Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2011 | June | Q#5
The polynomial is given by .
a. Use the Remainder Theorem to find the remainder when is divided by .
b. Use the Factor Theorem to show that is a factor of .
i. Express in the form , where b and c are integers.
ii. Hence show that the equation has exactly one real root.
Remainder theorem states that if is divided by then;
For the given case is divided by .
Here and . Hence;
Factor theorem states that if is a factor of then;
For the given case is factor of .
We can write the divisor in standard form as;
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 (b) 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) is .
We may divide the given polynomial by this factor .
Therefore, we get the quadratic factor for given polynomial.
Hence, can be written as product of linear and quadratic factors as follows;
We are required to show that the equation has only one real root.
We have found in (c:i) that can be expressed as;
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 root and, therefore, has only one real root which is -1.