Polynomial Functions - Pre-Calculus

In order to determine the positive number of real zeroes, we must count the number of sign changes in the coefficients of the terms of the polynomial. The number of real zeroes can then be any positive difference of that number and a positive multiple of two.

For the function

there are four sign changes. As such, the number of positive real zeroes can be

In order to determine the positive number of real zeroes, we must count the number of sign changes in the coefficients of the terms of the polynomial after substituting for The number of real zeroes can then be any positive difference of that number and a positive multiple of two.

After substituting we get

and there is one sign change.

As such, there can only be one negative root.

One of the roots is because if we plug in 1, we get 0. We can factor the polynomial as

So now we solve the roots of .

The root will not be real.

The roots of this polynomial are .

So, the roots are

If this polynomial has a real zero at 1.5, that means that the polynomial has a factor that when set equal to zero has a solution of . We can figure out what this is this way:

multiply both sides by 2

is the factor

Now that we have one factor, we can divide to find the other two solutions:

To finish solving, we can use the quadratic formula with the resulting quadratic, :

Since we know that one of the zeros of this polynomial is 3, we know that one of the factors is . To find the other two zeros, we can divide the original polynomial by , either with long division or with synthetic division:

This gives us the second factor of . We can get our solutions by using the quadratic formula:

We know that the real zero of this polynomial is 3, so one of the factors must be . To find the other factors, we can divide the original polynomial by , either by long division or synthetic division:

This gives us a second factor of which we can solve using the quadratic formula:

First, factorize the equation using grouping of common terms:

Next, setting each expression in parenthesis equal to zero yields the answers.

First, pull out the common t and then factorize using quadratic factoring rules:

This equation has solutions at two values: when and when

Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, and respectively. The last expression must be broken up into two equations:

which are then set equal to zero to yield the expressions and

Finally, we multiply together all of the parenthesized expressions, which multiplies out to

First, factor the expression by grouping:

To find the complex zeroes, set the term equal to zero:

First, factorize the equation using grouping of common terms:

Next, setting each expression in parentheses equal to zero yields the answers.

First, pull out the common t and then factorize using quadratic factoring rules:

This equation has a solution as two values: when , and when . Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, and respectively. The last expression must be broken up into two equations: which are then set equal to zero to yield the expressions and

Finally, we multiply together all of the parenthesized expressions, which multiplies out to

First, factor the expression by grouping:

To find the complex zeroes, set the term equal to zero:

To find the potential rational zeros by using the Rational Zero Theorem, first list the factors of the leading coefficient and the constant term:

Constant 24: 1, 2, 3, 4, 6, 8, 12, 24

Leading coefficient 2: 1, 2

Now we have to divide every factor from the first list by every factor of the second:

Removing duplicates \[for example, and are both equivalent to 1\] gives us the following list:

The only choice not on this list is .

The potential zeros must have a factor of -15 as their numerator and a factor of 6 as their denominator. This eliminates as a possibility since 6 is not a factor of -15.

Now we need to test which of these values actually give zero when plugged into the polynomial.

First, :

Now :

Finally :

Since this one doesn't give us zero, it is not a solution of the polynomial.

To use Rational Zeros Theorem, take all factors of the constant term and all factors of the leading coefficient. That gives you:

Constant Term = 12, so Factors: 1, 2, 3, 4, 6, 12

Leading Coefficient = 3, so Factors: 1, 3

Then divide every number from the first list by every number from the second -- keeping in mind that both positive and negative values are possible. Of course, since this is a multiple-choice question you can also use process-of-elimination to get rid of any values that wouldn't be possible from such division. Note that the only prime factors you have are 2 and 3, meaning that there is no way to produce a 5 or 7. That eliminates (or at least makes very suspect) several choices, leaving the choice:

If you perform division from the two lists you should see how you can arrive at each value from this choice: dividing 12 by 3 would give you 4 (or negative 4, again as both + and - values are possible); dividing 2 by 3 would give you , and dividing 2 by 1 or 6 by 3 would give you 2. Therefore this choice only includes possible zeros.

To use Rational Zeros Theorem, express a polynomial in descending order of its exponents (starting with the biggest exponent and working to the smallest), and then take the constant term (here that's 6) and the coefficient of the leading exponent (here that's 4) and express their factors:

Constant: 6 has as factors 1, 2, 3, and 6

Coefficient: 4 has as factors 1, 2, and 4

Then all possible rational zeros must be formed by dividing a factor from the Constant list by a factor from the Coefficient list (and note that the results should be considered as both positive and negative values).

Here if you then look at the answer choices, note that since the divisor - which must be a factor from the Coefficient list of 1, 2, and 4 - cannot then be a multiple of 3, you know that is not a possible zero of this polynomial

To apply Rational Zero Theorem, first organize a polynomial in descending order of its exponents. Then take the constant term and the coefficient of the highest-valued exponent and list their factors:

Constant: 2 has factors of 1 and 2

Coefficient: 2 has factors of 1 and 2

Then the potential rational zeros need to be formed by dividing a factor from the Constant list by a factor from the Coefficient list. Here you only have 1s and 2s, so your options are 1, 2, and 1/2 (and note that both positive and negative values will work). There is no way to get to 1/4, meaning that that is the correct answer.

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

We mulitply what's below the line by 1 and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

with remainder

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

Our first step is to list the coefficiens of the polynomials in descending order and carry down the first coefficient.

We multiply what's below the line by and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficient.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

with reminder

This can be rewritten as: