Algebra Flashcards
When to use –> and for an example see 2008 4ii
Spot when we have an identity rather than an equality so compare coefficients
d = a^x * b^y * c^z therefore d has how many factors:
(x+1)(y+1)(z+1)
(A+1)(B+1) =
AB + A + B + 1
(A+B)(A+B) =
A^2 + 2AB + B^2
(A-B)(B-C)(C-A) =
A^2 + B^2 + C^2 - AB - AC - CA
Arithmetic mean > geometric mean
0.5(x+y) > root(xy)
k/k-k/k-k/k-k/k… therefore
x = k/k-x
Finding solutions
factorise, reason about the graph, consider the discriminant
99^2 =
100-1)^2 = 100^ - 200 +1
Divide by x^2 +3x + 2 means
F(-2) = F(-1)
Identity allows us to
compare coefficients rather than equality
With limits, constants…
become inconsequential
a^2 + b^2 = 1
a is greatest when b = 0
For a quadratic to have maximum value,
the x^2 coeffecient must be negative
a^x > cb^y –>
find counterexamples, use logs
Complete the square
maximum when leading coefficient is -ve
Use algebra to show things…:
(x-y)^2
Dividing by algebra
think of factor theorem or subbing in values
When finding powers, check if coefficients cancel
Compare coefficients
x^3 - x^2 - x + 1 = 0 –>
x^2(x-1)-1(x-1) = (x^2 - 1)(x-1)
Max/min –>
complete the square
a^4 - a^2 =
(a+1)(a-1)(a^2 + 1)
x^3 + 6yx^2 + 12xy^2 + 8y^3 =
(x+2y)^3
Don’t have fractions in equations of lines i.e. 1/a as
prevents a from equaling 0
For factorizing, if coefficient = 2
try +/- 0.5
(a+b)^6 =
a^6 + 6ba^5 + (6,2)a^4b^2 + (6,3)a^3b^3 + … + b^6
SinXcosX <= 0.5 use
(sinX - cosX)^2 > 0
a^2 - a^1.5 - 8 = 0
solutions
Expand (ax-by)^n
using choose function and pascals triangle
C(n,0)x^0(-by)^n \+ C(n,1)x^1(-by)^n-1 \+ ... \+ C(n,n)x^n(-by)^0
Pascals triangle
Row 0: 1 Row 1: 1 1 Row 2: 1 2 1 Row 3: 1 3 3 1 Row 4: 1 4 6 4 1 Row 5: 1 5 10 10 5 1 .... 1's on the outside, add both numbers above to get number below
therefore (ax-by)^n would follow with decreasing power of ax from n to 0, increasing power of -by from 0 to n, and coefficients reading left to right of the nth row of pascals triangle
(ax-by)^4, coefficients would b 1, 4, 6, 4 ,1
C(n,r)
n!/r!(n-r)!
X^2+1 is a factor of p(x)
sub x^2 = -1 into it then p(x^2=-1) = 0
Base cases: a and b are positive integers
let a, b =1
Expand (x-1)^3
x^3 - 3x^2 + 3x - 1
2x +3y < a
Form y = mc + c
