# Additional maths revision notes

- Created by: Sophie
- Created on: 15-05-14 18:49

First 389 words of the document:

Addit ional Mat hs

Remainder theorem

- f(x) ÷ g(x) = q(x) with a remainder of r(x) can be rearranged to

- so if you make the divisor equal to 0,

seeing as it is multiplied with the quotient, the quotient will also be equal to 0, and you'll be left

with the remainder

- So if f(x) = x3 + 3x² - x - 3, and g(x) = x + 1 :

- To make x + 1 (the divisor) = 0, x = -1

- So then (-1)3 + 3(-1)² - (-1) - 3 = r(x)

- -1 + 3 + 1 - 3 = r(x)

- R(x)=0 so there is a remainder of 0 when x3 + 3x² - x - 3 is divided by x + 1

- This can also be used to see if g(x) is a factor, as factors would give a remainder of 0

- Other linear factors can be found too by trial and error but they must be in the form (x - a)

where a is a factor of the constant term (in this case, -3):

- E.g. F(-3) = (-3)3 + 3(-3)² - (-3) - 3 = r(x)

- -27 + 27 + 3 - 3 = r(x) = 0

- Therefore (x + 3) is a factor

Binomial expansion

Quadratic Inequalities

The solution here is X < -1 or X > 1

- Pascal's Triangle gives us the coefficients for the expansion,

We know that as we are looking

e.g.:

for when y < 0, so when the curve

is below the x-axis.

Method: factorise, sketch (to see

when curve is below/above axis), if

y > ax² + bx + c, answer is where

curve is above x-axis

-You can use 5C0, 5C1, 5C2, 5C3, 5C4, 5C5 to find the coefficients (when it's to the power of 5)

Solving equations

- Quadratic formula:

- Factorisation:

- ax² + bx + c = (x + ½b)²

- ½b² must be equal to c

- When factorising a cubic equation, use the factor theorem then long division to split it

into a linear and quadratic first, then factorise the quadratic

- Completing the square:

- ax² + bx + c = (x + ½b)² - (½b)² + c

- E.g. x² + 6x + 3 = (x + 3)² - 9 + 3 = (x + 3)² - 6

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