Explanation:

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 ofif 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 ofif 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 arenoodd-degree coefficients, the sum is the same:

is also a zero.

Explanation:

来确定the zeros of this equation, we will need to factorize the polynomial.

The only common factors ofthat will give us a middle term of negativeby addition or subtraction is:

Set each binomial equal to zero and solve.

The zeros are:

Explanation:

First we need to factor the denominator.

Now we can rewrite it as such

Now we need to get a common denominator.

Now we set up an equation to figure outand.

To solve for, we are going to set.

To find, we need to set

Thus the answer is:

Explanation:

Now we can rewrite it as such

Now we need to get a common denominator.

Now we set up an equation to figure outand.

To solve for, we are going to set.

To find, we need to set

Thus the answer is:

Explanation:

To add rational expressions, you must find the common denominator. In this case, it's.

Next, you must change the numerators to offset the new denominator.

becomesandbecomes.

Now you can combine the numerators:.

Put that over the denomiator and see if you can simplify/factor further. In this case, you can't.

Therefore, your final answer is:

.

Explanation:

To subtract rational expressions, you must first find the common denominator, which in this case is. That means we only have to adjust the first fraction since the second fraction has that denominator already.

Therefore, the first fraction now looks like:

.

Now that the denominators are the same, combine numerators:

.

Now, put that over the denominator and see if you can simplify any further.

In this case, you can't, so your final answer is:
.

Explanation:

In order to add the numerators of the fractions, we need to find the least common denominator.

The least common denominator is:

We will need to multiply the numerator and denominator byto match the denominators of both fractions.

Simplify the fraction.

Combine the two fractions.

The answer is:

Explanation:

The rules for adding fractions containing unknownsare 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 thatandare 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,.

Explanation:

(1.a)

1) First factor the denominator as much as possible; characterize the denominator and write the appropriate expansion:

The denominator is a product of linear terms, so the partial fraction expansion will have the form,

(1.b)

2) Write a system of 3-equations and 3-unknowns in order to determine A,B,and C in the partial fraction expansion (equation 1.b).

If we were to take equation (1.b) and add each fraction under a common denominator, the numerator would have the form,

(2.a)

Distribute and multiply the's,

(2.b)

3) Find the constants A, B, and C.

To find,, andsimply expand and collect like terms (there are,, and constant terms) then compare to the original numerator.

For the terms withwe must have,

So forwe have,

(3.a)

For the-terms we have,

forwe have,

(3.b)

For the constant term we have,

(3.c)

马上我们可以读出的解决方案from equation (3.c)Substituteinto (3.a) and (3.b)

4) Solve for the remaining unknown constants B and C,

The system:

(4.a)

(4.b)

In order to remove the fraction it would be convenient to solve this after multiplying both equations by:

(4.c)

(4.d)

In order to make even more simple, multiply equation (4.c) byand solve forin terms ofas follows,

Substitute into (4.d),

Now we can use this value forto find that.

Finally, plug in the values for,, andwe obtained into equation (1.b).

Explanation:

Factor the denominator:

Multiply both sides of the equation by

Let:

Let: