18843047
Descripción
Mapa Mental por Charlotte Saul, actualizado hace más de 1 año
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Creado por Charlotte Saul
hace más de 5 años
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Algebraic
Techniques
- 1.01: Index Laws
- a^M * a^N = a^m+n
- a^m + a^n = a ^ m-n
- (a^m)^n = a^mn
- (ab)^n = a^n * b^n
- (a/b)^n = (a^n)/(b^n)
- a^M * a^N = a^m+n
- 1.02: Zero and
Negative Indicies
- x^0 = 1
- x^-n = 1/x^n
- x^0 = 1
- 1.03: Fractional Indicies
- Fractional Indices turn the
power into the root of that
number:
- For Example;
4^1/2 = square
root of 4
- For Example;
4^1/2 = square
root of 4
- Power of 1/n
- a ^1/n = n root a
- a ^1/n = n root a
- further
fractional
indices
- a ^-1/n = 1/ n root a
- a ^ m/n = n
root (a^m)
- (a/b)^-n =
(b/a)^n
- a ^-1/n = 1/ n root a
- Fractional Indices turn the
power into the root of that
number:
- 1.04:
Simplifying
Algebraic
Expressions
- Group like terms
- For Example;
4x^2 - 3x^2 +
6x^2 = x^2 + 6x^2
= 7x^2
- For Example;
4x^2 - 3x^2 +
6x^2 = x^2 + 6x^2
= 7x^2
- Group like terms
- 1.05: Expansion
- Expanding involves
removing the
brackets from an
equation
- a(b+c) = ab + ac
- a(b+c) = ab + ac
- Expanding involves
removing the
brackets from an
equation
- 1.06: Binomial Products
- Binomial expressions
consist of 2 terms.
Binomial products are
two terms multiplied
by each other
- (x+a)(x+b) = x^2 + bx + ax + ab
- (x+a)(x+b) = x^2 + bx + ax + ab
- Binomial expressions
consist of 2 terms.
Binomial products are
two terms multiplied
by each other
- 1.07: Special Products
- Difference of two squares
- (a+b)(a-b) = a^2 - b^2
- (a+b)(a-b) = a^2 - b^2
- Perfect Squares
- (a+b)^2 = a^2 + 2ab + b^2
- (a-b)^2 = a^2 - 2ab + b^2
- (a+b)^2 = a^2 + 2ab + b^2
- Difference of two squares
- (a+b)(a-b) = a^2 - b^2
- (a+b)(a-b) = a^2 - b^2
- Perfect squares
- (a+b)^2 = a^2 + 2ab + b^2
- (a-b)^2 = a^2 - 2ab + b^2
- (a+b)^2 = a^2 + 2ab + b^2
- Difference of two squares
- 1.08: Factorisation
- To factorise an
expression, we use the
distributive law in the
opposite way from
when we expand
brackets
- ax + bx = x (a+b)
- ax + bx = x (a+b)
- To factorise an expression, we use
the distributive law in the opposite
way from when we expand
brackets
- ax + bx = x(a + b)
- ax + bx = x(a + b)
- To factorise an
expression, we use the
distributive law in the
opposite way from
when we expand
brackets
- 1.09 Factorising by grouping in pairs
- If an expression has
4 terms, it can
sometimes be
factorised in pairs.
- If an expression has 4
terms, it can sometimes
be factorised in pairs.
- ax + bx + ay + by =
x(a+b) + y(a+b) =
(a+b)(x+y)
- ax + bx + ay + by =
x(a+b) + y(a+b) =
(a+b)(x+y)
- If an expression has
4 terms, it can
sometimes be
factorised in pairs.
- 1.10 Factorising Trinomials
- FInd the values for a and b so
that the sum a + b is the middle
term and the product ab is the
last term
- x^2 + (a+b)x + ab = (x+a) (x+b)
- x^2 + (a+b)x + ab = (x+a) (x+b)
- FInd the values for a and b so
that the sum a + b is the middle
term and the product ab is the
last term
- 1.11 Further trinomials
- When the coefficient of the first term is not 1, for
example 5x^2 - 13x + 6 , we need to use a different
method to factorise the trinomial
- This Method still involves finding 2
numbers that five a required sum and
product but it also involves grouping in
pairs.
- For example, 5x^2 - 13x +6 = 5x^2
- 10x - 3x + 6 = 5x(x-2) - 3(x-2) =
(x-2)(5x-3)
- First, multiply the coefficient of the first term by
the last term; 5 X 6 = 30. Now a + b = -13 and ab =
30. Since the sum is negative and the product is
positive, a and b must both be negative. 2
numbers with the product 30 and sum -13 are - 10
and -3. Now write the trinomial with the middle
term split into 2 terms, -10x and -3x, and then
factorise by grouping in pairs
- First, multiply the coefficient of the first term by
the last term; 5 X 6 = 30. Now a + b = -13 and ab =
30. Since the sum is negative and the product is
positive, a and b must both be negative. 2
numbers with the product 30 and sum -13 are - 10
and -3. Now write the trinomial with the middle
term split into 2 terms, -10x and -3x, and then
factorise by grouping in pairs
- For example, 5x^2 - 13x +6 = 5x^2
- 10x - 3x + 6 = 5x(x-2) - 3(x-2) =
(x-2)(5x-3)
- This Method still involves finding 2
numbers that five a required sum and
product but it also involves grouping in
pairs.
- When the coefficient of the first term is not 1, for
example 5x^2 - 13x + 6 , we need to use a different
method to factorise the trinomial
- 1.12 perfect squares
- You have to look at expanding
(a+b)^2 = a^2 + 2ab + b^2 and
(a-b)^2 = a^2 - 2ab + b^2. These
are perfect squares. When
factorising, use these results the
other way around
- You have to look at expanding
(a+b)^2 = a^2 + 2ab + b^2 and
(a-b)^2 = a^2 - 2ab + b^2. These
are perfect squares. When
factorising, use these results the
other way around
- 1.13 Difference of two squares
- 1.14 Mixed factorisation
- 1.15 simplifying
algebraic fractions
- 1.16 Operations with Algebraic fractions
- 1.17 Substitution
- 1.18 Simplifying surds
- 1.19 operations with surds
- 1.20 rationalising the denominator
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