A:

We may also proceed as follows

Now complete the solution and check your result with

A:

We may also proceed as follows

Now complete the solution and check your result with

A:

This is a trivial case and we may proceed thats
 recall

A:

Here we carry out long division first

Therefore

We now express  in partial fractions thus
At this level dear student; I am convinced you can solve or the constants  and .
Solve and see whether your result is

A:

Equivalent form for the partial fraction is
.
Dear student simplify and see whether you will anive at the following equations to solve for , and

Solve and see check your final solution with

A:

Equivalent form of the partial fractions is:







By the cover up method;
when
From equation
From equation (2),

A:

Equivalent form of the partial fractions is

By the cover method;
When
From equation (1)
From equation (3)
Hence

A:

Equivalent on of partial fractions is
.
The constants  and  can be obtained by the cover up metiod.
When
When
To get  we can substitute  and solve for

A:

Equivalent on of partial fractions is
.
The constants  and  can be obtained by the cover up metiod.
When
When
To get  we can substitute  and solve for

A:

Equivalent form of the partial fraction is

Here we can only get  by the cover up method thus:
To get  and  we can proceed thus


At this point you may expand and compare coefficients or you substitute simple values of , form equations and then solve them.
When  we have
Putting  we have

When  we have
Putting  we have

Equation
From equation (2)

A:

Equivalent form of partial fraction is
.
In this case we can get the constants  and  by cover up method but not .
When there are repeated factors, the cover up method gives the constant which has the higher or highest power at the denominator.
When
When
Hawing obtained the values of  and , we use any value of  not already used to solve for . For example let  then we get

A:

Here we see that the degree of the numerator and denominator is the same. So we carry out long

Therefore, we can now say

We now express  in partial fractions thus

When
When

A:

We can write  as

Equivalent form of the partial fraction is

By the cover up metliod
When
when

A:

Equivalent form of partial fraction is

By the cover up method when

A:

In the case of linear factors we can solve by two methods.
Method 1
Let

Comparing the coefficient and the constants we have:

From equation 
Substituting  in  we have

Thus

Method 2: The cover up method.
This method is strictly limited to linear factors and to a lesser extent to repeated factors at the denominator.
To get the constant , we equate the denominator of A to zero ie. , we get . Then we Substitute  in the left hand side; covering the factor under , that is .
Hence
Similarly to get  we put  or . We substutute  in the LHS, covering .
Thus
Again we see that

A:

Let
By the cover up method when  we have
When  we have
Thus

A:

 or

A:

In this example there are just wo ideas needed to solve the problems. We need just be acquainted with the definition of the modulus of a function.
Recall that
This the idea we shall use in solving problem 17
a)

Here we see that if the modulus of  is the same as  then  must be greater than zero
Thus we solve the inequality
Dear student at this level you can soive the inequality .
Solve and check whether your solution set is
b)

Here also we see that the modulus of  is  itself. Hence  must begreaterthanzero or positive.
In a similar mamer we solve the inequality
Here again complete the solution and check your solution with .
c)

Surely you can judge and see the inequality to be solved. Clearly the modulus of  is the same as . Hence  is greater than zero or is positive.
So we solve the inoululy  The inequality  is a quotient inequality. For it to be positive both the numerator and denominator must be positive or must be negative. In this particular case the
denominator is always positive. Hence the numerator  must be positive.
Thus we solve the inequality  
Get the zeros, solve and see whether you solution is
d)
Here the modulus of  is .
Here we see that the nodulus of  is the negative of ; that is  
Thus we conclude that  is negative or less than zero.
So we solve
Solve and check whether your solution is
d) In a similarjudgement if

The zeros are .

The solution set is

A:

In this example there are just wo ideas needed to solve the problems. We need just be acquainted with the definition of the modulus of a function.
Recall that
This the idea we shall use in solving problem 17
a)

Here we see that if the modulus of  is the same as  then  must be greater than zero
Thus we solve the inequality
Dear student at this level you can soive the inequality .
Solve and check whether your solution set is
b)

Here also we see that the modulus of  is  itself. Hence  must begreaterthanzero or positive.
In a similar mamer we solve the inequality
Here again complete the solution and check your solution with .
c)

Surely you can judge and see the inequality to be solved. Clearly the modulus of  is the same as . Hence  is greater than zero or is positive.
So we solve the inoululy  The inequality  is a quotient inequality. For it to be positive both the numerator and denominator must be positive or must be negative. In this particular case the
denominator is always positive. Hence the numerator  must be positive.
Thus we solve the inequality  
Get the zeros, solve and see whether you solution is
d)
Here the modulus of  is .
Here we see that the nodulus of  is the negative of ; that is  
Thus we conclude that  is negative or less than zero.
So we solve
Solve and check whether your solution is
d) In a similarjudgement if

The zeros are .

The solution set is

A:

 or