# Maths C1

- Created by: BenThomas17
- Created on: 10-05-15 20:32

## 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

## 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.

## 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

## 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

## 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

## 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

## 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

## 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

## 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

## 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

## 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)

## 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)

## Graph Transformations

Translate

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

Stretching

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

Reflection

-f(x) reflects in the x axis

f(-x) reflects in the y axis

## Indices and Surds

## Indices and Surds

## Indices and Surds

## Indices and Surds

Rationalise

•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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