Algebra and Geometry Review – Exercise 06.1

LESSON 1: Factoring a quadratic with leading coefficient greater than 1 (Problem type 1)

Factor: 2x^2+17x+15

EXPLANATION: Method 1 - We will factor using a method often called Trial and Error. 
The coefficient of x^2 is 2
Using whole numbers, only 1 and 2 have a product of 2
So we'll look for integers n and m that satisfy the following.
2x^2+17x+15 = (1x+m)(2x+n) 
Using FOIL to expand the right-hand side of this equation, we get the following
nm=15 and n+2m=17
We'll find all integers m and n such that mn=15
Then we'll check to see if n+2m=17

m
n
n+2m
1
15
17
-1
-15
-17
15
1
31
-15
-1
-31
3
5
11
-3
-5
-11
5
3
13
-5
-3
-13
From the table, we find that m=1 and n=15 work
So we have the following:
2x^2+17x+15 = (x+m)(2x+n)
=(x+1)(2x+15)

ANSWER: (x+1)(2x+15)

EXPLANATION: Method 2 - To factor 2x^2+17x+15, we'll first find integers p and q that satisfy the following.
pq=2*15=30 where 2 is the coefficient of x^2 and 15 is the constant term
p+q=17, where 17 is the coefficient of x
We'll find all integers p and q such that pq=30
Then we'll check to see if p+q=17

p
q
p+q
1
30
31
-1
-30
-31
2
15
17
-2
-15
-17
3
10
13
-3
-10
-13
5
6
11
-5
-6
-11
We find that 2 and 15 have a product of 30 and a sum of 17
Since 2 + 15 = 17, we have the following
2x^2+17x+15=2x^2+(2+15)x+15
=2x^2+2x+15x+15
We now factor by grouping
2x^2+2x+15x+15=2x(x+1)+15(x+1) Gathering terms and factoring out common factors
=(x+1)(2x+15) Factoring out

ANSWER: (x+1)(2x+15)

LESSON 2: Factoring out a monomial from a polynomial (Univariate)

Factor: 14w^2-15w^3

EXPLANATION: We first find the GCF of 14w^2 and 15w^3
The GCF is w^2
We then factor out the GCF using the distributive property
14w^2-15w^3=w^2 (14)-w^2 (15w)
=w^2 (14-15w)

ANSWER: w^2 (14-15w)


LESSON 3: Complex fraction without variables (Problem type 1)

Simplify (5/6)/(3/4) 
EXPLANATION: This is a complex fraction. Its numerator is 5/6 and its denominator is 3/4
A fraction bar means division. So (5/6)/(3/4) mean 5/6  ÷3/4  
Doing the division, we get the following 5/6  ÷3/4=5/6×4/3
5/3×2/3=10/9
The fraction 10/9 is in simplest form. It can also be written as 1 1/9

ANSWER: The answer is 10/9, which can be written as 1 1/9

LESSON 4: Introduction to square root addition or subtraction

Simplify 6√5+4√5

EXPLANATION: Note that 6√5 and 4√5 have the same radical part, √5
This means they are like radicals, we combine like radicals in the same way we combine like terms.
6√5+4√5=10√5

ANSWER: 10√5


LESSON 5: Power of a power rule with negative exponents

Simplify and write your answer without using negative exponents: (x^(-3) )^6 

EXPLANATION: We'll be using the following rules for exponents
Power of a power rule: For any number a and any integers m and n, we have the following
(a^m )^n=a^mn
Negative exponent rule: For any nonzero number a and any integer m, we have the following
a^(-m)=1/a^m 
Here is what we get:  (x^(-3) )^6=x^((-3)*6)
=x^(-18)=1/x^18 

ANSWER:  1/x^18 

Algebra and Geometry Review – Exercise 06.1





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