Maths C1


Algebra and Long Division

Demonstrate – Show that (x ± p) is a factor

Will be given a factor e.g. (x + 3) Substitute the solution (inverse value) e.g. sub x = -3 If a factor will equal zero

Factorise or Finding the Factor

Substitute by trial and error to find a value that gives an answer of zero Inverse of this is a factor Used to factorise cubic functions Divide whole function by this factor (then factorise quadratic)

Finding Unknowns

Substitute inverse of the factor(s) putting = 0 Rearrange to solve the unknown variable If finding 2 unknowns solve as a simultaneous equation

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Algebra and Long Division

1.Show that (x – 2) is a factor of x3+ x2 – 4x – 4

      Substitute x = 2 into x3+ x2 – 4x – 4

      answer should be zero.

2.(x ± p) is a factor of  x3+ 2x2  x – 2 find the

      value of p and hence factorise completely

      substitute x = 1, x = -1, x = 2, etc into

      x3+ 2x2  x – 2 until the answer is zero. Then

      inverse is a factor e.g. if x = 2 gave an

      answer of zero then the factor is ( x - 2 )

      Then dividex3+ 2x2  x – 2  by ( x - 2 ) will give

      a quadratic which can then be factorised.

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Remainder Theorem

To find one unknown

Substitute inverse of the given amount Put equal to the remainder Solve to find unknown

To find 2 unknowns

Set up 2 equations using the 2 given amounts Solve as a simultaneous equations

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Remainder Theorem

1.Find the remainder when  x3 – 20x + 3 is divided by (x – 4)

     Substitute x = 4 into x3 – 20x + 3

      the answer will be the remainder

2.   f(x) = 2x3- 5x2 – 16x + 10 show that f(4) = -6

      Substitute x = 4 into 2x3- 5x2 – 16x + 10

      the answer will be -6

3.Given that when 8x4- 4x3 + ax2 -1 is divided by

     (2x + 1)  the remainder is 3.  Find the value of a

     Substitute x = -½ into 8x4- 4x3 + ax2 -1 = 3

     Solve to find a

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Binomial Expansion

The formula is provided on the exam paper and is

used when you are given:

(1 + term)power  or  (term + term)power

  ( 1 + x )n             ( a + b )n

Determine what a, b and n are then substitute

into the formula provided

Must understand factorial e.g. 3! = 3 x 2 x 1

The coefficient means the number in front of

each term

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Binomial Expansion

Take care to apply the power to all parts of the expansion e.g. if b = px then the third term (px)2 = p2x2 NOT px2 If an unknown is in the expansion oExpand to the number of required terms oSubstitute given coefficient oSolve to find unknown e.g. if p2x2 = 16x2 then p = 4 If 2 unknowns are in the expansion oExpand to the number of required terms oSubstitute given coefficients oSolve as a simultaneous equation to find unknowns For estimation split into (1 + x) then expand

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Co-ordinate Geometry

To find the gradient (m)

difference in y co-ordinates ÷ difference in x co-ordinates

To find the intercept (c)

substitute known co-ordinates into y = mx + c

Gradient of a perpendicular  line

gradient = -1/m

To find the co-ordinates of midpoint

add 2 co-ordinates then half

To find the length of a line

Use Pythagoras (difference in x co-ordinates)2 + (difference in y co-ordinates)2 Square root

If 2 lines intersect

Their equations are equal To find where they intersect solve as a simultaneous equation

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Co-ordinate Geometry

Equation of a circle  ( x – a )2 + ( y – b )2 = r2

(a, b) are the co-ordinates at the centre of the circle (x, y) are co-ordinates on the circle r is the radius May have to rearrange the equation grouping x terms together & y terms together and then use completing the square

To find co-ordinates at the centre of circle

midpoint between to co-ordinates = add 2 co-ordinates then half

To find the radius

Use Pythagoras (difference in x co-ordinates)2 + (difference in y co-ordinates)2 Square root

To find where a circle meets a linear line or the axis

Use substitution of simultaneous equations

Circle Theorems

Tangent meets the radius at a right angle The angle at the circumference of a semi circle is a right angle A line drawn from the centre of a circle perpendicular to a chord bisects the chord

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Sketching Graphs

y = mx + c

  GRADIENT  INTERCEPT                           

  how steep  where intercepts y axis

To find intercept

look for where it cuts the y axis

To find gradient

gradient are equal if the lines are parallel find where the line cuts a corner move 1 unit out count how many units before you reach the line again difference in y co-ordinates ÷ difference in x co-ordinates

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Sketching Graphs

1.Know the shape 

        x2             - x2

2.Intercept - Where intersect y axisx = 0 3.Where or whether it intersect x axis

       y = 0.  Use the discriminant to check.

       Factorise and solve to find values of x

4.   Where its maximum or minimum is

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Sketching Graphs

To determine where the graph intercepts the y axis substitute x = 0 (the constant) To determine whether  the quadratic graph intersects the x axis (use the discriminant) To determine where the graph intersects the y axis substitute y = 0 and then solve (solve quadratic) To determine where the minimum/maximum – line of symmetry (use completing the square)

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Sketching Graphs

b2 – 4ac is called the discriminant

It allows us to determine whether we can solve

an equation or not

b2 – 4ac = 0 there is one solution

b2 = 4ac there are 2 solutions which are equal (equal roots)

b2 – 4ac > 0 there are 2 solutions

b2 – 4ac < 0

there are no solutions (no real roots)

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Graph Transformations


f(x)+ n moves vertically e.g. f(x) – 2 moves 2 units down

f(x + n) moves horizontally but the inverse

        e.g. f(x + 3) moves 3 units left


nf(x) stretches vertically

         e.g. 3f(x) vertical units (y co-ordinates) times 3

f(nx) narrows horizontally

        e.g. f(2x) horizontal units (x co-ordinates) divide 2


-f(x) reflects in the x axis

f(-x) reflects in the y axis

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Indices and Surds

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Indices and Surds

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Indices and Surds

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Indices and Surds


means remove the surd from the denominator of a fraction to remove a single surd multiply top and bottom by the surd because a surd times by itself becomes the number e.g. √3 x √3 = 3 to remove more than one term times top and bottom by inverse of the denominator

    e.g. 1         5                5 x (3 - √2)             

                  3 + √2              (3 + √2)(3 - √2)

    e.g. 2    3 + √5         (3 + √5) x (7 + √6)

                   7 - √6              (7 - √6) x (7 + √6)

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