( ) FACTORING. x In this polynomial the only variable in common to all is x.

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1 FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5*. The multiples are 5 and. In a polynomial it is the same way, however, the procedure is somewhat more complicated since variables, not just numbers, are involved. There are different ways of factoring an equation depending on the complexity of the polynomial. Factoring out the greatest common factor The first thing to do when factoring is to look at all terms and break up each term into its multiples: Examples: Factor: 8x + 4x + 10x ( 8x + 4x + 10) x In this polynomial the only variable in common to all is x. x(4x+x+5) The two is also common to all terms. Therefore, this is as far as the polynomial can be factored. Factor: 6x + 8x + 16x 6x + 8x + 16x The common variable is x, and the smallest exponent is 1. The common multiple is. x x + 4x + 8 Therefore, the greatest common factor is x. ( ) x x = 6x You can also look at it this way: ( x)( 4x) = 8x ( x)( 8) = 16x

2 Factoring by grouping There will be occasions when there are no common factors for all terms, but there will be terms that have a variable in common while other terms have a different variable in common. In this case, we can factor by grouping together these common terms. Examples: Factor: ax + ay + bx + by ax + ay + bx + by Notice how there are two variables, x and y. Group ax + bx + ay + by together the x s and the y s and then x a + b + y a + b factor out the common factor in each group. ( ) ( ) Factor: + 4 x x y y + There are two variables present, x and y. 4 x x y y Start by grouping two terms at a time. ( x ) + y ( x ) Notice how the two polynomials inside the parentheses are similar. To make them the same change the sign in the second polynomial. ( x ) y ( ) x Now that they are the same, factor out the common term. x y Now it is completely factored out. Factoring a difference of squares For a difference of squares, a b, the factors will be ( a b)( a + b). Examples: x y = ( x y)( x + y) ( ) ( ) ( ) 1 x = 1 x = 1 x 1+ x = 1 x 1+ x 1+ x 4

3 Factoring the sum and difference of two cubes For the addition or subtraction of two cubes, the following formulas apply: a + b = a + b a ab + b a b = a b a + a + bb Examples: x + 8 ( x) ( ) ( x ) x ( x)( ) = + Manipulate to be in a + b form. ( ) ( x )( x x 4) = + + follow formula. = + + 8y 7 Manipulate to be in a b = y follow formula. ( ) ( ) ( y ) ( y) ( y)( ) ( ) = + + ( y )( 4y 6y 9) = + + form. Factoring trinomials Factoring trinomials is based on finding the two integers whose sum and product meet the given requirements. Example: Find two integers whose sum is 11 and whose product is 0. Step 1 The product is positive, the sum is negative, therefore, both integers must be =+ and + = negative. ( ) ( ) Step Possible pairs of factors that will give a positive answer: 10 = 0 1. ( 1)( 0) = 0.. ( )( 15) = = 0 Step The pair whose sum is 11 is 5 and 6. Therefore, the critical integers are = 11. and 6, ( )

4 Factoring trinomials of the form x + bx + c To factor trinomials of the form x + bx + c follow the same principle described above. Find two integers whose sum equals the middle term and whose product equals the last term. Examples: Factor: x x 6 Step 1 Find two integers whose sum is equal to 1 and whose product is equal to 6. The product is negative; therefore, we need integers that have different sign. The sum is negative, so the bigger integer is negative. Step Possible pairs of factors that will give us 6: 1. ( )( ) = 6. ( )( ) = 6. ( 1)( 6) = 6 4. ( 1)( 6) = 6 Step The pair whose sum is 1 is and, ( + = 1) The critical integers are and. Step 4 The factors for x x 6 are ( x )( x + ).. Factor: x + 16x + 60 Step 1 Find two integers whose sum is 16 and whose product is 60. Both integers must be positive since their sum and product are positive. Step Possible pairs of factors: 1. ( 1)( 60). ( )( 0) 5.. ( )( 0) 4. ( 4)( 15) Step The pair whose sum is 16 is (6, 10). Therefore, the critical integers are 6, 10. Step 4 The factors for x + 16x+ 60 are ( x + 6)( x + 10 ).

5 Factoring trinomials of the form ax + bx + c In this type of trinomials the coefficient of x is not 1, therefore we must consider the product of Examples: Factor: x 7x 4 Step 1 Find two integers whose sum is 7 and whose product is ( )( ) 4 = 8. The a c. integers must be of different signs since their product is a negative number. The larger integer is negative since their sum is negative. Step Possible pairs of factors: 1. ( 1)( 8) = 8.. ( 1)( 8) = = 8 4 = 8 Step The pair whose sum is 7 is 1 and 8. The critical integers are 1 and 8. Step 4 Use the critical integers to break the first degree term into two parts: x 7x 4 = x 8x + x 4 Step 5 Factor by grouping the first two and last two terms: x 8x + x 4 = x( x 4) + 1( x 4) There are now two terms in the expression. The binomial factor in each of these two terms should be the same; otherwise, there is an error. In this case, the common binomial factor is( x 4). Step 6 Factor out the common binomial factor: x( x 4) + 1( x 4) = ( x 4)( x + 1) Factor: 6x x + 0 Step 1 Find two integers whose sum is and whose product is 6(0) = 10. Since the product is positive and the sum is negative, the two integers are negative. Step Possible pairs of factors: 1. ( 1)( 10) = ( 4)( 0) = ( )( 60) = ( 5)( 4) = ( )( 40) = ( 6)( 0) 8 15 = = 10 Step The sum of 8 and 15 is. Therefore, the critical integers are 8 and 15.

6 Step 4 Use the critical integers to break the first degree term into two parts: 6x x + 0 = 6x 15x 8x + 0 Step 5 Factor, separately, the first two and last two terms: 6x 15x 8x + 0 = x( x 5) 4( x 5) Step 6 Factor out the common binomial factor. x( x 5) 4( x 5) = ( x 5)( x 4)

7 FACTORING EXERCISES 1. x 10x 8. x ( x ) 5x( x ) ( x ). x + x 9x x 6 x 10x + 5 y y 1 a ( x + 5) + 4( x + 5) t + t 4 x 18x + 7x x 100x x x x + xy + y 1. x ( x + ) + 1x( x + ) + 15( x + ) 14. x + x x y + y + x a x + 8x x 0x 50x 19. 9x + x x + x

8 FACTORING ANSWER TO EXERCISES x. x ( x ) 5x( x ) ( x ) 1. 10x 8 x 1x + x 8 ( x )(x 5x ) x( x 4) ( x 4) ( x )( x 4) + ( x )(x 6x + x ) + ( x )(x( x ) + 1( x )) ( x )( x+ 1)( x ) = ( x ) ( x+ 1). x + x 9x x x ( x + ) 9( x + ) ( x 9)( x + ) 10x + 5 y ( x 10x + 5) y ( x 5) y x 9 = ( x + )( x ) Difference of squares: ( a b ) = ( a b)( a + b) ( )( x + )( x + ) x where ( ) ( x 5 y)( x 5 + y) a = x 5 and b= y x y 6. ( ) ( ) 1 a x y = x y Difference of cubes: Difference of squares: ( a b ) = ( a b)( a + b) where x y x y x y 6 6 a b = ( a b)( a + a + bb ) a x b y = and = a 1 a = and b = = a 1 = a a = + a a ( a) Sum & Difference of Cubes: a + b = a + b a ab + b a b = a b a + a + bb = ( x + y)( x xy + y )( x y)( x + xy + y ) = ( x + y)( x y)( x + xy + y )( x xy + y )

9 7. ( x+ 5) + 4( x+ 5) + 4= ( x+ 7)( x+ 7) = ( x+ 7) 8. 6 ( )( ) ( ) ( ) 1 t + t = t 1 t 1 = t 1 t + t ( ) ( ) x 18x + 7x = x x 6x+ 9 = x x x 100x 600 = 100( x )( x+ ) 11. 8x x 15 = ( 4x+ 5)( x ) 1. x + xy + y = ( x + y)( x + y) 1. x ( x+ ) + 1x( x+ ) + 15( x+ ) = ( x+ )( x+ )( x+ 5) 14. x + x 1 = ( x+ 4)( x ) 15. xy + y+ x + 6= ( x + )( y+ ) a 50 = ( a+ 5)( a 5) 17. x + 8x+ 15 = ( x+ 5)( x+ ) x 0x 50x= x( x 5)( 4x+ 5) 19. 9x + x+ 0 = ( x+ )( x+ 5) 0. 1x + x = ( x 1)( x+ 1)

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