青藜苑教育 Find the value of x in eah of the following: 6 x 6 x a) b) ) (7 6 ) x 7 d) 7 x 0 Examination S

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1 青藜苑教育 Revision Topi 1: Algebra Indies: At Grade B and C levels, you should be familiar with the following rules of indies: a b a b y y y i.e. add powers when multiplying; y a a b a b y y i.e. subtrat powers when dividing; b ab ( y ) y i.e. when you have a power of a power, multiply powers together. y 0 1 i.e. anything to the power 0 is 1. Example: Simplify eah of the following, if possible: a) a b) ) 0 ( x ) d) e e) a b a) b) a (add powers together). simplifyin g top subtrating powers 0 ) (x) 1 (anything to the power 0 is 1) 6 d) e e (multiply powers together) e) a b this annot be simplified as the base numbers (a and b) are different letters. Example : Work out the value of eah of the following: a) 0 b) ) a) 81 b) (anything to the power 0 is 1) ) Example : Simplify: 7 15y a) x x b) ) 6a b b y 5 a) x x x x 1x (i.e. multiply together the numbers and add the powers) b) ) 15y y 7 15 y 7 5y 5 (i.e. divide numbers and subtrat powers) 6a b b (6 ) ( a ) ( b b ) b. 5 6 Examination Style Question 1

2 青藜苑教育 Find the value of x in eah of the following: 6 x 6 x a) b) ) (7 6 ) x 7 d) 7 x 0 Examination Style Question : Simplify fully eah of these expressions. Leave your answers in power form. 5 6 a) b) ) 5 5 d) 9 9 e) 6. Examination Style Question : Simplify eah of the following expressions. a) a 8 b) 6x x ) x y 5x y Expanding brakets Expanding out a single braket: You an remove a single braket by multiplying everything inside the braket by the number, or expression, on the outside. Example Expand the following brakets: a) 6(7d ) b) y(8y x + 1) ) 5x (x x ) d) xy(x y) a) 6(7d ): Multiply both the 7d and the by the number on the outside: We get 6(7d ) = d - b) y(8y x + 1): Multiply everything in the braket by y: We get: y(8y x + 1) = 8y xy + y ) 5x (x x ) : Multiply all the terms inside the braket by 5x: We get 5x(x x ) 10x 15x 10x d) xy(x y): Multiply the x and y by xy: This gives: xy(x y) = 8x y xy. You need to take are when there is a minus sign in front of a braket. Example : Expand and simplify: a) -(x y) b) 6x(x + ) (5x ) a) Here we multiply everything in the braket by -. This gives: -(x y) = -8x + 1y b) If we multiply out the first braket we get: 6x(x + ) = 18x + 1x If we multiply out the seond braket, we get: (5x ) = -15x + 6. Putting it all together: 6x(x + ) (5x ) = 18x + 1x - 15x + 6 = 18x x y = +1y as two minuses multiply to make a plus!

3 青藜苑教育 Examination Question: a) Multiply out: t ( t t ). b) Multiply out and simplify: (a + 6) (a 6) 1a b ) Simplify:. ab Expanding out double brakets: When there is a pair of brakets multiplied together, you need to multiply everything in the first braket by everything in the seond. Example: Multiply out the following brakets: a) (x )(x + ) b) (x y)(x ) ) (x + y) a) (x )(x + ) We an expand these brakets diretly, multiplying everything in the first braket by the terms in the seond braket. This gives: (x )(x + ) = x + 1x x 8 = x + 10x 8. Alternatively, you an draw a grid to help expand the brakets: x - x x -x 1x -8 Adding the numbers inside the grid gives: (x )(x + ) = x + 1x x 8 = x + 10x 8. b) (x y)(x ) The grid for this would look like: x -y x x -6xy - -8x +1y Adding the numbers inside the grid gives: (x y)(x ) = x 6xy 8x + 1y This annot be simplified! ) (x + y) To square a braket, you multiply it by itself! Drawing a grid: x y x x 6xy y 6xy 9y Adding the expressions inside the grid: (x + y)(x + y) = x + 6xy + 6xy + 9y = x + 1xy + 9y Examination Questions: Expand and simplify: 1) (x y)(x + y) ) (x 5)(x + ) ) (x y) Fatorising Fatorising is the reverse of multiplying out brakets, i.e. when you fatorise an expression you need to put brakets bak into an expression.

4 青藜苑教育 Common fators: Some expressions an be fatorised by finding ommon fators. Example: Fatorise the following expressions. a) 1e + 18 b) xy + 5x ) x 6x d) x + x e) 10x y 15xy. a) We look for a ommon fator of 1e and 18. We notie that 6 goes into both of them. We therefore write 6 outside a braket: 1e + 18 = 6(e + ) 6 goes into 1 twie 6 goes into 18 times b) We notie that x appears in both xy and 5x. This an be taken outside a braket: xy + 5x = x y + 5 x = x( y + 5) = x(y + 5) ) As x is x x, both x and 6x have x as a fator. So, x 6x = x x 6x = x(x 6) d) Looking at the number parts, we notie that is a ommon fator of both and. x + x = (x + x). This hasn t been ompletely fatorised yet, as both x and x also ontain an x. We therefore an an x outside the braket. x + x = (x + x) = x(x + 1). x is 1x, so when x is taken outside the braket, we are left with 1 inside. e) 10x y 15xy : Looking at the numbers, we see that both 10 and 15 have 5 as a fator. Both terms also have an x and a y in ommon. We an therefore fatorise by writing 5xy in front of a braket. 10x y 15xy = 5 x x y 5 x y y = 5xy(x y) Note: You an hek your answers by expanding out the brakets. Examination Question Fatorise ompletely: (a) x x (b) p q + pq. Examination Question : a) Expand and simplify: 5(x + 1) (x ). b) Expand and simplify: (x + 5)(x ). ) Fatorise ompletely: 6a 9ab. Fatorising quadratis Simple quadratis like x x or x 7x 1 an often be fatorised into two brakets. General steps for fatorising x bx Example: Fatorise x 9x 18

5 青藜苑教育 Step 1: Find two numbers that multiply to make and add to make b. Step : Write these two numbers in the brakets: (x )(x ) Step 1: Find two numbers that multiply to make 18 and add to give 9. These numbers are 6 and. Step : The fatorised expression is (x + 9)(x + ) Example : Fatorise x 7x 1 We need to find two numbers that multiply to make 1 and add to give -7. These numbers are - and -. So the answer is (x )(x ). Example : a) Fatorise x 8x 0 b) Solve x 8x 0 0 a) We have to find two numbers that multiply to make -0 and add to give -8. These are -10 and. The fatorised expressions is (x 10)(x + ). b) To solve the equation x 8x 0 0 we use our fatorised expression: (x 10)(x + ) = 0. We have two brakets that multiply together to make 0. The only way this an happen is if one of the brakets is 0. If the first braket is 0, then x 10 = 0, i.e. x = 10. If the seond braket is 0, then x + = 0, i.e. x = -. So the solutions are x = 10 and x = -. Examination question Fatorise x x 1. Hene or otherwise solve x x 1 0. Examination question: a) Fatorise x + 8y. b) Fatorise ompletely a 6a. ) Fatorise x 9x 18. Examination question: a) Expand and simplify (x + 5)(x ). b) Fatorise x 5x 1. ) Solve x 5x 1= 0. Differene of two squares When you expand out the brakets for (x + a)(x a) you get x + ax ax a whih simplifies to x a. The result x a = (x + a)(x a) is alled the differene of two squares result. Examples: 1) y 16 = y = (y + )(y ). ) z 5 = z 5 5 = (z + 5)(z 5). ) 9x y = (x) (y) = (x + y)(x y).

6 青藜苑教育 Examination question: a) Fatorise x y. b) Use your answer to a) to work out the EXACT answer to