QUANT Flashcards
(92 cards)
UNFOIL: (A + B)^2 Unfoiled
(A + B) (A+B) = A^2 + 2AB + B^2
UNFOIL: (A - B)^2
(A - B) (A - B) = A^2 - 2AB +B^2
[Note neg x neg = Pos, but when factored some will not have neg sign so will be neg x pos = neg]
UNFOIL: A^2 - B^2
DIFFERENCES OF SQUARES: (A + B) (A - B)
Negative - Negative
-5 - -5 = -5+5 = 0 -4–5= -4+5 = 1.
Positive - Negative
5 - -6 = 5 + 6 = 11
What is the square root of 2
1.4
what is the square root of 3
1.7
What is the square root of 4
2.2
Negative x Negative
Positive (Same signs = positive)
Negative x Positive
Negative (Opposite signs = Negative)
Negative / Negative
Positive (Same signs = Positive)
Negative / Positive
Negative (Opposite signs = Negative)
Negative + Negative
Negative (Keep sign)
What can be learnt from (k^3 - k)
Consecutive integers when FOILED: K(K^2 - 1) ===> K^2-1 is same as (K+1) (K-1). (Another consecutive series of 2)
What can be learnt form (K^2 - 1)
Consecutive integers of 2: (K - 1) (K + 1) = (K^2 -1)
Solve x^2y^2 - xy= 6
Factor: xy(xy-1)=6. meaning 6 = xy x xy-1 (I.e. 1 less than xy) figure out which two numbers 1 number apart multiply to get six
2 and 3 positive or negative but same sign as 6 is positive.
Solve X^2 - Y^2
Difference of squares: (x - y) (x + y). opposite signs = negative thus + and minus
When I see a question with numbers raised to the 4th power I will…
Show these numbers as squares of squares in order to better manipulate the equation. i.e. 9x^4 = (3x^2)^2
When I see divisibility problems I will…
Prime factorize and know that the smallest integers that are divisible will have prime factors contained by no more than needed (I.e. the exact amount of factors)
List the first 10 prime numbers
2, 3, 5, 7, 11, 13, 17,19, 23, 29
Which is greater 3/4 or 31/42 ….
to find this you cross multiply and keep the anchor on the numerators: 3 x 42 = 124. and 31 x 4 = 126. 126 > 124. so 3/4 > 31/42
How can you tell how many trailing zeros there are in a number based on their prime factorisation? I.e. How many trailing zeros are there in an interger with the prime factorisation of:
2^16 x 3^15 x 4^11 x 5^19
Trailing zeros= ,0000 the only numbers with a 0 at the end are multiples of 10.
10 has the prime factorisation of 2 x 5
so you can get the number of 0’s by counting the number of 2 x 5 pairs.
In this case the number of trailing zeros is 19 because the max pairs are limited to 5^19 as 4^11 has some 2’s in their as well
If a set of numbers (D) are all multiples of 3 what will this mean for their prime factorization?
All the numbers in that set will have a prime factorisation that includes the number 3 if they are multiples of 3. I.e. 24 = 2 x 2 x 2 x 3
If a set of numbers (E) are all factors of 400 what does this tell us about the prime factorisation of all the numbers in set E.
The prime factorisation of 400 = 2^4 x 5^2. All the factors of 400 will have some combination of these same prime factors. I.e. 40 = 2^3 x 5 ….the prime factors will always be the same.