Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2008 | June | Q#6

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Question

The polynomial p(x) is given by  .

a.   Use the Remainder Theorem to find the remainder when p(x) is divided by x-1 .

b.    

                    i.       Use the Factor Theorem to show that x+2 is a factor of p(x).

                  ii.       Express p(x) as the product of linear factors.

c.    

                    i.       The curve with equation  passes through the point (0,k). State the  value of k.

                   ii.       Sketch the graph of   , indicating the values of x where the curve  touches or crosses the x-axis.

Solution

a.
 

Remainder theorem states that if  is divided by  then;

For the given case  is divided by .

Here  and . Hence;

b.
 

         i.
 

Factor theorem states that if  is a factor of   then;

For the given case  is factor of .

We can write the factor in standard form as;

Here  and . Hence;

Hence,  is factor of .


ii.
 

If  is a polynomial of degree  then  will have exactly  factors, some of which may repeat. 

We are given a polynomial of degree ,

Therefore, it will have 03 factors some of which may repeat.

From (b:i) we already have 01 factor  of .

We may divide the given polynomial with any of the already known linear factor(s) to get a quadratic  factor and then factorize the obtained quadratic factor to find the 2nd and 3rd linear  factors.

For the given case;

Already known factor from (b:i) is .

We may divide the given polynomial by factor .

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Therefore, we get the quadratic factor  for given polynomial. Now we factorize this quadratic factor.

Hence,  can be written as product of three linear factors as follows;

It is evident that factor  is a repeating factor.

 

c.    

                            i.
 

If a curve or line passes through a point then equation of that curve or line must satisfy the  coordinates of that point.

We are given that the curve with equation  passes through the point (0,k).

Substituting  and  in the equation of the curve;

                          ii.
 

We are required to sketch a cubical polynomial given as;

ü Find the sign of the coefficient of . This gives the shape of the graph at the extremities. If the  coefficient of  is such that  cubic graph is increasing from left to right at the extremities and  when  the cubic graph is inverted.

The coefficient of  in the given polynomial is positive. Therefore it will cause normal cubic graph  increasing from left to right at the extremities.

ü Find the point where the graph crosses y-axis by finding the value of  when

Therefore for

Hence, the cubic graph intercepts y-axis at .

ü Find the point(s) where the graph crosses the x-axis by finding the value of  when . If  there is repeated root the graph will touch the x-axis.

From (b:ii) we know that given polynomial can be written as;

It is evident from factors that the function  will cross/intersect horizontal axis when;

The curve will touch x-axis at the point found from repeating factor. 

It is evident cubic graph will intersect x-axis at  and will touch x-axis at .

ü Calculate the values of  for some value of . The is particularly useful in determining the  quadrant in which the graph might turn close to the y-axis.

ü Complete the sketch of the graph by joining the sections.

Sketch should show the main features of the graph and also, where possible, values where the  graph intersects coordinate axes.

desmos-graph (1).png

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