All IGCSE AddMaths

Factor and Remainder Theorem

• The factor and remainder theorem is a fundamental topic in IGCSE Additional Mathematics.
• When a polynomial p(x) is divisible by x - a, then p(a) = 0.
• When p(x) is divided by x - a, the remainder is p(a).

Example Question

• Given polynomial p(x) = ax^3 + bx^2 + 6x + 4, where a and b are integers, and p(x) is divisible by x - 2.
• The derivative of p(x), denoted as p'(x), when divided by x + 1, gives a remainder of -7.
• Find the values of a and b.

Solution

1. Divisibility: Since p(x) is divisible by x - 2, then p(2) = 0.
   p(2) = a(2)^3 + b(2)^2 + 6(2) + 4 = 8a + 4b + 12 + 4 = 8a + 4b + 16
   Therefore, 8a + 4b + 16 = 0.

2. Derivative: Find the derivative of p(x).
   p'(x) = \frac{d}{dx}(ax^3 + bx^2 + 6x + 4) = 3ax^2 + 2bx + 6

3. Remainder Theorem: When p'(x) is divided by x + 1, the remainder is -7.
   p'(-1) = 3a(-1)^2 + 2b(-1) + 6 = 3a - 2b + 6
   So, 3a - 2b + 6 = -7.
   Which simplifies to 3a - 2b = -13.

4. Simultaneous Equations: Solve the simultaneous equations:
   8a + 4b = -16
   3a - 2b = -13

*Multiplying the 2nd equation by 2:

6a - 4b = -26

*Adding the two equations:

14a = -42

Therefore, a = -3.

*Substituting a into the first equation:

8(-3) + 4b = -16
-24 + 4b = -16
4b = 8

Therefore, b = 2.

Verification

• a = -3 and b = 2.
• p(x) = -3x^3 + 2x^2 + 6x + 4

Part B

• Find the remainder when the double derivative, p''(x), is divided by x.
• p'(x) = -9x^2 + 4x + 6
• p''(x) = -18x + 4
• When p''(x) is divided by x, the remainder is 4.

Example Question 2

• Given polynomial p(x) = 6x^3 + ax^2 + bx + 2, where a and b are integers, has a factor of x - 2.
• Also, p(1) = -2p(0).
• Find the values of a and b.

Solution

1. Factor Theorem: Since x - 2 is a factor, p(2) = 0.
   p(2) = 6(2)^3 + a(2)^2 + b(2) + 2 = 48 + 4a + 2b + 2
   Therefore, 4a + 2b + 50 = 0.

2. Condition: Given p(1) = -2p(0).

   • First calculate p(1).
     p(1) = 6(1)^3 + a(1)^2 + b(1) + 2 = 6 + a + b + 2 = 8 + a + b
   • Next calculate p(0).
     p(0) = 6(0)^3 + a(0)^2 + b(0) + 2 = 2
   • Apply the condition.
     8 + a + b = -2(2)
     8 + a + b = -4
     a + b = -12

3. Simultaneous Equations: Solve:
   4a + 2b = -50
   a + b = -12

• Multiply the 2nd equation by 2:
  2a + 2b = -24

• Subtract 2a + 2b = -24 from 4a + 2b = -50.:
  2a = -26
  So, a = -13.

• Substitute a = -13 in to the second equation:
  -13 + b = -12
  b = 1.

Verification

• a=-13 and b=1

Follow Up Question

*With a and b subbed in, find the remainder when p is divided by 2x-1:
*The key here is algebraic long divison!

1. Using your values a and b, find the remainder when p(x) is divided by 2x - 1.
   p(x) = 6x^3 - 13x^2 + x + 2.

2. Algebraic Long Division: Divide p(x) by 2x - 1.

                  3x^2 - 5x - 2
        2x - 1 | 6x^3 - 13x^2 + x + 2
                - (6x^3 - 3x^2)
                  ------------------
                       -10x^2 + x
                       - (-10x^2 + 5x)
                         ----------------
                                -4x + 2
                                - (-4x + 2)
                                  --------
                                        0

*The remainder is 0.
*Sub question. Factorise P(x) completely:
*First of all, we know it can be written as (2x-1)(3x^2-5x-2)
*We also know that x-2 is a factor, meaning we can work backwards to factorise fully to (2x-1)(x-2)(3x+1)

Quadratics and Indices

• Quadratics and indices are often interconnected, especially in exponential equations.

Indices

Example Question

• Simplify the expression

\frac{q^{-2} \sqrt{pr}}{\sqrt[3]{r}(pq^{-3})}

in the form p^aq^br^c and find the values of a, b, c.

Solution

1. Convert to Index Form: Rewrite the expression using index notation.
   \frac{q^{-2}(pr)^{\frac{1}{2}}}{r^{\frac{1}{3}}(pq^{-3})}

2. Expand: Apply the power to each term.
   \frac{q^{-2}p^{\frac{1}{2}}r^{\frac{1}{2}}}{r^{\frac{1}{3}}pq^{-3}}

3. Rearrange: Put the terms in alphabetical order.
   \frac{p^{\frac{1}{2}}q^{-2}r^{\frac{1}{2}}}{p^{-3}q^{-3}r^{\frac{1}{3}}}

4. Simplify: Simplify by adding or subtracting indices.
   p^{\frac{1}{2} - (-3)}q^{-2 - (-3)}r^{\frac{1}{2} - \frac{1}{3}}
   p^{\frac{7}{2}}q^{1}r^{\frac{1}{6}}

• Therefore, a = \frac{7}{2}, b = 1, c = \frac{1}{6}.

Quadratics

Identifying Quadratics

• Look for the relationship between exponents.
• Having 3 terms with the variable.

Putting in Quadratic Form

• If 4/5 and 2/5 are present, consider \left(x^{\frac{2}{5}}\right)^2=x^{\frac{4}{5}}

Example Question

• Solve 3x^{\frac{4}{5}} - 8x^{\frac{2}{5}} + 5 = 0.

Solution

1. Rewrite: Rewrite in quadratic form.
   3(x^{\frac{2}{5}})^2 - 8x^{\frac{2}{5}} + 5 = 0

2. Substitute: Let y = x^{\frac{2}{5}}.
   3y^2 - 8y + 5 = 0

3. Factorize: Factorize the quadratic equation.
   (3y - 5)(y - 1) = 0

4. Solve Factorised Equation: Solve for y.
   y = \frac{5}{3} or y = 1

5. Back Substitution: Substitute back to solve for x.

   • Case 1
     x^{\frac{2}{5}} = \frac{5}{3} therefore x=(\frac{5}{3})^{\frac{5}{2}}=3.59

   • Case 2
     x^{\frac{2}{5}} = 1 therefore x=1

   Verification and Answer

   Answers are x=1 or x = 3.59

Example Question 2

*If you have fractions like this, expand, put in power form, and simplify:
\frac{\sqrt[3]{xy}z^2y^2}{xz^{-3}z^{\frac{1}{2}}}
is equal to x^ay^bz^c.
\frac{(xy)^{\frac{1}{3}}z^2y^2}{xz^{-3}z^{\frac{1}{2}}}

Solution Part 1

First step combine what we can so we can make variables cancel:
\frac{x^{\frac{1}{3}}y^{\frac{1}{3}}y^2z^2}{xz^{-\frac{5}{2}}}
Bring it all together in top:
x^{\frac{1}{3}-1}y^{\frac{1}{3}+2}z^{2+\frac{5}{2}}
Simplify our indices, \frac{1}{3}-1 = -\frac{2}{3}, \frac{1}{3}+2 = \frac{7}{3}, and 2+\frac{5}{2} = \frac{9}{2}
This gives x^{-\frac{2}{3}}y^{\frac{7}{3}}z^{\frac{9}{2}}
Therefore a=-\frac{2}{3}, b=\frac{7}{3}, and c=\frac{9}{2}

Example Question 3

*It is important to rearrange the powers, so we can easily work out quadratics!

• Solve 10 \cdot 2^{2p} - 17 \cdot 2^p + 3 = 0.

Solution

1. Rewrite: Express 2^{2p} as (2^p)^2.
   10(2^p)^2 - 17(2^p) + 3 = 0

2. Substitute: Let y = 2^p.
   10y^2 - 17y + 3 = 0

3. Factorize: Factorize the quadratic equation.
   (5y - 1)(2y - 3) = 0

4. Solve: Solve for y.
   y = \frac{1}{5} \text{ or } y = \frac{3}{2}

5. Back Substitute: Substitute back to solve for p.

   • Case 1.

     2^p = \frac{1}{5}
     p = \frac{\log(\frac{1}{5})}{\log(2)}
     p = -2.32

   • Case 2

     2^p = \frac{3}{2}
     p=\frac{\log(\frac{3}{2})}{\log(2)}
     p=0.585

Therefore, p is equal to -2.32 or 0.585.

Vectors

• Vectors on the IGCSE AdMaths syllabus requires a solid foundation.
• Use of position vectors and unit vectors, and vector algebra.

Vector Algebra

• Consider the shape OABC, where OA=a, OB=b and OC=c. P is the midpoint in the area.
• The ratio of AP:PC is 3:2.

Find Vector OP in Terms of a and c

Use what we know abouit the given variables to obtain a formula for this

1. Find Vector AC: Determine the vector from A to C.

*To do this go from A to O, and then O to C, which can be written as: \vec{AO}+\vec{OC}=-a+c

2. Calculate Vector AP: Calculate vector AP

*The ratio of AP:PC is 3:2, this means we only want to progress \frac{3}{5}th of the way, i.e. \vec{AP}=\frac{3}{5}\vec{AC}
Meaning: \vec{AP}=\frac{3}{5}(-a+c)

3. Calculate Vector OP: Calculate vector OP

*To do this go from O to A, and then A to P, or: \vec{OA}+\vec{AP}=a+\frac{3}{5}(-a+c)
Expand to give:a-\frac{3}{5}a+\frac{3}{5}c
Group: \frac{2}{5}a+\frac{3}{5}c

• So finally we can conclude, that \vec{OP}=\frac{2}{5}a+\frac{3}{5}c

Given The Ratio OP:BP Is 2:3, Show 2b=3c+2a

1. State Vector OP in terms of B: Determine the vector from O to P in terms of B
   Meaning that we want \frac{2}{5} of the way along B, which means: \vec{OP}=\frac{2}{5}b
2. Equate to Current Vector OP Formula: Previously we said \vec{OP}=\frac{2}{5}a+\frac{3}{5}c, as they are equal \vec{OP}=\frac{2}{5}a+\frac{3}{5}c=\frac{2}{5}b
3. Rearrange to form Required Form: In this case, simply times 5 by everything, and rearrange!
   *Times all by 5: 2b=3c+2a

Position Vectors And Unit Vectors

These formulas are used to show a position relative to the origin.

Example Question:

*Let's say we have the points: O,A,B,C and D and \vec{OA}=\begin{pmatrix}6\-5\end{pmatrix}, \vec{OB}=\begin{pmatrix}10\3\end{pmatrix}, \vec{OD}=\begin{pmatrix}12\7\end{pmatrix}
What is A to B and what is the unit vector of A to B?

Method For Finding AB And The Unit Vector

If we want AB, we first go to AO and then OB

• \vec{AB}=\vec{AO}+\vec{OB}=-\vec{OA}+\vec{OB}=\begin{pmatrix}-6\5\end{pmatrix}+\begin{pmatrix}10\3\end{pmatrix}=\begin{pmatrix}4\8\end{pmatrix}
  *The unit vector refers to one ''unit'' of AB so each of the vectors will be proportional.
  *If the magnitude of AB is \sqrt{4^2+8^2}=\sqrt{16+64}=\sqrt{80}, then we can factor out \frac{1}{\sqrt{80}}
  As the magnitude is proportional, we can times each of the top vectors by this to obtain the answer.
  *Thus the unit factor of AB=\begin{pmatrix}\frac{4}{\sqrt{80}}\\frac{8}{\sqrt{80}}\end{pmatrix}

Knowing the Vector BA=AC and A is \begin{pmatrix}6\-5\end{pmatrix}, Find the Coordinates Of C

*Let's say that what ever is at C is the coordinates \begin{pmatrix}x\y\end{pmatrix}

1. Rearrange to Make OC Subject: As we know A is the midpoint, then \vec{OB}+\vec{OC}=2\vec{OA}, and rearrange:
   *\vec{OC}=2\vec{OA}-\vec{OB}
   That is \vec{OC}=2\begin{pmatrix}6\-5\end{pmatrix}-\begin{pmatrix}10\3\end{pmatrix}=\begin{pmatrix}2\-13\end{pmatrix}
   Thus the coordinates of C is \begin{pmatrix}2\-13\end{pmatrix}

Tricky Vector Question

*If the point E lies on OD such that \vec{OE}:\vec{OD} is \frac{1}{1+\lambda}, meaning E is a point that lies somewhere on \vec{OD} (\begin{pmatrix}12\7\end{pmatrix})
If BE is parallel to the X-Axis, find the value of \lambda
If BE is parallel to the X-Axis, then the y value will have to be similar! In this case will be 3.
If \vec{OE}= \frac{1}{1+\lambda}\vec{OD}=\begin{pmatrix}\frac{12}{1+\lambda}\\frac{7}{1+\lambda}\end{pmatrix}, this means \frac{7}{1+\lambda}=3
Rearranging, \lambda=\frac{4}{3}

Trig Equations and Trig Identities

• Students often find this topic tricky, particularly when radians are present.
• Working between radians and degrees is essential.

Identities To Consider

*When approaching these problems you want to have a list of trignometric identites to work with:

1. \sec(\alpha)=\frac{1}{\cos(\alpha)}
2. \cosec(\alpha)=\frac{1}{\sin(\alpha)}
3. \frac{\sin(\alpha)}{\cos(\alpha)}=tan(\alpha)

Example Question

• Solve \sin(\alpha)\cosec^2(\alpha)+\cos(\alpha)\sec^2(\alpha)=0
  where -\pi<x<\pi
  **It is vital to convert cosec and sec to sin and cos respectively.
  If remember the identities, it is vital you convert then to Cosec^2=\frac{1}{sin^2}

1. Conversion: We can now go ahead with the Conversion, which leads us to obtain:
   \frac{sin(\alpha)}{sin^2(\alpha)}+\frac{cos(\alpha)}{cos^2(\alpha)}=0
2. Simplify: Simplify by factorising like terms and take to the top to ensure all letters are on numerator.
   *This leaves: \frac{1}{\sin(\alpha)}+\frac{1}{\cos(\alpha)}=0
3. Make The Fraction In Terms of one: Add to make one fraction, i.e:
   *\frac{\cos(\alpha)+\sin(\alpha)}{\sin(\alpha)\cos(\alpha)}=0
4. Rearrange: Times it to the other side to make denominator disappear, meaning we only have to deal with \cos(\alpha)+\sin(\alpha)=0
   We know want to get sinx = c, divide both side by cos. \tan(\alpha)=-1
5. Solve: Solve tan, first using \tan^{-1} to obtain first number.
   It is essential that you keep the calculator in radians here. In this case: #\alpha =-0.07854
   With trig/tan, it repeats very 180 degrees, thus \alpha =-0.07854 + \pi= 2.356
   Thus final answers are \alpha =-0.07854 and \alpha = 2.356

Example Question 2

*Show that \frac{cos(\theta)}{1-sin(\theta)}+\frac{1+sin(\theta)}{cos(\theta)}=2sec(\theta)

Method Of Solving The Identity

*Multiply denominator, and use common multiplication (i.e. a/b + c/d becomes (ad+bc)/bd

1. Common Multiplication: Leads to:
   \frac{cos^2(\theta)+(1-sin(\theta))(1+sin(\theta))}{Cos(\theta)(1-sin(\theta))}
   Important thing is now to expand the numerator, to have 1-sin^2(\theta)
   \frac{Cos^2(\theta)+1+sin(\theta)-sin(\theta)-sin^2(\theta))}{Cos(\theta)(1-sin(\theta))}
   And this equation cancels out to \frac{Cos^2(\theta)+1-sin^2(\theta))}{Cos(\theta)(1-sin(\theta))}
   Rearrange:
   \frac{Cos^2(\theta)-sin^2(\theta)+1}{Cos(\theta)(1-sin(\theta))}
   But key identity cos^2(\theta) + sin^2(\theta)=1, thus cos^2(\theta) -sin^2(\theta)=1-2sin^2(\theta)
   Thus sub in to give\frac{2cos^2(\theta)}{Cos(\theta)(1-sin(\theta))}
   This leaves: *\frac{2}{cos(\theta)}
   As the identity says, switch round as: 2sec(\theta)

Important Point: