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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