Algebra

Question 1

Modulus function definition; denoted as…

Answer 1

|x| means all y values will be positive
y ≥ 0

Question 2

Vertex of a modulus function in linear form

Answer 2

y = |mx + c|
vertex: x = -c/m

Question 3

Steps to sketch nonlinear modulus function

Answer 3

1. Sketch graph of f(x)
2. Reflect the part of the graph below the x-axis

Question 4

Modulus properties

Answer 4

|a x b| = |a|x|b|
|a / b| = |a| / |b| where b ≠ 0
|a + b|≠ |a|+|b|

y =|sin(x)| is not the same as
y = sin|x|

Question 5

Modulus on a number line

Answer 5

Distance between 2 points a and b on a number line is |b – a| as it is…
b – a where b – a ≥ 0
or
– (b – a) where b – a ≤ 0

If a ≥ 0, |x|≤ a is equivalent to –a ≤ x ≤ a
If a ≥ 0, |x – k|≤ a is equivalent to
k – a ≤ x ≤ k + a

Question 6

Methods to solve modulus function

Answer 6

Sketching graph
Modulus definition (x ≥ 0, x ≤ 0)
From results,
a ≥ 0, |x|≤ a is equivalent to –a ≤ x ≤ a

Question 7

Squares, square root and moduli

Answer 7

If x is any real number then x² ≥ 0 hence |x²|= x²
|x|= a² then, x² = a²
|x|> a² then, x² > a²
|x|< a² then, x² < a² , if a ≠ 0

Question 8

Division of polynomials

Answer 8

Example: 10/ 3 = 3 R(1)
Dividend: f(x), 10
Divisor: (x – a), 3
Quotient: q(x), 3
Remainder: R, 1

f(x) ≡ (x – a) * q(x) + R
dividend ≡ divisor x quotient + remainder

Where the degree of the remainder is less than the degree of the divisor. The degree of the quotient is equal to the degree of f(x) – the degree of (x – a).

Degree is the highest power present

Question 9

Division of polynomial steps

Answer 9

1. Lay out equation as long division
2. Ensure all powers of x are written including 0 values
3. To begin filling in the quotient or the answer multiply the divisor by a value that when subtracted by the dividend will cancel that power
4. Continue to do this till you are left with the remainder
5. To check remainder, if its degree is less than the degree of the divisor it is correct

Question 10

Remainder theorem

Answer 10

If f(x) is divided by (x – a) such that the quotient is q(x) and the remainder is R then f(a) = R

If f(x) is divided by (ax – b) such that the quotient is q(x) and the remainder is R then f(b/a) = R

To find the remainder when a dividend is divided by the divisor. Equate the divisor to 0 and solve for x. Sub the value of x into the the equation then calculate for the remainder

Question 11

Factor theorem

Answer 11

If f(x) is divided by (x – a) and the remainder is 0 then the division identity
f(x) ≡ (x – a) * q(x) + R , shows that (x – a) is a factor of f(x).

List the factors of the constant term in the polynomial to figure out what its factors are.
Sub each factor into the equation and if the answer is zero then that is an answer, take it’s negative for the factor in brackets.
Show working for all correct answers, include 1 or 2 incorrect factors as working.

If the degree of the polynomial is ≠ 1 then note that Ax – a is no different to –Ax + a, so only positive values are needed to be considered for values A, B and C
eg.
(Ax – a)(Bx – b)(Cx – c)

If there is no common factor of the coefficient of f(x) then none of the factors can share a common factor to be removed from the term in brackets.

Question 12

Simplifying partial fractions

Answer 12

For partial fractions the partial fraction is placed over the denominator and x values are substituted to find constants (A, B, C etc)

Denominator containing linear factors only:
An expression of the form
ax+b / (px + q)(rx + s)
can be split into partial fractions of the form
|A / (px + q)| + |B / (rx + s)|
Note: x values chosen for this should be in attempt to make one of the brackets equal to 0 and check answer by testing it with a simple value of x

Denominator containing repeated linear factors:
An expression of the form
ax² + bx + c / (px + q)(rx + s)²
can be split into partial fractions of the form
|A / (px + q)| + |B / (rx + s)²|+ … |C / (rx + s)|
For this try to make any of the brackets 0, after done to as many as possible choose a simple value of x eg 0 and solve for the other constant

Denominator contains a quadratic factor that cannot be factorised:
ax² + bx + c / (px + q)(rx² + s)
can be split into partial fractions of the form
|A / (px + q)| + |B / (rx² + s)|
Repeat same process as for repeated linear factor

Question 13

Binomial expansion

Answer 13

When n is a rational but not positive integer and |x|< 1, (1 + x)ⁿ can be written as a series, using the binomial expansion:
(1 + x)ⁿ = 1 + nx + [n(n - 1) / 2! ]x² +
[n(n – 1)(n – 2) / 3! ]x³ + …

Note:
If n is a positive integer, the series terminates at the term in xⁿ
The series 1 + x² + x³ + … only converges to 1 / 1 – x if |x|< 1

Question 14

Binomial expansions (approximate)

Answer 14

One use of binomial expansion is to find numerical approximations to square roots, cube roots and other calculations

1. Complete binomial expansion as required
2. Coefficient of x written into modulus form
3. Equate binomial approximate to given equation to find x
4. Sub x into binomial expansion to find what the approximate is

Question 15

Expanding other expressions

Answer 15

The binomial series can also be used to expand powers of expressions more complicated than 1+x or 1+ax

The use of both binomial expansion and working with partial fractions allows to solve these types of expressions

Note:
Rational functions where the degree of the numerator is greater than or equal to the denominator are called improper fractions. Improper fractions may not be the most convenient form to work with so by using long division express them in another form.