also has no common factors, if it were a +3 instead of a -3 then I could do this $$6a^4-3a^3+9a^2$$
$$3a^2(2a^2-a+3)$$

hmm can you give me a pointer? I assume I have to get rid of the remaining squares?

4. Jul 21, 2013

### verty

What about $8x(x + 3/2)$? No one said anything about needing the polynomial factor to have integer coefficients. The question is ambiguous; I would skip it.

One more thing, if x is negative, the factor 4x is smaller than the factor 2x.

5. Jul 21, 2013

### HallsofIvy

Staff Emeritus
I disagree with verty about $8x(x+ 3/2)$ being a solution. In factoring problems, it is generally understood that the coefficients are to be integers in order to have a unique solution.

I have no idea what "+3" or "-3" you are talking about. $3a^2(2a^2- a+ 3)$ is a perfectly good solution to the original problem.

6. Jul 21, 2013

### MathJakob

"Write each polynomial as the product of it's greatest common monomial factor and a polynomial"

So this answer satisfies the original question?

7. Jul 21, 2013

### QuantumCurt

That should do it. The negative doesn't make a difference in this case for the purposes of factoring. What's important is that the two expressions remain equal. If you now distribute the 3a^2 back into the parentheses, you will end up with exactly the same expression as the original problem. You just have to make sure to keep your signs straight when you're factoring.

I'd also agree that it should remain in integer form. Typically, "simplest form" of an expression is going to involve integer coefficients. There's no rule that you can't factor fractions out, but in general, you don't unless you have to.

8. Jul 21, 2013

### micromass

Staff Emeritus
I agree with verty. If the question doesn't specifically state that you want to factor it as integer polynomials, then any factorization goes. A question should state this.

9. Jul 21, 2013

### QuantumCurt

I agree. A fractional factorization certainly wouldn't be "wrong." However, for simplicities sake, I'd imagine a factorization with integer coefficients would be preferred.

You could take it further though.

$$6a^4-3a^3+9a^2$$
$$3a^2(2a^2-a+3)$$
From this point, you could still factor a 2 out.

$$6a^2(a^2-\frac{1}{2}a+\frac{3}{2})$$

But, that strikes me as unnecessary.

10. Jul 21, 2013

### Staff: Mentor

I am quite sure that the solution is supposed to have integers only.
Even more, I don't think those other proposed ideas are solutions: They assume an additional structure which is not given in the problem statement. You can study polynomials outside of real and complex numbers. It could be that 1/2 or 1/3 are not even well-defined there.

11. Jul 21, 2013

### micromass

Staff Emeritus
Although I agree, I think the problem statement should have clarified this. Right now we're only guessing.

12. Jul 21, 2013

### Staff: Mentor

If you are asked to factorize 12, do you write it as 12=2*2*2*3*(1/2) or 12=2*2*2*(3/2)?

13. Jul 21, 2013

### micromass

Staff Emeritus
It's rather clear that we work in the integers there. If we work with polynomials, then it is not at all clear to me where we work in. There are many very natural rings to work in. Unlike factorizing 12, where there is really only one interesting ring.

14. Jul 21, 2013

### QuantumCurt

I agree that they are likely just asking for integer factorization. It's the simplest and clearest factorization of the given polynomials.
$$6a^2(a^2-\frac{1}{2}a+\frac{3}{2})$$

We could factor another 2 of it from this point...

$$12a^2(\frac{1}{2}a^2-\frac{1}{4}a+\frac{3}{4})$$

But, that seems unnecessary, and we could continue taking it infinitely further if we wanted...but that seems silly, given the problem statement.

What class is this for? If it's a pre-calc algebra class, it's likely safe to assume that they're asking for integer factorization.

15. Jul 21, 2013

### verty

I apologize, I didn't see the significance of the word "common" in "greatest common monomial factor", I read it as "greatest monomial factor common to each term", equivalent to "greatest monomial factor", equivalent to "greatest monomial that divides evenly". On this reading, the question seemed vague to me.

If it had said "greatest common monomial divisor", I would have seen the connection. I would have assumed that the coefficient of the monomial is the GCD of the coefficients and the index is the minimum of the indices.

16. Jul 21, 2013

### MathJakob

It isn't for any class, it's just from a Algebra II + Trig eBook that I'm studying from. I'm learning math by myself as a hobby.

17. Jul 21, 2013

### QuantumCurt

Gotcha. Within the context of algebra II, we can basically be certain that they're asking for integer factorization. The Greatest Common Monomial Factor would imply the greatest factor that is common to all of the terms in the given polynomial. In the case of the first expression, the greatest common monomial factor is $2x$, and in the second expression, it is the $3a^2$