1. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 7.3 - 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 7.3 - 2 Factoring Chapter 7

3. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 7.3 - 3 7.3 Special Factoring

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 4 7.3 Special Factoring Objectives 1. Factor a difference of squares. 2. Factor a perfect square trinomial. 3. Factor a difference of cubes. 4. Factor a sum of cubes.

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 5 7.3 Special Factoring The Difference of Squares Difference of Squares x 2 – y 2 = ( x + y )( x – y )

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 6 7.3 Special Factoring EXAMPLE 1Factoring Differences of Squares Factor each polynomial. There is a common factor of 2. 2 n 2 – 50 = 2( n 2 – 25)Factor out the common factor. (a) 2 n 2 – 50 = 2( n + 5)( n – 5) Factor the difference of squares.

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 7 7.3 Special Factoring EXAMPLE 1Factoring Differences of Squares Factor each polynomial. (b) 9 g 2 – 16 9 g 2 – 16= (3 g ) 2 – (4) 2 A2A2 B2B2 = (3 g + 4)(3 g – 4) ( A + B )( A – B ) (c) 4 h 2 – ( w + 5) 2 4 h 2 – ( w + 5) 2 = (2 h ) 2 – ( w + 5) 2 A2A2 B2B2 = (2 h + w + 5)(2 h – [ w + 5]) ( A + B ) ( A – B ) – – = (2 h + w + 5)(2 h – w – 5)

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 8 7.3 Special Factoring Caution CAUTION Assuming no greatest common factor except 1, it is not possible to factor (with real numbers) a sum of squares, such as x 2 + 16.

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 9 7.3 Special Factoring Perfect Square Trinomial x 2 + 2xy + y 2 = ( x + y ) 2 x 2 – 2xy + y 2 = ( x – y ) 2

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 10 7.3 Special Factoring EXAMPLE 2Factoring Perfect Square Trinomials Factor each polynomial. Here 9 g 2 = (3 g ) 2 and 49 = 7 2. The sign of the middle term is –, so if 9 g 2 – 42 g + 49 is a perfect square trinomial, the factored form will have to be (3 g – 7) 2. (a) 9 g 2 – 42 g + 49 Take twice the product of the two terms to see if this is correct. 2(3 g )(–7) = –42 g This is the middle term of the given trinomial, so 9 g 2 – 42 g + 49 = (3 g – 7) 2.

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 11 7.3 Special Factoring EXAMPLE 2Factoring Perfect Square Trinomials Factor each polynomial. If this is a perfect square trinomial, it will equal (5 x + 8 y ) 2. By the pattern described earlier, if multiplied out, this squared binomial has a middle term of 2(5 x )(8 y ), which does not equal 60 xy. Verify that this trinomial cannot be factored by the methods of the previous section either. It is prime. (b) 25 x 2 + 60 xy + 64 y 2

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 12 7.3 Special Factoring EXAMPLE 2Factoring Perfect Square Trinomials Factor each polynomial. since 2( n – 4)9 = 18( n – 4), the middle term. (c) ( n – 4) 2 + 18( n – 4) + 81 = [ ( n – 4) + 9 ] 2 = ( n + 5) 2,

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 13 7.3 Special Factoring EXAMPLE 2Factoring Perfect Square Trinomials Factor each polynomial. The result is the difference of squares. Factor again to get (d) c 2 – 6 c + 9 – h 2 = ( c – 3 + h )( c – 3 – h ). ( c 2 – 6 c + 9) – h 2 = ( c – 3) 2 – h 2 Since there are four terms, we will use factoring by grouping. The first three terms here form a perfect square trinomial. Group them together, and factor as follows.

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 14 7.3 Special Factoring Difference of Cubes x 3 – y 3 = ( x – y )( x 2 + xy + y 2 )

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 15 –5 a a3a3 –125 7.3 Special Factoring EXAMPLE 3Factoring Difference of Cubes Factor each polynomial. Recall, x 3 – y 3 = ( x – y )( x 2 + xy + y 2 ). = a 3 – 5 3 = ( a – 5)( a 2 + 5 a + 5 2 ) = ( a – 5)( a 2 + 5 a + 25) (a) a 3 – 125 Check:= ( a – 5)( a 2 + 5 a + 25) Opposite of the product of the cube roots gives the middle term.

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 16 7.3 Special Factoring EXAMPLE 3Factoring Difference of Cubes Factor each polynomial. Recall, x 3 – y 3 = ( x – y )( x 2 + xy + y 2 ). = (2 g ) 3 – h 3 = (2 g – h ) [ (2 g ) 2 + (2 g )( h ) + h 2 ) ] = (2 g – h )(4 g 2 + 2 gh + h 2 ) (b) 8 g 3 – h 3

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 17 7.3 Special Factoring EXAMPLE 3Factoring Difference of Cubes Factor each polynomial. Recall, x 3 – y 3 = ( x – y )( x 2 + xy + y 2 ). = (4 m ) 3 – (3 n ) 3 = (4 m – 3 n ) [ (4 m ) 2 + (4 m )(3 n ) + (3 n ) 2 ] = (4 m – 3 n )(16 m 2 + 12 mn + 9 n 2 ) (c) 64 m 3 – 27 n 3

18. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 18 7.3 Special Factoring Sum of Cubes x 3 + y 3 = ( x + y )( x 2 – xy + y 2 )

19. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 19 7.3 Special Factoring Note on Signs NOTE Difference of Cubes x 3 – y 3 = ( x – y )( x 2 + xy + y 2 ) Sum of Cubes x 3 + y 3 = ( x + y )( x 2 – xy + y 2 ) The sign of the second term in the binomial factor of a sum or difference of cubes is always the same as the sign in the original polynomial. In the trinomial factor, the first and last terms are always positive; the sign of the middle term is the opposite of the sign of the second term in the binomial factor.

20. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 20 7.3 Special Factoring EXAMPLE 4Factoring Sums of Cubes Factor each polynomial. Recall, x 3 + y 3 = ( x + y )( x 2 – xy + y 2 ). = n 3 + 2 3 = ( n + 2)( n 2 – 2 n + 2 2 ) = ( n + 2)( n 2 – 2 n + 4) (a) n 3 + 8 = (4 v ) 3 + (3 g ) 3 = (4 v + 3 g ) [ (4 v ) 2 – (4 v )(3 g ) + (3 g ) 2 ] = (4 v + 3 g ) (16 v 2 – 12 gv + 9 g 2 ) (b) 64 v 3 + 27 g 3

21. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 21 7.3 Special Factoring EXAMPLE 4Factoring Sums of Cubes Factor each polynomial. Recall, x 3 + y 3 = ( x + y )( x 2 – xy + y 2 ). = 2( k 3 + 125) = 2( k 3 + 5 3 ) = 2( k + 5)( k 2 – 5 k + 25) (c) 2 k 3 + 250 =

22. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 22 7.3 Special Factoring Factoring Summary Special Types of Factoring (Memorize) Difference of Squares x 2 – y 2 = ( x + y )( x – y ) Perfect Square Trinomial x 2 + 2xy + y 2 = ( x + y ) 2 x 2 – 2xy + y 2 = ( x – y ) 2 Difference of Cubes x 3 – y 3 = ( x – y )( x 2 + xy + y 2 ) Sum of Cubes x 3 + y 3 = ( x + y )( x 2 – xy + y 2 )