Polynomial Functions - College Algebra

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0:

is not a zero.

1 is the only rational zero.

By the Rational Zeroes Theorem, any rational zero of a polynomial with integer coefficients must be equal to a factor of the constant divided by a factor of the leading coefficient, taking both positive and negative numbers into account.

The constant 17, being prime, has only two factors, 1 and 17; the leading coefficient is 1, which only has 1 as a factor. Thus, the only possible rational zeroes of the given polynomial are given in the set

,

a set of four elements. This makes 4 the correct choice.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

Both values can be tested as follows:

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is indeed a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0. However, as their are no odd-degree coefficients, the sum is the same:

is also a zero.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0:

is not a zero.

1 is the only rational zero.

By the Rational Zeroes Theorem, any rational zero of a polynomial with integer coefficients must be equal to a factor of the constant divided by a factor of the leading coefficient, taking both positive and negative numbers into account.

The constant 17, being prime, has only two factors, 1 and 17; the leading coefficient is 1, which only has 1 as a factor. Thus, the only possible rational zeroes of the given polynomial are given in the set

,

a set of four elements. This makes 4 the correct choice.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

Both values can be tested as follows:

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is indeed a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0. However, as their are no odd-degree coefficients, the sum is the same:

is also a zero.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0:

is not a zero.

1 is the only rational zero.

By the Rational Zeroes Theorem, any rational zero of a polynomial with integer coefficients must be equal to a factor of the constant divided by a factor of the leading coefficient, taking both positive and negative numbers into account.

The constant 17, being prime, has only two factors, 1 and 17; the leading coefficient is 1, which only has 1 as a factor. Thus, the only possible rational zeroes of the given polynomial are given in the set

,

a set of four elements. This makes 4 the correct choice.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

Both values can be tested as follows:

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is indeed a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0. However, as their are no odd-degree coefficients, the sum is the same:

is also a zero.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0:

is not a zero.

1 is the only rational zero.

By the Rational Zeroes Theorem, any rational zero of a polynomial with integer coefficients must be equal to a factor of the constant divided by a factor of the leading coefficient, taking both positive and negative numbers into account.

The constant 17, being prime, has only two factors, 1 and 17; the leading coefficient is 1, which only has 1 as a factor. Thus, the only possible rational zeroes of the given polynomial are given in the set

,

a set of four elements. This makes 4 the correct choice.

Factor:

Double check by factoring:

Add together:

Therefore:

The rules for adding fractions containing unknowns are the same as for fractions containing explicit numbers, so you can guide yourself by recalling how you would proceed adding fractions such as,

As you know you need to write them with a common denominator. In this case the least common denominator is . So simply multiply the numerator and denominator of each fraction by the denominator of the other fraciton.

Notice that and are equal to one, this ensures that we are not changing the value of the fractions, we are changing only the representation of the value.

Similairily, the procedure for an algebraic expression containing unknowns parallels this idea,

Now we can add the numerators directly since we now have both terms expressed with a common denominator, .

First, factor the numerator of the quotient term by recognizing the difference of squares:

Cancel out the common term from the numerator and denominator:

FOIL (First Outer Inner Last) the first two terms of the equation:

Combine like terms:

We can accomplish this division by re-writing the problem as a fraction.

The denominator will distribute, allowing us to address each element separately.

Now we can cancel common factors to find our answer.

Divide the leading coefficients to get the first term of the quotient:

, the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

Repeat these steps with the differences until the difference is an integer. As it turns out, we need to repeat only once:

, the second term of the quotient

, the remainder

Putting it all together, the quotient can be written as .

First, rewrite this problem so that the missing term is replaced by

Divide the leading coefficients:

, the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

Repeat this process with each difference:

, the second term of the quotient

One more time:

, the third term of the quotient

, the remainder

The quotient is and the remainder is ; this can be rewritten as a quotient of

Simplify the following expression:

First, let's multiply the 3x through:

Next, divide out the x from the bottom:

So our answer is:

Simplify the following expression:

To begin, we need to recognize the bottom as a difference of squares. Rewrite it as follows.

So our answer is: