Demonstrate Basic Algebra Skills US20659 Demonstrate Basic Algebra Skills

LIKE TERMS LIKE terms: 2x, 3x, 31x UNLIKE terms: 2x, 3 4ab, 7ab - LIKE terms are those with exactly the same letter, or combination of letters and powers LIKE terms: 2x, 3x, 31x UNLIKE terms: 2x, 3 4ab, 7ab 5x, 6x2 2ab, 2ac e.g. Circle the LIKE terms in the following groups: a) 3a 5b 6a 2c b) 2xy 4x 12xy 3z 4yx While letters should be in order, terms are still LIKE if they are not.

ADDING/SUBTRACTING TERMS - We ALWAYS aim to simplify expressions from expanded to compact form - Only LIKE terms can be added or subtracted - When adding/subtracting just deal with the numbers in front of the letters e.g. Simplify these expanded expressions into compact form: a) a + a + a 1 1 1 = (1 + 1 + 1)a b) 5x + 6x + 2x = (5 + 6 + 2)x = 3a = 13x c) 3p + 7q + 2p + 5q = (3 + 2)p (+ 7 + 5)q d) 4a + 3a2 + 7a + a2 1 = 5p + 12q = (4 + 7)a (+ 3 + 1)a2 = 11a + 4a2 - For expressions involving both addition and subtraction take note of signs e.g. Simplify the following expressions: a) 4x + 2y – 3x = (4 – 3)x + 2y b) 3a – 4b – 6a + 9b = x + 2y = (3 - 6)a (- 4 + 9)b = -3a + 5b c) 3x2 - 9x + 6x2 + 8x - 5 = (3 + 6)x2 (- 9 + 8)x - 5 = 9x2 - x - 5 If the number left in front of a letter is 1, it can be left out d) 4ab2 +2a2b – 5ab2 + 3ab = -ab2 + 2a2b + 3ab

POWER RULES a) p10 × p2 = p(10 + 2) b) a3 × a2 × a = a(3 + 2 + 1) 1 1. Multiplication - Does x2 × x3 = x × x × x × x × x ? YES - Therefore x2 × x3 = x5 - How do you get 2 3 = 5 ? + - When multiplying index (power) expressions with the same letter, ADD the powers. No number = 1 i.e. p = 1p1 e.g. Simplify a) p10 × p2 = p(10 + 2) b) a3 × a2 × a = a(3 + 2 + 1) 1 = p12 = a6 - Remember to multiply any numbers in front of the variables first e.g. Simplify a) 2x3 × 3x4 = 2 × 3 x(3 + 4) b) 2a2 × 3a × 5a4 1 = 6 x7 = 2 × 3 × 5 a(2 + 1 + 4) = 30 a7

2. Division - Does 6 = 1 ? 6 YES - Therefore x = 1 x - Does x5 = x × x × x × x × x ? x3 x × x × x YES = x × x × 1 × 1 × 1 - Therefore x5 = x2 x3 - How do you get 5 3 = 2 ? - - When dividing index (power) expressions with the same letter, SUBTRACT the powers. e.g. Simplify a) p5 ÷ p = p(5 - 1) b) x7 x4 = x(7 - 4) 1 = p4 = x3 - Remember to divide any numbers in front of the variables first e.g. Simplify a) 12x5 ÷ 6x4 = 12 ÷ 6 x(5 - 4) b) 5a7 15a2 ÷ 5 = 1 5 a(7 - 2) = 2 x = 1 5 a5 or a5 5 If the power remaining is 1, it can be left out of the answer

a) (c4)6 = c(4 × 6) b) (a3)3 = a(3 × 3) = c24 = a9 a) (3d2)3 = 33 3. Powers of powers - Does (x2)3 = x2 × x2 × x2 ? YES - Does x2 × x2 × x2 = x6 ? YES - Therefore (x2)3 = x6 - How do you get 2 3 = 6 ? × - When taking a power of an index expression to a power, MULTIPLY the powers e.g. Simplify a) (c4)6 = c(4 × 6) b) (a3)3 = a(3 × 3) = c24 = a9 - If there is a number in front, it must be raised to the power, not multiplied e.g. Simplify a) (3d2)3 = 33 × d(2 × 3) b) (2a3)4 = 24 × a(3 × 4) = 27 d6 = 16 a12 - If there is more than one term in the brackets, raise all to the power e.g. Simplify b) (4b2c5)2 = 42 b(2 × 2) c(5 × 2) a) (x3y z4)3 = x(3 × 3) y(1 × 3) z(4 × 3) 1 = x9 y3 z12 = 16 b4 c10

EXPANDING - Does 6 × (3 + 5) = 6 × 3 + 6 × 5 ? YES 6 × 8 = 18 + 30 48 = 48 - The removal of the brackets is known as the distributive law and can also be applied to algebraic expressions - When expanding, simply multiply each term inside the bracket by the term directly in front e.g. Expand a) 6(x + y) = 6 × x + 6 × y b) -4(x – y) = -4 × x - -4 × y = 6x + 6y = -4x + 4y c) -4(x – 6) = -4 × x - -4 × 6 d) 7(3x – 2) = 7 × 3x - 7 × 2 = -4x + 24 = 21x - 14 e) x(2x + 3y) = x × 2x 1 1 + x × 3y f) -3x(2x – 5) = -3x × 2x - -3x × 5 = 2x2 + 3xy = -6x2 + 15x Don’t forget to watch for sign changes!

- If there is more than one set of brackets, expand them all then collect any like terms. e.g. Expand and simplify a) 2(4x + y) + 8(3x – 2y) = 2 × 4x + 2 × y + 8 × 3x - 8 × 2y = 8x + 2y + 24x - 16y = 32x - 14y b) -3(2a – 3b) – 4(5a + b) = -3 × 2a - -3 × 3b - 4 × 5a + -4 × 1b = -6a + 9b - 20a - 4b = -26a + 5b

EXPANDING TWO BRACKETS - To expand two brackets, we must multiply each term in one bracket by each in the second Remember integer laws when multiplying e.g. Expand and simplify a) (x + 5)(x + 2) = x2 + 2x + 5x + 10 b) (x - 3)(x + 4) = x2 + 4x - 3x - 12 = x2 + 7x + 10 = x2 + 1x – 12 To simplify, combine like terms c) (x - 1)(x - 3) = x2 - 3x - 1x + 3 c) (2x + 1)(3x - 4) = 6x2 - 8x + 3x - 4 = x2 – 4x + 3 = 6x2 – 5x – 4

PERFECT SQUARES - When both brackets are exactly the same To simplify, combine like terms e.g. Expand and simplify Watch sign change when multiplying a) (x + 8)2 = (x + 8)(x + 8) b) (x - 4)2 = (x – 4)(x – 4) = x2 + 8x + 8x + 64 = x2 - 4x - 4x + 16 = x2 + 16x + 64 = x2 - 8x + 16 Write out brackets twice BEFORE expanding c) (3x - 2)2 = (3x – 2)(3x – 2) = 9x2 - 6x - 6x + 4 = 9x2 – 12 x + 4

DIFFERENCE OF TWO SQUARES - When both brackets are the same except for signs (i.e. – and +) e.g. Expand and simplify a) (x – 3)(x + 3) = x2 + 3x - 3x - 9 b) (x – 6)(x + 6) = x2 + 6x - 6x - 36 = x2 – 9 = x2 – 36 Like terms cancel each other out c) (2x – 5)(2x + 5) = 4x2 + 10x - 10x - 25 = 4x2 – 25

FACTORISING - Factorising is the reverse of expanding - To factorise: 1) Look for a common factor to put outside the brackets 2) Inside brackets place numbers/letters needed to make up original terms You should always check your answer by expanding it e.g. Factorise a) 2x + 2y = 2( ) x + y b) 2a + 4b – 6c = 2( ) a + 2b - 3c - Always look for the highest common factor e.g. Factorise a) 6x - 15 = 3( ) 2x - 5 b) 30a + 20 = 10( ) 3a + 2 - Sometimes a ‘1’ will need to be left in the brackets e.g. Factorise a) 6x + 3 = 3( ) 2x + 1 b) 20b - 10 = 10( ) 2b - 1

a) 5a2 – 7a5 = a2( ) 5 - 7a3 b) 4b2 + 6b3 = 2b2( ) 2 + 3b - Letters can also be common factors e.g. Factorise a) cd - ce = c( ) d - e b) xyz + 2xy – 3yz = y( ) xz + 2x - 3z c) 4ad – 8a = 4 ( ) a d - 2 - Powers greater than 1 can also be common factors e.g. Factorise a) 5a2 – 7a5 = a2( ) 5 - 7a3 b) 4b2 + 6b3 = 2b2( ) 2 + 3b

FACTORISING QUADRATICS The general equation for a quadratic is ax2 + bx + c When a = 1 - You need to find two numbers that multiply to give c and add to give b e.g. Factorise To check answer, expand and see if you end up with the original question! a) x2 + 11x + 24 = (x + 3)(x + 8) 1, 24 List pairs of numbers that multiply to give 24 (c) Check which pair adds to give 11 (b) Place numbers into brackets with x 2, 12 3, 8 4, 6 b) x2 + 7x + 6 = (x + 1)(x + 6) 1, 6 List pairs of numbers that multiply to give 6 (c) Check which pair adds to give 7 (b) Place numbers into brackets with x 2, 3

- Expressions can also contain negatives e.g. Factorise a) x2 + x – 12 = (x - 3)(x + 4) b) x2 – 6x – 16 = (x + 2)(x - 8) - 1, 12 1, 16 - As the end number (c) is -12, one of the pair must negative. As the end number (c) is -16, one of the pair must negative. - 2, 6 2, 8 - - 3, 4 4, 4 - Check which pair now adds to give b Make the biggest number of the pair the same sign as b Check which pair now adds to give b Make the biggest number of the pair the same sign as b c) x2 – 9x + 20 = (x - 4)(x - 5) d) x2 – 10x + 25 = (x - 5)(x - 5) - 1, 20 - - 1, 25 - = (x - 5)2 As the end number (c) is +20, but b is – 9, both numbers must be negative - 2, 10 - - 5, 5 - As the end number (c) is +25, but b is – 10, both numbers must be negative - 4, 5 - Check which pair now adds to give b

SPECIAL CASES 1. No end number (c) e.g. Factorise a) x2 + 6x + 0 b) x2 – 10x = x( ) x - 10 0, 6 = x(x + 6) Add in a zero and factorise as per normal OR: factorise by taking out a common factor 2. No x term (b) (difference of two squares) e.g. Factorise a) x2 - 25 + 0x = (x - 5)(x + 5) b) x2 – 100 = (x )(x ) - 10 + 10 -5, 5 c) 9x2 – 121 = (3x )(3x ) - 11 + 11 Add in a zero x term and factorise OR: factorise by using A2 – B2 = (A – B)(A + B)