Algebra+Y10


 * ==Expanding Brackets==

When expanding out brackets we need to make sure that the term outside of the bracket is multiplied by everything inside. a(b + c) = ab + ac This is called the distributive law.

When expanding out double brackets we need to apply the same principle. (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd

If the values of a,b,c and d allow then terms can be collected togther to **simplify** the expression.

There are 2 methods here than I like to use:

FOIL Method - remembering to multiply the First terms, Outside terms, Inside terms and Last terms in the double brackets.

E.g. (a + b)(c + d) F = ac O = ad I = bc L = bd

Smiley Face Method - create 2 eyebrows, and nose and a mouth to join all the terms, then Multiply along the lines to expand out the brackets.

E.g. (a + b)(c + d) left eyebrow = ac right eyebrow = bd nose = bc mouth = ad


 * ==Factorising Quadratic Expressions==

When we factorise quadratic expressions such as x 2 + 3x + 2 we are looking for 2 factors to put into double brackets (x )(x ).

To do this we have to find 2 numbers (our factors) that will multiply together to give us 2 and add toether to give us 3 (relating to the coefficient of x)

2 has factors 1 and 2, so when we multiply these we get 2 and when we add them together we get 3. We can now say that x 2 + 3x + 2 **//factorises//** to **(x + 1)(x + 2).**

__Example 2:__ solve x 2 + 7x + 12 = 0

12 has factors 1 and 12, 2 and 6, 3 and 4. 1 + 12 doesn't equal 7 so these aren't our factors. 2 + 6 doesn't equal 7 so these aren't our factors. 3 + 4 = 7 so we can say that 3 and 4 are our factors and...

x 2 + 7x + 12 **//factorises//** to **(x + 3)(x + 4)**

We can easily check this by expanding out the brackets by using either the FOIL or Smiley Face methods from earlier this term. See above to refresh.

Now we have factorised the equation - we need to solve it! This means that we need to find a value (or two) for x.

Rewriting our equations as: (x + 3)(x + 4) = 0 we can say that either (x + 3) = 0 or (x + 4) = 0 x + 3 = 0 x + 4 = 0
 * x = -3 is a solution**
 * x = -4 is also a solution**

Try out some questions on factorising. Remember your rules for negative numbers as these can get tricky. -2 x -3 = 6 -2 x 3 = -6 2 x -3 = -6 2 + 3 = 5 2 - 3 = -3 + 2 = -1 -2 + 3 = 3 - 2 = 1

On these questions, solve them by making the quadratic expression equal to 0.

When we factorise quadratics such as 2x 2 + 5x + 2 we have to be more careful as to the factors we choose.
 * ==​Factorising Quadratics for 2x 2 + ax + b==

We can see that our brackets will look like (2x )(x )

2 has the factors 1 and 2, but we then have to check that these work by putting them into the formula (F1 + (2 x F2))

So we can now say that 2x 2 + 5x + 2 **//factorises//** to **(2x + 1)(x + 2)**
 * Expression || Factor 1 || Factor 2 || Multiply to get b ||< (F1 + 2 x F2) to get a ||
 * 2x 2 + 5x + 2 || 2 || 1 || 2 ||< 2 + (2x1) = 4 as this is not 5 then this is not right. ||
 * 2x 2 + 5x + 2 || 1 || 2 || 2 ||< 1 + (2 x 2) = 5. As this equals the coefficient of x the factors are correct. ||

Again we can check this by expanding out the brackets!

Test your self on these using the tables to help you calculate:

(x + 2)(x - 2) = x 2 - 4
 * ==Solving using the difference of squares==

(x + 3)(x - 3) = x 2 - 9

(x + 4)(x - 4) = x 2 - 16

(x + a)(x - a) = x 2 - a 2

This is the difference of squares rule and can be used to help factorise certain equations that are in the form x 2 - a 2 = 0.

__Example__

x 2 - 81 = 0 (x + 9)(x - 9) = 0 x + 9 = 0 or x - 9 = 0
 * x = -9 or x = 9 are both solutions to this equation.**

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