REVIEW OF SIMPLE FACTORING

1. REVIEW OF SIMPLE FACTORING You are expected to be able to factor algebraic expressions. You will also need to be able to solve algebraic equations by factoring. When appropriate, you may use the quadratic formula rather than by factoring. However the quadratic formula only works for problems of the form ax2 + bx + c = 0. The basic types of expressions you will be factoring are of the type: Greatest Common Factor: G C F Difference of two squares: a2– b2= (a + b)(a– b) Trinomial: x2 + bx + c Trinomial: ax2 + bx + c

2. Greatest Common Factor: G C F The GCF must be found in all of the terms, not in a majority of the terms. If it is not found in every term then it does not exist or in other words the GCF is a 1. EXAMPLE: 1. 6x 3– 9x5y 2 + 12x2y 4 The only common term is 3x 2 3x 2(2x–3x3y 2 + 4y 4 ) The only common factor is 1. Therefore we say the expression is prime. 2.4x2– 8x + 5y The first step in factoring should always be to factor the GCF. The remainder of the problem may or may not be factorable.

3. Difference of two squares: a 2– b 2= (a + b)(a– b) Number Square Root 1 1 4 2 9 3 16 4 25 5 etc. • Recognition: • Two terms always separated by a minus sign • All numbers are perfect squares • All exponents are divisible by two • Solve: • The two parentheses will always be alike except one will be + the other –. ( + )( – ) • In order of the problem write the square root for the number and one half of the variable’s exponent. • Write the same thing in the second parenthesis. • The order of the signs is not important. Example: 1. 9x2 – 25y 4 (3x + 5y 2)(3x – 5y 2) 2. 28a2– 175c2 7(4a2 – 25c2) 7(2a + 5c)(2a – 5c)

4. Trinomial: x 2 + bx + c • I call this a plain trinomial because the coefficient of x 2 is a one. • Reading from left to right answer the following questions: • What two numbers multiply to give c • and at the same time (if +) add to give b or (if –) have a difference of b? • Do not worry about the sign of b yet. Put the signs in the answer last not first. What two numbers have a product of 20 and atthe same time add to give 12? Example: x2– 12x+ 20 The numbers are 10 and 2 because 10times 2 is 20 and 10 plus 2 is 12. ( x 10)( x 2) However we need – 12 so it must have been – 10 and – 2. ( x – 10)( x – 2)

5. What two numbers have a product of 33 and atthe same time have a difference of 8? 1. a2– 8a– 33 The numbers are 3 and 11 because3 times 11 is 33 and 11 minus 3 is 8. (a 3 )(a 11) (a + 3 )(a– 11) It must be – 11 and + 3 to be – 8 What two numbers have a product of 84 and atthe same time have a sum of 19? 2. c2 + 19c+ 84 1*84 = 84 but 84 + 1 ≠ 192*42 = 84 but 2 + 42 ≠ 193*28 = 84 but 3 + 28 ≠ 194*21 = 84 but 4 + 21 ≠ 196*14 = 84 but 6 + 14 ≠ 197*12 = 84 and 7 + 12 = 19 It must have been + 7 and + 12 to be + 19 (c 7 )(c 12) If the problem had been c2– 19c + 84,7 and 12 would still be the numbers used except it would be a – 7 and – 12. (c + 7 )(c + 12) (c– 7 )(c– 12) If you do not know what the numbers are, learn to find them in a systematic way. Stop all of the guessing! Try

6. 1. Multiply these two numbers including the sign. Trinomial: ax2 + bx + c The value of a adds to the difficulty of factoring the trinomial. If we use a concept that is taught in factoring what is called group factoring, the problem is simplified but only if we are consistently systematic about the procedure and not trying to guess the answer as many students try to do. – 120 8x2– 14x – 15 2. Now find two numbers that multiply to give 120 and at the same time have a difference of 14. 3. Write 8x2 – 15 Works6 & 20 Do not work 1 & 120 2 & 60 3 & 40 4 & 30 5 & 24 We need – 14x. So– 20x + 6x is – 14x 4. Write 8x2– 20x + 6x – 15 6*20 = 12020 – 6 = 14 5. Split the problem in half. Write this in step 4 8x2– 20x + 6x – 15 The underlined parentheses must be exactly alike. If not, a mistake was made. 4x(2x– 5) 3(2x– 5) 6. Factor each half + (2x– 5) (4x + 3) • Write the common factor in the first parenthesis Final answer 8. and the leading coefficients in the second.