Mathematics 9709/22 — October/November 2021

Cambridge AS Level · Pure Mathematics 2 · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Logarithmic and Exponential Functions · Algebra · Differentiation · Trigonometry · Integration · Numerical Solution of Equations

Q1AlgebraFree sample

The polynomial $p(x)$ is defined by

p(x) = ax^3 + bx - 10,

where $a$ and $b$ are constants. It is given that $(x + 2)$ is a factor of $p(x)$ and that the remainder is $-55$ when $p(x)$ is divided by $(x + 3)$ .

(a)

Find the values of $a$ and $b$ .

DifficultyMedium

Worked solution

Approach

Since $(x + 2)$ is a factor of $p(x)$ , the factor theorem gives $p(-2) = 0$ . Since the remainder is $-55$ when $p(x)$ is divided by $(x + 3)$ , the remainder theorem gives $p(-3) = -55$ . Substitute these values into $p(x) = ax^3 + bx - 10$ to obtain two linear equations in $a$ and $b$ , then solve them simultaneously.

Working

Using the factor theorem:

p(-2) = a(-2)^3 + b(-2) - 10 = 0

-8a - 2b - 10 = 0

Using the remainder theorem:

p(-3) = a(-3)^3 + b(-3) - 10 = -55

-27a - 3b - 10 = -55

So the equations are:

-8a - 2b = 10

-27a - 3b = -45

Divide the first equation by $-2$ and the second by $-3$ :

4a + b = -5

9a + b = 15

Subtract the first equation from the second:

5a = 20

a = 4

Substitute $a = 4$ into $4a + b = -5$ :

16 + b = -5

b = -21

Answer

a = 4, \quad b = -21

Final answer

a = 4, b = -21

Detailed explanation

Walkthrough

First, recognise that $(x + 2)$ being a factor means $x = -2$ makes $p(x) = 0$ . This is the factor theorem. Substitute $x = -2$ into $ax^3 + bx - 10$ to get $-8a - 2b - 10 = 0$ .

Next, when a polynomial is divided by $(x + 3)$ , the remainder is the value of the polynomial at $x = -3$ . This is the remainder theorem. Substitute $x = -3$ to get $-27a - 3b - 10 = -55$ .

Now simplify both equations. The first becomes $-8a - 2b = 10$ , and the second becomes $-27a - 3b = -45$ . Dividing the first by $-2$ gives $4a + b = -5$ , and dividing the second by $-3$ gives $9a + b = 15$ . Subtracting eliminates $b$ , giving $5a = 20$ , so $a = 4$ . Substituting back gives $b = -21$ .

Key Takeaways

The factor theorem and remainder theorem let us turn information about factors and remainders into equations involving unknown coefficients. Solving those equations then determines the polynomial.

Common Mistakes

Substituting $x = 2$ instead of $x = -2$ when $(x + 2)$ is a factor.
Substituting $x = 3$ instead of $x = -3$ when dividing by $(x + 3)$ .
Making sign errors when simplifying expressions such as $a(-2)^3$ .
Not showing both substitution equations, which the mark scheme requires.

Things to Be Careful About

Remember that $(x + c)$ being a factor means the root is $x = -c$ . Also, the remainder when dividing by $(x + c)$ is $p(-c)$ , not $p(c)$ . Check the final values by substituting them back into both original equations.

Techniques used

apply the factor theoremapply the remainder theoremsolve simultaneous linear equations

(b)

Hence factorise $p(x)$ completely.

DifficultyMedium

Worked solution

Approach

Substitute $a = 4$ and $b = -21$ into $p(x)$ to get $p(x) = 4x^3 - 21x - 10$ . Since $(x + 2)$ is a known factor, divide the cubic by $(x + 2)$ to obtain a quadratic factor, then factorise that quadratic completely.

Working

Write the cubic with a zero $x^2$ term:

4x^3 + 0x^2 - 21x - 10

Divide by $x + 2$ :

4x^3 + 0x^2 - 21x - 10 = (x + 2)(4x^2 - 8x - 5)

Now factorise the quadratic:

4x^2 - 8x - 5 = (2x + 1)(2x - 5)

Therefore:

p(x) = (x + 2)(2x + 1)(2x - 5)

Answer

p(x) = (x + 2)(2x + 1)(2x - 5)

Final answer

p(x) = (x + 2)(2x + 1)(2x - 5)

Detailed explanation

Walkthrough

First replace $a$ and $b$ with the values found in part (a), giving $p(x) = 4x^3 - 21x - 10$ . Notice there is no $x^2$ term, so write it as $4x^3 + 0x^2 - 21x - 10$ when dividing.

Because $(x + 2)$ is a factor, divide the cubic by $x + 2$ . The quotient is $4x^2 - 8x - 5$ . This quadratic can be factorised further: $4x^2 - 8x - 5 = (2x + 1)(2x - 5)$ . Therefore the complete factorisation is $(x + 2)(2x + 1)(2x - 5)$ .

Key Takeaways

Once one linear factor of a cubic is known, polynomial division reduces the problem to factorising a quadratic. A complete factorisation requires continuing until all factors are linear.

Common Mistakes

Forgetting to include the missing $x^2$ term as $0x^2$ during division.
Making sign errors when subtracting terms in polynomial division.
Stopping after finding the quadratic factor instead of factorising it completely.

Things to Be Careful About

The word "completely" means the final answer must be written as three linear factors. The mark scheme condones writing $= 0$ , but the factorisation itself should be fully simplified.

Techniques used

divide a cubic polynomial by a linear factorfactorise a quadratic expression

The rest of this paper

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Q7Trigonometry · Integration10M

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