# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2015 | Oct-Nov | (P2-9709/22) | Q#4

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Question

The polynomial  is defined by

where  is a constant. It is given that  is a factor of

i.       Use the factor theorem to show that .

ii.       When ;

a.  Factorise p(x) completely,

b.  Solve the equation  for .

Solution

i.

We are given that;

We are also given that is a factor of .

When a polynomial, , is divided by , and  is factor of , then the remainder is ZERO i.e. .

We can write the factor in standard form as;

Therefore;

ii.

a.

We are given that;

We have found in (i) that  therefore;

We are required to factorise   completely.

We are also given that is factor of .

When a polynomial, , is divided by , and  is factor of , then the remainder is ZERO i.e.

Therefore, division of with  factor will yield a quadratic factor with ZERO remainder.

We divide  by .

Therefore;

b.

We are required to solve the equation;

Let , then;

We have found in (i) that  therefore;

We have also found above that;

Therefore;

Now we have three options.

Since ;

provided that

We know that;

Therefore;

 NOT POSSIBLE!

Therefore, only possible solution is;

Using calculator;

To find the other solution of we utilize the odd/even property of .

 Properties of Domain Range Periodicity Odd/Even Translation/ Symmetry

We use odd/even property;

Therefore, we have two solutions (roots) of the equation;

To find all the solutions (roots) over the interval , we utilize the periodic property of    for both these values of .

Therefore;

It is evident that all solutions obtained through use of periodic property will be beyond desired interval, therefore, only possible solution within the desired interval is.