If -2 is a root, then x - (-2) = x + 2 is a factor.shreddinglicks said:Homework Statement
-x^3 - 6x^2 -12x -8
Homework Equations
The Attempt at a Solution
I don't know, I just know the roots are -2 with multiplicity 3.
Mark44 said:If -2 is a root, then x - (-2) = x + 2 is a factor.
Do you know how to do polynomial long division? If you don't, do an internet search using that as the search phrase.
hilbert2 said:Multiply ##(x+2)^2## with ##-(x+c)## and deduce from the result what ##c## must be to produce the polynomial in the question.
shreddinglicks said:-x^3 - 6x^2 -12x -8
I will look into this.RPinPA said:It can be tedious if there are a lot of possible roots to try, but the Rational Root Theorem will tell you all possible roots.
Here's your polynomial:The RRT says that any rational root must be of the form +-p/q where p is a root of the constant term (8, ignoring the sign) and q is a root of the leading coefficient.
That means the only possible rational roots, are +-1, +-2, +-4, +-8. So you evaluate the polynomial at each of those values, one by one.
When you see a final answer like a third order zero at -2, you can assume that someone started with a simple final answer of integer roots and worked backward to get the problem statement. That kind of final answer is very unusual otherwise. Other techniques are usually needed to find the roots.shreddinglicks said:Thanks FactChecker, I was just curious. I'm studying for a test that will cover systems of linear ODE's. The book I was reviewing had a problem with that characteristic polynomial. I wanted to see if I could get the eigenvalues without a calculator.
FactChecker said:When you see a final answer like a third order zero at -2, you can assume that someone started with a simple final answer of integer roots and worked backward to get the problem statement. That kind of final answer is very unusual otherwise. Other techniques are usually needed to find the roots.
FactChecker said:People who make up in-classroom tests usually make sure that the answer is "rigged" to be solvable using techniques available to students. If it's a take-home test, anything goes and it may be a lot harder.
symbolipoint said:The example is a very typical one which occurs as material for College Algebra or Pre-Calculus, in which students will learn about Rational Roots Theorem and the Zeros of Polynomials.
You have the theorem which tells you the possible roots to examine are -1, -2, -3, -4, -8, 1, 2, 3, 4, 8.
First, you can pull out a factor of -1, so you have
-1(x^3+6x^2+12x+8).
You will find that the first possible root which gives 0 remainder in Synthetic Division is -2.
Keep on checking possible roots using synthetic division.
You should find that the ONLY roots which will work are -2;
and you will find that the multiplicity for this root is 3.
I will assume that this is a very clear solution to your request for help on this factorization.
In case you want to see the full result:
-1(x+2)^3
Do not wait. Look into BEFORE your examination so you have more chance to learn what you need to know how to do!shreddinglicks said:Thanks, I'll will certainly look into this when I have a little more time after my exam.
hilbert2 said:If you don't know the roots and have to solve this kind of problem, you have to try dividing the polynomial with ##x-c##, where ##c## is a number that the constant factor ##-8## is divisible with (the possibilities are ##\pm 1, \pm 2,\pm 4## and ##\pm 8##). For a cubic equation with integer coefficients and three real roots, a solution can always be found that way.
Ray Vickson said:You don't need three real roots. Furthermore, real roots cannot always be found that way: only integer roots (if any) can be found using the rational root theorem. A a cubic equation can have integer coefficients and one or three real roots, but with all roots irrational, needing either a complicated solution formula or a numerical solving method.
So the OP doesn't misunderstand, you mean for this equation only integer roots can be found because the leading coefficient is ##-1##. More generally, the rational root can indeed find the rational roots, if any, of polynomials with integer coefficients.Ray Vickson said:You don't need three real roots. Furthermore, real roots cannot always be found that way: only integer roots (if any) can be found using the rational root theorem. A a cubic equation can have integer coefficients and one or three real roots, but with all roots irrational, needing either a complicated solution formula or a numerical solving method.
The original poster for the question is assumed to be a college algebra student trying to get some help on dealing with polynomial functions and the Rational Roots Theorem. Usually all one needs is the discussion in the textbook. Just this topic, some students find hard enough.LCKurtz said:So the OP doesn't misunderstand, you mean for this equation only integer roots can be found because the leading coefficient is ##-1##. More generally, the rational root can indeed find the rational roots, if any, of polynomials with integer coefficients.
shreddinglicks said:I will look into this.
and that is dead wrong. It's called the Rational Root Theorem because it tells you all possible rational roots. If there are any roots that are rational, it has to be one of those. It gives no information about other kinds of roots, complex or irrational. Except that you know any that aren't rational have to be irrational or complex.RPinPA said:the Rational Root Theorem will tell you all possible roots.
shreddinglicks said:I don't know, I just know the roots are -2 with multiplicity 3.
I think you are misunderstanding what the OP meant in the same way I did in an earlier post. My new interpretation is that the OP knows what the roots are from the posted answer, but is asking how one would find these roots. Several posters mentioned the rational root test as a starting point for doing this.neilparker62 said:Seems like they are pretty much giving you the answer - what factor results from a root at -2 ? Repeat it 3 times!
Oh - ok sorry for that. Well it depends on what one can assume. If one root of the cubic is rational and can be obtained from the question ( -2 in this case) - or by trial and error - you could write something like:Mark44 said:I think you are misunderstanding what the OP meant in the same way I did in an earlier post. My new interpretation is that the OP knows what the roots are from the posted answer, but is asking how one would find these roots. Several posters mentioned the rational root test as a starting point for doing this.
Yes (your f(x) example). Some other information may be needed, but you could use general solution formula of quadratic expression to find an expression for the roots - which may be irrational (or possibly, complex), depending on what is given. I am not sure about your second expression of 2kx+4x=12x.neilparker62 said:Oh - ok sorry for that. Well it depends on what one can assume. If one root of the cubic is rational and can be obtained from the question ( -2 in this case) - or by trial and error - you could write something like:
f(x) = -(x+2)(x^2 + kx + 4) where the first and last terms in the quadratic factor are obtained by inspection and k by simple equation (2kx + 4x = 12x). Again , assuming one rational root, the nature of the other two can be determined from the coefficients in the quadratic factor. If it is a 'pre-cooked' example, the quadratic will most likely have rational roots.
Comes from obtaining terms of order x from the product. Alternatively terms of order x^2 in which case 2x^2 + kx^2 = 6x^2 using the OP's given cubic.symbolipoint said:Yes (your f(x) example). ... I am not sure about your second expression of 2kx+4x=12x.
A polynomial is cubic if it has a degree of 3, meaning the highest exponent of the variable is 3. For example, x^3 + 2x^2 - 5x + 1 is a cubic polynomial.
The general form of a cubic polynomial is ax^3 + bx^2 + cx + d, where a, b, c, and d are constants and a ≠ 0.
To factor a cubic polynomial by grouping, you first need to group the terms in pairs. Then, factor out the greatest common factor from each pair. Finally, factor out the common binomial from the resulting terms. This method works best when the polynomial has four terms.
Yes, a cubic polynomial can have complex or imaginary roots. This means that the solutions to the polynomial will involve complex numbers, which are numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).
The rational root theorem states that if a polynomial with integer coefficients has a rational root, it will be in the form of p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. To factor a cubic polynomial using this theorem, you can list all possible rational roots and use synthetic division to determine if they are actual roots of the polynomial.