Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2011 | January | Q#5
i. Sketch the curve with equation
ii. Show that the equation can be expressed as
b. The polynomial is given by .
i. Find the remainder when is divided by .
ii. Use the Factor Theorem to show that is a factor of .
iii. Express in the form , where b and c are integers.
c. Hence show that the equation has only one real root and state the value of this root.
We are required to sketch;
It is evident that it represents a cubical polynomial.
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.
It is evident from factors that the function will cross/intersect horizontal axis when;
It is evident cubic graph will intersect 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.
Remainder theorem states that if is divided by then;
For the given case is divided by .
We can write the divisor in standard form as;
Here and . Hence;
Factor theorem states that if is a factor of then;
For the given case is factor of .
Here and . Hence;
Hence, is factor of .
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 (a:ii) 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:ii) is .
We may divide the given polynomial by factor .
Therefore, we get the quadratic factor for given polynomial.
Hence, can be written as follows;
We are required to show that the equation has only one real root and find the value of this root.
We have found in (a:ii) that can be expressed as;
We have also shown in (b:iii) that this equation can be written as;
We have found in (b:ii) that is a factor of .
Therefore, one root of the given equation can be found from this factor;
Now we need to verify that given equation has only one real root.
Any other root, if exists will come from;
We recognize that it is a quadratic equation.
For a quadratic equation , the expression for solution is;
Where is called discriminant.
If , the equation will have two roots.
If , the equation will have two identical/repeated roots.
If , the equation will have no roots.
If there is no other real root of the given equation then for this quadratic factor of given equation should not have any real root. Therefore;
Hence, has no other root and, therefore, has only one real root which is 3.