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Chapter 6 Polynomials Cheat Sheet

A Education › Mathematics cheat sheet in English by liv.skreka

Cambridge Chapter 6 Polynomials Unit 1 - AOS 2- Polynomials

Maths Polynomials Cubics Quartics
6,103 views 961 downloads 30 Mar 2024 History
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6A The Language of Polyno­mials

6A The Language of Polynomials
- A polynomial function is a function that can be written in the form:
P(x) = ax^n + ax^n-1 + ... + ax + a

- The leading term, ax^n, of a polynomial is the term of the highest index among those terms with a non-zero coeffi­cient.

The function f(x)=x^1/3

The function f(x)=x^1/3
Inverse functions
f(x)=x1/3 is the inverse of f(x)= x3

6E - Cubic Functions of f(x) = a(x-h)^3 + k

6E - Cubic Functions of f(x) = a(x-h)^3 + k

6E - Cubic Functions of f(x) = a(x-h)^3 + k

6E - Cubic Functions of f(x) = a(x-h)^3 + k

6E - Cubic Functions of f(x) = a(x-h)^3 + k

Point of Inflection (POI)
Vertical Transl­ations
Horizontal Transl­ations
A point of zero gradient
by adding or subtra­cting a constant term to y-x^3, the graph moves either up or down
The graph of y=(x-h)^3 is moved h units to the right (for h>0)
The 'flat point' of the graph
e.g. y=x^3 + k moves the graph k units up (for k>0)
the POI is at (h,0)
The POI of graph of y=x^3 is at (0,0)
The POI becomes (0,k)
In this case, the graph of y=x^3 is translated h units in the positive direction of the x-axis.
The implied domain of all cubics is R and the range is also R

6D - Solving Cubic Equations

6D - Solving Cubic Equations

6D - Solving Cubic Equations

6D - Solving Cubic Equations
In order to solve a cubic equation, the first step is often to factorise.
Factorise by identi­fying a factor, then using polynomial division.
Factor the quadratic factor
Then, use null factor law.

6B - Division of Polyno­mials

6B - Division of Polynomials
- When we divide the polynomial P(x) by the polynomial D(x) we obtain two polyno­mials, Q(x) the quotient and R(x) the remainder, such that
P(x) = D(x)Q(x) + R(x)
and either R(x) =0 or R9x) has degree less than D(x)
Here P(x) is the dividend and D(x) is the divisor

6B - Division of Polyno­mials

6B - Division of Polynomials
Dividing Polyno­mials involving fractions

6B - Division of Polyno­mials

6B - Division of Polynomials
Dividing polyno­mials when we have a remainder

6B - Division of Polyno­mials

6B - Division of Polynomials
Equating Coeffi­cients Methods instead of dividing

6F - Graphs of factorised cubic functions

6F - Graphs of factorised cubic functions

6F - Graphs of factorised cubic functions

6F - Graphs of factorised cubic functions
Repeated roots/­factors

6F - Graphs of factorised cubic functions

6F - Graphs of factorised cubic functions
Cubic equations with one x intercept

6G - Solving Cubic Inequa­lities

6G - Solving Cubic Inequalities

6H - Families of cubic polynomial functions

y=ax^3
y=a(x-h)^3 + k
y=a(x-­a)(­x-b­)(x-c)
y=ax^3­+bx­^2+cx+d

6H - Families of cubic polynomial functions

6H - Families of cubic polynomial functions

6H - Families of cubic polynomial functions

6H - Families of cubic polynomial functions
Quartic Functions

6B - Division of Polyno­mials

6B - Division of Polynomials
Equating Coeffi­cients Methods instead of dividing

6C - Special Cases Differ­ences of Cubes

6C - Special Cases Differences of Cubes

6C - Rational Root Theorem

6C - Rational Root Theorem

6C - Rational Root Theorem

P(x) = 2x3 - x2 - x - 3
Choose factors of -3, which are ±1 and ±3.
However, P(1) ≠ 0, P(-1) ≠ 0, P(3) ≠ 0, and P(-3) ≠ 0
Therefore, we must use the Rational Root Theorem.
We must use P(± factors of consta­nt/­factors of leading coeffi­cient)
factors of constant (of -3) = ±1, ±3 factors leading coeffi­cient (of 2) = ±1, ±2
e.g. P(±3/2), P(±1/2)
We now have to check these factors --> P(3/2) = 2(3/2)3 - (3/2)2 -(3/2) -3 = 0
therefore, x-3/2, which equates to 2x-3 as a factor

6C - Factor­isation of Polyno­mials

6C - Factorisation of Polynomials
For a polynomial P(x)
- If P(a) = 0, then x-a is a factor of P(x)
- Conver­sely, if x-a is a factor of P(x), then P(a) =0

6C - Factor­isation of Polyno­mials

6C - Factorisation of Polynomials
Remainder Theorem:
When P(x) is divided by bx+a, the remainder is P(-a/b).

For example, if P(x) is divided by x-1, let x-1=0, x=1.
P(1) = Remainder (R(x))

For example, if P(x) is divided by 3x-2, let 3x+2=0, x=-2/3.
P(-2/3) = Remainder (R(x))

6J - Applic­ations

6J - Applications

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