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find all rational zerosh(x) = 2x^3 + x^2 + 8x +4

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2 Answers

First, you need to factor the equation. Group 2x3 and 8x together, and x2 and 4 together:

(2x3 + 8x) + (x2 + 4)

Factor the first part of the equation:
2x(x2 + 4) + (x2 + 4)

Notice that we now have the same thing in parenthesis twice. We can think of this as:
2x(x2 + 4) + 1(x2 + 4) = (2x + 1)(x2 + 4)

By the Zero Product Property, our zeros will come from:
2x + 1 = 0
and
x2 + 4 = 0

2x+1=0               subtract 1 from both sides, then divide by 2
x = -1/2

x2 + 4 = 0           subtract 4 from both sides, then take the square root. note that we will have to use i since we are taking the square root of a negative number.
x = 2i

Comments

The problem asks us to find all rational zeros of the function. Therefore, imaginary numbers are not in the solution set.

For polynomials like this - which are easily factored - you will find that factoring the polynomial and taking all real zeros will be the easiest method. However, for polynomials which are not easily factored, there is another method - a method that you should probably use for all problems of this type given that it can sometimes be difficult to tell whether a polynomial factors or not.

Using the Rational Zero Theorem, we find all possible zeros of the function. We start doing this by listing all of the factors of the constant term and all factors of the leading coefficient, then making all possible combinations of fractions. In this case, our constant term is 4, and our leading coefficient is 2. Therefore, our factors of these two numbers are:

4: ±4, ±2, ±1

2: ±2, ±1

We form all combinations of fractions of the factors of the constant over the leading coefficient - but not the leading coefficient over the constant. It is good to, also in this step, simplify each possible zero. For this problem, we have:

±4/±2 = ±2

±2/±2 = ±1

±1/±2 = ±(1/2)

±4/±1 = ±4

±2/±1 = ±2

±1/±1 = ±1

Notice that ±2 and ±1 show up twice. This is not of any predictive significance (that is, it doesn't mean they are more or less likely to be our real zeros). However, for zeros that show up twice, we need only consider each of them once.

So, our possible rational zeros are ±4, ±2, ±1, ±(1/2). Notice that we have 8 different possibilities here - the positive and the negative of 4 different numbers.

From this point, you do the rather laborious task of substituting each of these numbers for x in the polynomial and checking which of them calculate out to give you a function value of 0 - since we're looking for the rational zeros.

This doesn't mean, however, that you must check all 8. Notice that the degree (the highest power of any variable term in the function) of this function is 3. That means that out of the 8 possibilities we have found using the Rational Zero Theorem, at most 3 of them will be solutions. If you find three solutions before you calculate all 8 possibilities, you know you've found all the zeros out of the possibilities. If you go through all 8 possibilities and only find 1 or 2 actual zeros (as you will in this problem), it doesn't mean you're wrong - only that the function doesn't have its maximum number of zeros.

Again, as I stated at the beginning of the solution, for expressions which are easily factored, I would go ahead and factor and solve for all rational solutions (if the problem is only asking for rational zeros), but for those polynomial functions which are not easily factored or are not readily obvious as factorable, I hope this explanation of the Rational Zero Theorem helps.

Sorry. I just realized I posted in the wrong place. See my answer below.

For polynomials like this - which are easily factored - you will find that factoring the polynomial and taking all real zeros will be the easiest method. However, for polynomials which are not easily factored, there is another method - a method that you should probably use for all problems of this type given that it can sometimes be difficult to tell whether a polynomial factors or not.

Using the Rational Zero Theorem, we find all possible zeros of the function. We start doing this by listing all of the factors of the constant term and all factors of the leading coefficient, then making all possible combinations of fractions. In this case, our constant term is 4, and our leading coefficient is 2. Therefore, our factors of these two numbers are:

4: ±4, ±2, ±1

2: ±2, ±1

We form all combinations of fractions of the factors of the constant over the leading coefficient - but not the leading coefficient over the constant. It is good to, also in this step, simplify each possible zero. For this problem, we have:

±4/±2 = ±2

±2/±2 = ±1

±1/±2 = ±(1/2)

±4/±1 = ±4

±2/±1 = ±2

±1/±1 = ±1

Notice that ±2 and ±1 show up twice. This is not of any predictive significance (that is, it doesn't mean they are more or less likely to be our real zeros). However, for zeros that show up twice, we need only consider each of them once.

So, our possible rational zeros are ±4, ±2, ±1, ±(1/2). Notice that we have 8 different possibilities here - the positive and the negative of 4 different numbers.

From this point, you do the rather laborious task of substituting each of these numbers for x in the polynomial and checking which of them calculate out to give you a function value of 0 - since we're looking for the rational zeros.

This doesn't mean, however, that you must check all 8. Notice that the degree (the highest power of any variable term in the function) of this function is 3. That means that out of the 8 possibilities we have found using the Rational Zero Theorem, at most 3 of them will be solutions. If you find three solutions before you calculate all 8 possibilities, you know you've found all the zeros out of the possibilities. If you go through all 8 possibilities and only find 1 or 2 actual zeros (as you will in this problem), it doesn't mean you're wrong - only that the function doesn't have its maximum number of zeros.

Again, as I stated at the beginning of the solution, for expressions which are easily factored, I would go ahead and factor and solve for all rational solutions (if the problem is only asking for rational zeros), but for those polynomial functions which are not easily factored or are not readily obvious as factorable, I hope this explanation of the Rational Zero Theorem helps.

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