# Factors of a polynomial

1. Feb 23, 2007

### danago

Given that $$x^5 - 2x^4 + 2x^3 - 4x^2 - 3x + 6$$ is exactly divisible by $$x^2 + 3$$, find the remaining real factors of the polynomial.

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I could easily do that question using a calculator, just by graphing it and finding the roots. Another thing i can do is multiply the factor by the general cubic equation, expand the brackets, and then compare the coefficients. But then that leaves me with the polynomial in terms of the product of only 2 of its factors.

If i was given a question like this in a test, how can i show working? How can i do it without even touching a calulator?

Thanks,
Dan.

Last edited: Feb 23, 2007
2. Feb 23, 2007

### dextercioby

Did you find the ratio, that 3-rd order polynomial ?

3. Feb 23, 2007

### danago

$$x^5 - 2x^4 + 2x^3 - 4x^2 - 3x + 6= (x^2 + 3)(x^3 - 2x^2 - x + 2)$$

Is that what you mean?

4. Feb 23, 2007

### Gib Z

Rational Root Theorem. Look it up. After that its a nice quadratic.

And yes, thats what he meant. O btw, graphing on your calculator does not give complex roots.

5. Feb 23, 2007

### danago

Hmm never heard of the rational root theorem. Looks good though. So according to that, the potential roots to my cubic are 1 and 2. On testing them, it appears they are both roots. So then i can use one of those roots and use the same coefficient comparison method on the cubic, and keep factorizing it like that, until i have it as the product of linear factors?

Is that how you would have done it?

Thanks for the help by the way.

6. Feb 23, 2007

### dextercioby

You can factor the cubic w/o knowing the theorem.

$$x^3 -2x^2 -x +2 = x^2 (x-2)-(x-2)=(x-2)(x^2 -1)=(x-2)(x-1)(x+1)$$

Sometimes it works, sometimes it doesn't. When it doesn't, you can apply the theorem. A cubic polynomial always has a real root.

7. Feb 23, 2007

### danago

Ahh ok. Never thought about doing that. Thanks :)

8. Feb 23, 2007

### dextercioby

Actually, the theorem concerns polynomials with "integer ($\in\mathbb{Z}$) coefficients" and consequently all integer (so both the negative) divisors of the "free term" could be roots.

9. Feb 25, 2007

### Feldoh

The theorem provides a complete list of possible rational roots of the polynomial equation: $$a_{n}x^n + a_{n–1}x^n–1 + ··· + a_{2}x^2 + a_{1}x + a_{0} = 0$$ where all coefficients are integers.

This list consists of all possible numbers of the form p/q, where p and q are integers. p must divide evenly into the constant term $$a_{0}$$. q must divide evenly into the leading coefficient $$a_{n}$$.

In the simplest terms, say you have:
$$1x^3-2x^2-x-2$$

You take q, the 1 in $$1x^3$$, and find all the integer factors of it, in this case, ±1.

Then you take p, the constant term in the cubic, in this case 2. Once again find all the integer factors of p: ±1, ±2.

Now, all possible rational roots of the equation is defined as p/q, or: $$\pm 1, \pm 2$$

Sure enough the roots are $$(x-2)(x-1)(x+1)$$

Last edited: Feb 25, 2007
10. Jul 8, 2010

### Domun3000

take the question and start with P(x)=the question. this is an easy question that requires 3 steps. Step 1: find a zero; Step 2: divide the question with the zero using Remainder or factorization therom; Step 3: factorize the quotiont. il show you

P(x)=$$x^5 - 2x^4 + 2x^3 - 4x^2 - 3x + 6$$

Then subsutue x with +1. you should get a zero.
take x-1(By the factorization therom) and divide by the original question
now take the quotion( if done right, their should be no remainer)and factorize

you should get all the zeros of the question.