Before looking at factoring a sum of two cubes lets look at the possible

# Before looking at factoring a sum of two cubes lets

This preview shows page 13 - 15 out of 19 pages.

Before looking at factoring a sum of two cubes, let’s look at the possible factors. It turns out that a 3 + b 3 can actually be factored as ( a + b )( a 2 ab + b 2 ). Let’s check these factors by multiplying. Does ( a + b )( a 2 ab + b 2 ) = a 3 + b 3 ? ( a )( a 2 ab + b 2 ) + ( b )( a 2 ab + b 2 ) Apply the distributive property. ( a 3 a 2 b + ab 2 ) + ( b )( a 2 - ab + b 2 ) Multiply by a . ( a 3 a 2 b + ab 2 ) + ( a 2 b ab 2 + b 3 ) Multiply by b . a 3 a 2 b + a 2 b + ab 2 ab 2 + b 3 Rearrange terms in order to combine the like terms. a 3 + b 3 Simplify Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial a 3 + b 3 . So, the factors are correct. You can use this pattern to factor binomials in the form a 3 + b 3 , otherwise known as “the sum of cubes.” The Sum of Cubes A binomial in the form a 3 + b 3 can be factored as ( a + b )( a 2 ab + b 2 ). Examples: The factored form of x 3 + 64 is ( x + 4)( x 2 – 4 x + 16). The factored form of 8 x 3 + y 3 is (2 x + y)(4 x 2 – 2 xy + y 2 ). Example Problem Factor x 3 + 8 y 3 . x 3 + 8 y 3 Identify that this binomial fits the sum of cubes pattern: a 3 + b 3 . a = x , and b = 2 y (since 2 y • 2 y • 2 y = 8 y 3 ). ( x + 2 y )( x 2 x (2 y ) + (2 y ) 2 ) Factor the binomial as ( a + b )( a 2 ab + b 2 ), substituting a = x and b = 2 y into the expression. ( x + 2 y )( x 2 x (2 y ) + 4 y 2 ) Square (2 y ) 2 = 4 y 2 . Answer ( x + 2 y )( x 2 – 2 xy + 4 y 2 ) Multiply − x (2 y ) = −2 xy (writing the coefficient first. And that’s it. The binomial x 3 + 8 y 3 can be factored as ( x + 2 y )( x 2 – 2xy + 4 y 2 )! Let’s try another one. You should always look for a common factor before you follow any of the patterns for factoring.
Example Problem Factor 16 m 3 + 54 n 3 . 16 m 3 + 54 n 3 Factor out the common factor 2. 2(8 m 3 + 27 n 3 ) 8 m 3 and 27 n 3 are cubes, so you can factor 8 m 3 + 27 n 3 as the sum of two cubes: a = 2 m , and b = 3 n . 2 (2 m + 3 n )[(2 m ) 2 – (2 m )(3 n ) + (3 n ) 2 ] Factor the binomial 8 m 3 + 27 n 3 substituting a = 2 m and b = 3 n into the expression ( a + b )( a 2 ab + b 2 ) . 2(2 m + 3 n )[ 4 m 2 – (2 m )(3 n ) + 9 n 2 ] Square: (2 m ) 2 = 4 m 2 and (3 n ) 2 = 9 n 2 . Answer 2(2 m + 3 n )(4 m 2 – 6 mn + 9 n 2 ) Multiply −(2 m )(3 n ) = −6 mn. Factor 125 x 3 + 64. (5 x + 4)(25 x 2 – 20 x + 16) Correct. 5 x is the cube root of 125 x 3 , and 4 is the cube root of 64. Substituting these values for a and b , you find (5 x + 4)(25 x 2 – 20 x + 16). Difference of Cubes Having seen how binomials in the form a 3 + b 3 can be factored, it should not come as a surprise that binomials in the form a 3 b 3 can be factored in a similar way. The Difference of Cubes A binomial in the form a 3 b 3 can be factored as ( a b )( a 2 + ab + b 2 ). Examples: The factored form of x 3 – 64 is ( x – 4)( x 2 + 4 x + 16). The factored form of 27 x 3 – 8 y 3 is (3 x – 2 y )(9 x 2 + 6 xy + 4 y 2 ). Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the + and – signs. Take a moment to compare the factored form of a 3 + b 3 with the factored form of a 3 b 3 .

#### You've reached the end of your free preview.

Want to read all 19 pages?

• Spring '12
• george
• Quadratic equation, Elementary algebra, Negative and non-negative numbers