Download Math 115 MAP 2 Perform the indicated operation(s

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 .


(
)
(
)