How can I solve a polynomial problem without using a calculator?

In summary, the conversation discusses finding the remaining real factors of a polynomial that is exactly divisible by x^2 + 3. Various methods are suggested, such as using a calculator to graph the polynomial and find its roots, multiplying by the general cubic equation and comparing coefficients, or using the Rational Root Theorem to find potential roots. The theorem provides a list of possible rational roots based on the polynomial's coefficients, and these can be used to find the remaining factors of the polynomial. The conversation also includes a step-by-step explanation of how to find the remaining factors using the factorization theorem.
  • #1

danago

Gold Member
1,123
4
Given that [tex]
x^5 - 2x^4 + 2x^3 - 4x^2 - 3x + 6[/tex] is exactly divisible by [tex]x^2 + 3[/tex], find the remaining real factors of the polynomial.

__________________

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:
Physics news on Phys.org
  • #2
Did you find the ratio, that 3-rd order polynomial ?
 
  • #3
dextercioby said:
Did you find the ratio, that 3-rd order polynomial ?

[tex]
x^5 - 2x^4 + 2x^3 - 4x^2 - 3x + 6=
(x^2 + 3)(x^3 - 2x^2 - x + 2)
[/tex]

Is that what you mean?
 
  • #4
Rational Root Theorem. Look it up. After that its a nice quadratic.

And yes, that's what he meant. O btw, graphing on your calculator does not give complex roots.
 
  • #5
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
You can factor the cubic w/o knowing the theorem.

[tex] x^3 -2x^2 -x +2 = x^2 (x-2)-(x-2)=(x-2)(x^2 -1)=(x-2)(x-1)(x+1) [/tex]

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
Ahh ok. Never thought about doing that. Thanks :)
 
  • #8
Actually, the theorem concerns polynomials with "integer ([itex] \in\mathbb{Z} [/itex]) coefficients" and consequently all integer (so both the negative) divisors of the "free term" could be roots.
 
  • #9
dextercioby said:
Actually, the theorem concerns polynomials with "integer ([itex] \in\mathbb{Z} [/itex]) coefficients" and consequently all integer (so both the negative) divisors of the "free term" could be roots.
The theorem provides a complete list of possible rational roots of the polynomial equation: [tex]a_{n}x^n + a_{n–1}x^n–1 + ··· + a_{2}x^2 + a_{1}x + a_{0} = 0[/tex] 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 [tex]a_{0}[/tex]. q must divide evenly into the leading coefficient [tex]a_{n}[/tex].

In the simplest terms, say you have:
[tex]1x^3-2x^2-x-2[/tex]

You take q, the 1 in [tex]1x^3[/tex], 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: [tex]\pm 1, \pm 2[/tex]

Sure enough the roots are [tex](x-2)(x-1)(x+1)[/tex]
 
Last edited:
  • #10
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)=[tex]

x^5 - 2x^4 + 2x^3 - 4x^2 - 3x + 6
[/tex]

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.
 

What are the factors of a polynomial?

The factors of a polynomial are the expressions that can be multiplied together to get the original polynomial. For example, the polynomial 2x^2 + 4x + 6 has factors of (2x + 3)(x + 2).

How do you find the factors of a polynomial?

To find the factors of a polynomial, you can use a variety of methods such as factoring by grouping, using the quadratic formula, or using the difference of squares formula. It is important to first check if the polynomial has any common factors that can be factored out.

What is the difference between a factor and a term in a polynomial?

A factor is an expression that can be multiplied with other factors to get the original polynomial, while a term is a single number, variable, or combination of the two. For example, in the polynomial 4x^2 + 8x + 12, the factors are (4x + 6)(x + 2), while the terms are 4x^2, 8x, and 12.

Can a polynomial have more than one set of factors?

Yes, a polynomial can have multiple sets of factors. For example, the polynomial 6x^2 + 12x + 18 can be factored as (6x + 6)(x + 3) or as (2x + 9)(3x + 2). Both sets of factors will give the same result when multiplied together.

Why is factoring polynomials important?

Factoring polynomials is important because it allows us to simplify complex expressions and solve equations. It also helps us identify patterns and relationships between numbers and variables, which can be useful in solving more advanced mathematical problems.

Suggested for: How can I solve a polynomial problem without using a calculator?

Replies
36
Views
4K
Replies
22
Views
1K
Replies
9
Views
1K
Replies
15
Views
1K
Replies
23
Views
2K
Back
Top