# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | June | Q#4

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**Question**

**a. **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.

** iv. **The equation has one root equal to -2. Show that equation has no other real roots.

**b. **The curve C with equation , sketched below.

The curve cuts the x-axis at the point and the y-axis at point B.

** i. **State the y-coordinate of the point B.

** ii. **Find

** iii. **Hence find the area of the shaded region bounded by the curve and the line AB.

**Solution**

**a.
**

** i.
**

We are required to find the remainder when is divided by .

Remainder theorem states that if is divided by then;

For the given case is divided by .

Here and . Hence;

** ii.
**

We are required to to show that is a factor of .

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 .

** iii.
**

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 2^{nd} and 3^{rd} linear factors.

For the given case;

Already known factor from (a:ii) 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;

** iv.
**

We are required to show that equation has only one real root and that is the given one -2.

We are given that;

From (iii) we have also found that can be expresses as;

Since ;

From (ii) we have found that, is factor of and therefore is a root of .

It is evident that one root of is obtained from

For other roots we use;

It is quadratic factor of and solution of this quadratic equations would yield other roots of .

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.

As we are given that has no real roots other than -2; in this case;

Therefore, has no real roots.

Hence, has only one real root -2.

**b. **

** i.
**

We are required to find y-coordinate of point B.

It is evident that point B is the y-intercept of the curve with equation .

The point at which curve (or line) intercepts y-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

Therefore;

Hence, y-coordinate of point B is 6.

** ii.
**

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

** iii.
**

To find the area of region under the curve , we need to integrate the curve from point to along x-axis.

For the given case;

First we find area under curve ;

We have found from (b:ii) that;

Now we find area under the line AB.

We need equation of line AB with points A(-2,0) and B(0,6).

Two-Point form of the equation of the line is;

Therefore;

Therefore;

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

Finally, we can calculate area of shaded region;

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