Download Math 101 Study Session Spring 2016 Quiz 6 Sections 10.1, and 10.2

Math 101 Study Session Spring 2016 Quiz 6 Sections
10.1, and 10.2, 10.3, 10.4
March 31, 2016
Chapter 10 Section 1: Addition and Subtraction of Polynomials
A monomial is a number, a variable, or a product of numbers and variables.
A polynomial is a variable expression in which the terms are monomials.
A polynomial of two terms is a binomial.
A polynomial of three terms is a trinomial.
The degree of a polynomial in one variable is the value of the largest exponent on the
variable.
The degree of 4x3 − 3x2 + 6x − 1 is 3, the degree of 5y 4 − 2y 3 + y 2 − 7y + 8 is 4, the degree
of a nonzero constant (number) is 0, and the number zero has no degree.
Polynomials can be added using a vertical or horizontal format.
The opposite of a polynomial is the polynomial with the sign of every term changed, i.e.,
−(x2 − 2x + 3) = −x2 + 2x − 3.
Chapter 10 Section 2: Multiplication of Monomials
Rule for Multiplying Exponential Expressions
If m and n are integers,then xm · xn = xm+n .
Rule for Simplifying Powers of Exponential Expressions
If m and n are integers, then (xm )n = xmn
Rule for Simplifying Powers of Products
If m, n, and p are integers, then (xm y n )p = xmp y np
Chapter 10 Section 3: Multiplication of Polynomials
FOIL Method
(A + B) (C + D) = AC + AD + BC + BD
1
Product of the Sum and Difference of Two Terms
(a + b) (a − b) = a2 − b2
The Square of a Binomial
(a + b)2 = (a + b) (a + b) = a2 + 2ab + b2
(a − b)2 = (a − b) (a − b) = a2 − 2ab + b2
Chapter 10 Section 4 Integer Exponents and Scientific
Notation
Zero as an Exponent
If x 6= 0, then x0 = 1. The expression 00 is not defined.
Definition of Negative Exponents
1
1
If n is a positive integer and x 6= 0, then x−n = n and −n .
x
x
Rules for Dividing Exponential Expressions
xm
If m and n are integers and x 6= 0, then n = xm−n .
x
Below are some examples for us to try with solutions at the end of the study
guide:
Solve each of the following:
1. Add the two polynomials using a vertical format.
(6y 3 + y 2 + 1) + (−7y 3 − 8y − 6)
2. What polynomial must be added to −9x3 + 6x − 3 so that the sum is x2 − x − 1?
3. Subtract the two polynomials using a vertical format.
(2x − x − 3x2 ) − (−3 + 7x − x3 )
4. What polynomial must be subtracted from 2x3 − x2 + 6x − 2 so that the difference is
x3 + 2x − 7.
5. Simplify
−5y 4 z
−9y 6 z 5
4a4 b4
−9ab7
6. Simplify
7. Simplify
(−3y) −4x2 y 3
2
3
8. Simplify
ab2
2
(ab)2
9. Simplify
(y + 3) y 3 + 3y 2 − 8y + 1
10. Simplify
50a3 b8 c9
60a7 b4 c
11. Simplify
6a−6
12. Write the 8, 350, 000 in scientific notation.
13. Write 0.000096 in scientific notation.
3
a9 b−1
6
Solutions to Study Guide Questions
Solution to # 1:
6y 3 y 2
+ −7y 3
−8y
3
2
−y
y −8y
1
-6
−5
Solution to # 2:
We are trying to determine the polynomial P such that P + (−9x3 + 6x − 3) = x2 − x − 1
We will solve the equation for P by moving all of the other terms to the right. Remember,
when you move a term from one side of an equation to the other, it changes sign.
P + (−9x3 + 6x − 3) = x2 − x − 1
P = x2 − x − 1 + 9x3 − 6x + 3
P = 9x3 + x2 − x − 6x + 3 − 1
P = 9x3 + x2 − 7x + 2
Solution to #
−3x2
+ x3
x3 −3x2
3:
−x 2
−7x 3
−8x 5
Solution to # 4:
To determine the polynomial P such that (2x3 − x2 + 6x − 2) − P = x3 + 2x − 7. We will solve
the equation for P by moving terms from one side of the equation to the other. Remember,
when you move a term from one side of the equation to the other you must change the sign
of the term:
2x3 − x2 + 6x − 2 − P = x3 + 2x − 7
2x3 − x2 + 6x − 2 = x3 + 2x − 7 + P
2x3 − x2 + 6x − 2 − x3 − 2x + 7 = P
2x3 − x3 − x2 + 6x − 2x + 7 − 2 = P
x3 − x2 + 4x + 5 = P
Solution to # 5:
−5y 4 z −9y 6 z 5
= (−5)(−9)(y 4 )(y 6 )(z)(z 5 )
= 45y 4+6 z 1+5
= 45y 10 z 6
4
Solution to # 6:
4a4 b4 −9ab7
= (4)(−9)(a4 )(a)(b4 )(b7 )
= −36a4+1 b4+7
= −36a5 b11
Solution to # 7:
3
(−3y) −4x2 y 3
= (−3y)(−4)3 (x)3 (y 3 )3
= (−3y)(−64)x3 y 3·3
= (−3)(−64)x3 yy 9
= 192x3 y 1+9
= 192x3 y 10
Solution to # 8:
(ab2 )2 (ab)2
= a2 (b2 )2 a2 b2
= a2 b 4 a2 b 2
= a2 a2 b 4 b 2
= a2+2 b4+2
= a4 b 6
Solution to # 9:
(y + 3) y 3 + 3y 2 − 8y + 1
= y y 3 + 3y 2 − 8y + 1 + 3 y 3 + 3y 2 − 8y + 1
= y(y 3 ) + y(3y 2 ) + y(−8y) + y(1) + 3(y 3 ) + 3(3y 2 ) + 3(−8y) + 3(1)
= y 4 + 3y 3 − 8y 2 + y + 3y 3 + 9y 2 − 24y + 3
= y 4 + 6y 3 + y 2 − 23y + 3
Solution to # 10:
5
50a3 b8 c9
60a7 b4 c
50
= a3−7 b8−4 c9−1
60
5 −4 4 8
= a bc
6
5 1 4 8
=
bc
6 a4
5b4 c8
=
6a4
Solution to # 11:
6
6a−6 a9 b−1
1
= 6 6 (a9 )6 (b−1 )6
a
1
= 6 6 (a9·6 )(b−1·6 )
a
1
= 6 6 a54 b−6
a
1
1
= 6 6 a54 6
a
b
54
6a
= 6 6
ab
6a54−6
=
b6
6a48
= 6
b
Solution to # 12:
To write 8, 350, 000 in scientific notation, observe that 8, 350, 000 = 8350000.0. We will move
the decimal point to the left until there is only one nonzero number in front of the decimal
point, i.e., we will move the decimal point to the left until it is between the 8 and the 3. To
do this we must move the decimal point to the left 6 places. Hence,
8, 350, 000 = 8.35 × 106
Solution to # 13:
To write 0.000096 in scientific notation, we move the decimal to the right 5 places. Hence,
0.000096 = 9.6 × 10−5
6