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Math 115 MAP 2 Perform the indicated operation(s) and simplify. 1. ( 8a 4 − 9a3 + 10a + 6 ) − ( −a + 15a 2 + 3a3 − 12 ) + ( 4a3 − 5a 2 + 3a − 7 ) Solution: Let’s show the vertical format. Remember to change the signs when distributing a negative sign and to align like terms. 8a 4 − 9a3 + 10a + 6 3 2 − 3a − 15a + a + 12 4a 3 − 5a 2 + 3a − 7 8a 4 − 8a3 − 20a 2 + 14a + 11 2. (4 x 2 + 3 x − 2)(3 x 2 − 5) Solution: Distribute each term of the trinomial to each term of the binomial (can also be done the other way) and apply the product rule for exponents. (4 x 2 + 3 x − 2)(3 x 2 − 5) = 12x 4 − 20 x 2 + 9x3 − 15 x − 6x 4 distribute 4 x 2 3 2 distribute 3 x + 10 2 12 x + 9 x − 26 x − 15 x + 10 distribute − 2 combine like terms 3. ( 3a + 8 )( 4a − 3 ) Solution: Do FOIL. ( 3a + 8 )( 4a − 3 ) = 12a2 − 9a + 32a − 24 FOIL = 12a2 + 23a − 24 Combine like terms. 4. ( 8 − 9w )( 8 + 9w ) Solution: Either do FOIL: ( 8 − 9w )( 8 + 9w ) = 64 + 72w − 72w − 81w 2 = 64 − 81w 2 ( 8 − 9w )( 8 + 9w ) 2 OR do “fast” multiplication: = 82 − ( 9w ) = 64 − 81w 2 5. ( 8a + 5b ) Solution: 3 ( 8a + 5b )3 = ( 8a + 5b )( 8a + 5b )( 8a + 5b ) ( = ( 64a ) = 64a 2 + 40ab + 40ab + 25b 2 ( 8a + 5b ) 2 ) + 80ab + 25b 2 ( 8a + 5b ) Do FOIL on the first two factors. Combine like terms. = 512a3 + 320a 2 b Distribute 64a2 . + 640a2 b + 400ab2 Distribute 80ab. 2 + 200ab + 125b 3 Distribute 25b 2 . 512a3 + 960a 2 b + 600ab 2 + 125b3 Combine like terms. 6. ( 24a 6 b 4 + 30a 2 b3 − 12 a 4 b5 + 6 ) ÷ 6a 2 b3 Solution: The divisor is a monomial so we can split the problem into 4 mono/mono problems. ( 24a b 6 6 4 ) + 30a 2 b 3 − 12 a 4 b 5 + 6 ÷ 6a 2 b 3 4 2 3 4 5 24a b 30a b 12 a b 6 + − + 2 3 2 3 2 3 2 3 6a b 6a b 6a b 6a b 1 = 4a 4 b + 5 − 2a 2 b 2 + 2 3 Divide the coefficients and apply the quotient rule for exponents. a b = 7. 9 y − 4 y 2 + 4y 3 − 7 2y − 1 Solution: The divisor is a binomial so we need to do long division. Remember to arrange both dividend and divisor in descending order before starting the DMSB process. 2y 2 − y + 4 2y − 1 4 y 3 − 4 y 2 + 9 y − 7 − ( 4y 3 − 2y 2 ) − 2y 2 + 9 y − 7 ( − −2y 2 + y ) 8y − 7 − ( 8y − 4 ) −3 Thus, the answer is 2y 2 − y + 4 − 3 . 2y − 1 Translate and then perform the operation. 8. The square of the sum of three times a number and 2 Solution: The square of the sum of three times a number and 2 3 x 3x +2 ( 3 x + 2 )2 2 So the translation is ( 3 x + 2 ) . We can do the binomial expansion either slow (FOIL) or fast (special product formula). 2 Do FOIL: ( 3 x + 2 ) = ( 3 x + 2 )( 3 x + 2 ) = 9 x 2 + 6 x + 6 x + 4 = 9 x 2 + 12x + 4 OR 2 2 Use the formula: ( 3 x + 2 ) = ( 3 x ) + 2 ( 3 x )( 2 ) + 22 = 9 x 2 + 12 x + 4 9. The difference of the product of a number and the quantity 3 less than the number, and the product of the number and the quantity 2 more than three times the number Solution: The difference of the product of a number and the quantity 3 less than the number , x −3 x ( x −3 ) and the product of the number and the quantity 2 more than thr ee times the number 3x 3x +2 x ( 3 x +2) “Difference” means subtract “first minus second” so the translation is x ( x − 3 ) − x ( 3 x + 2) . First perform the multiplications before performing the subtraction and combine like terms: x ( x − 3) − x ( 3x + 2) = x 2 − 3 x − 3 x 2 − 2x = −2 x 2 − 5 x Factor completely. 10. n 2 + n − 6 Solution: This is a trinomial square with a leading coefficient of 1, so we think of factors of -6 (the constant term) that will add up to 1 (the coefficient of n). The integers we need are 3 and -2. Thus, the factored form is ( n + 3 )( n − 2 ) . 11. 45n 2 p2 − 20 p4 Solution: We note that the two terms have a GCF so we factor out this GCF first. 45n2 p2 − 20p4 ( = 5 p2 9n 2 − 4 p2 ) Next, we note that the two terms inside the parentheses look like they are perfect squares, and indeed 2 2 they are because 9n 2 − 4 p2 = ( 3n ) − ( 2p ) and thus can be factored as the difference of squares. The complete factored form is then 5 p 2 ( 3n − 2p )( 3n + 2p ) . 12. c 2 − 7cd + 12d 2 Solution: There are 3 terms so we suspect that this is a trinomial square. To convince ourselves, we look at the degrees in c: 2, 1, 0; we also check the degrees in d: 0, 1, 2. Both are good! The leading coefficient is 1 so we again look for the factors of 12 that add up to -7. These are -3 and -4 and thus the factored form of the problem is ( c − 3d )( c − 4d ) . 13. n 3 + 1 Solution: There are 2 terms and each term is a perfect cube, so we factor using the formula for the sum of cubes. ( ) n 3 + 1 = n 3 + 13 = ( n + 1) n 2 − n + 1 14. 20y 2 + 13 y − 21 Solution: There are 3 terms and this is a trinomial square since the degrees of the terms are 2, 1, 0. The leading coefficient is not 1 but 20 so this is when we use either grouping or bottoms-up. No matter which method we want to use, we have to start the same way which is we want to find two numbers that multiply to 20 ⋅ ( −21) and that add up to 13. Let’s break down the numbers: 20 ⋅ ( −21) = − 2 ⋅ 2 ⋅ 5 ⋅ 3 ⋅ 7 = 28 ⋅ ( −15 ) We see that 28 and -15 are the numbers that we need. Those who like grouping will do the following: 20 y 2 + 13 y − 21 = 20 y 2 + 28 y − 15 y − 21 ( ) = 20 y 2 + 28 y − (15 y + 21) = 4 y ( 5 y + 7 ) − 3 ( 5y + 7 ) = ( 5 y + 7 )( 4 y − 3 ) Those who like bottoms-up will divide 28 and -15 by 20 and reduce the fractions: − 15 3 =− . 20 4 The first fraction 28 7 and = 20 5 3 7 gives the factor 5y + 7 and − gives the factor 4y − 3 . 4 5 15. xz2 − yz2 − 16 x + 16y Solution: There are 4 terms so we know we have to group. Let’s group the first two and the last two. xz 2 − yz 2 − 16 x + 16y ( ) = xz 2 − yz 2 − (16 x − 16 y ) 2 = z ( x − y ) − 16 ( x − y ) ( ) = z 2 − 16 ( x − y ) = ( z − 4 )( z + 4 )( x − y ) Group 1st 2 and last 2. Note the sign changes! Factor out z 2 from 1st grp, 16 from 2nd grp. Factor out x − y from both groups. Factor the difference of squares z 2 − 16. Translate and then factor the resulting expression. 16. Four times a number, subtracted from the difference of the square of the number and 5 Solution: Four times a number square of the number and 5 , subtracted from the difference of the 4x x2 x 2 −5 Now, “a subtracted from b” translates to b − a and so the translation is x 2 − 5 − 4 x . Before we can factor this expression, we first have to rearrange the terms: x 2 − 4 x − 5 . This is a trinomial square with leading coefficient of 1, so we think of 2 numbers that multiply to -5 and add up to -4 and these are -5 and 1. Thus, the factored form is ( x − 5 )( x + 1) . 17. Three times the cube of one number, less twenty-four times the cube of another number Solution: Three times the cube of one number cube of another number , less twenty − four times the 3 x y3 3x3 24 y 3 Now, “a less b” translates to a − b and so the translation is 3 x 3 − 24y 3 . We note that there are 2 terms and that the terms have a GCF which is 3. Factor out 3 to get 3 x 3 − 8 y 3 . Now, each term inside the parentheses is a perfect cube so we can factor as a difference ( ) of cubes: 3 3 x 3 − 8 y 3 = 3 x 3 − ( 2y ) = 3 ( x − 2 y ) x 2 + 2 xy + 4 y 2 . ( ) ( )