2.6.2 Cramer’s Rule

Determinants can be used to solve a linear system of equations using Cramer’s Rule.

Cramer’s Rule for Two Equations in Two Variables

Given the system

This system has the unique solution

where

When solving a system of equations using Cramer’s Rule, remember the following:

1. Three different determinants are used to find x and y. The determinants in the denominators are identical.

2. The elements of D, the determinant in the denominator, are the coefficients of the variables in the system; coefficients of x in the first column and coefficients of y in the second column.

3. , the determinant in the numerator of x, is obtained by replacing the x-coefficients, , in D with the constants from the right sides of the equations, .

4. , the determinant in the numerator for y, is obtained by replacing the y-coefficients, yD

, in D with the constants from the right side of the equation, .

Example 2.4.7. Use Cramer’s Rule to solve the system:

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5 x – 4y = 2

6 x – 5y = 1

Solution. We begin by setting up and evaluating the three determinants , x yD D and D

From Cramer’s Rule, we have

The solution is ( ,x y ) = (6,7).

Cramer’s Rule does not apply if D = 0. When D = 0, the system is either inconsistent or dependent. Another method must be used to solve it.

Example 2.4.8. Solve the system:

3 62 4 3x yx y− = −⎧

⎨ + =⎩

1

Solution. We begin by finding D:

Since D = 0, Cramer’s Rule does not apply. We will use elimination to solve the system.

3 x + 6y = -1

2 x + 4y = 3

2(3 x + 6y) = 2(-1)

-3(2 x + 4y) = -3(3)

Multiply both sides of equation 1 by 2 and both sides of equation 2 by –3 to eliminate x

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6 x + 12y = -2

-6 x – 12y = -9

Simplify

0 = -11 Add the equations

The false statement, 0 = -11, indicates that the system is inconsistent and has no solution.

Cramer’s Rule can be generalized to systems of linear equations with more than two variables.

Suppose we are given a system with the determinant of the coefficient matrix D. Let denote the determinant of the matrix obtained by replacing the column containing the coefficients of "n" with the constants from the right sides of the equations. Then we have the following result:

If a linear system of equations with variables x, y, z, . . . has a unique solution given by the

formulas

Example 2.4.9: Use Cramer’s Rule to solve the system:

4 52 2 3 15 2 6 1

x y zx y zx y z

− + = −⎧⎪ + + =⎨⎪ − + =⎩

0

Solution. We begin by setting up four determinants: :

D consists of the coefficients of x, y, and z from the three equations

is obtained by replacing the x-coefficients in the first column of D with the constants from the right sides of the equations.

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is obtained by replacing the y-coefficients in the second column of D with the constants from the right sides of the equations

is obtained by replacing the z-coefficients in the third column of D with the constants from the right sides of the equations

Next, we evaluate the four determinants:

= 4(12 – (-6)) + 1(12 – 15) + 1(-4 – 10)

= 4(18) + 1(-3) + 1(-14)

= 72 – 3 – 14

= 55

= -5(12 – (-6)) + 1(60 – 3) + 1(-20 – 2)

= -5(18)+1(57) + 1(-22)

= -90 + 57 – 22

= -55

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= 4(60 – 3) + 5(12 – 15) + 1(2 – 50)

= 4(57) + 5(-3) + 1(-48)

= 228 - 15 – 48

= 165

= 4(2 – (-20)) + 1(2 – 50) – 5(-4 – 10)

= 4(22) + 1(-48) – 5(-14)

= 88 – 48 + 70

= 110

Substitute these four values into the formula from Cramer’s Rule:

The solution is (-1, 3, 2).

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2.5 Inequalities

2.5.1: Interval and inequalities

An inequality compares two unequal quantities. If we consider part of a number line which is the subset of the real numbers, such subsets are called INTERVALS. There is a relationship between the two concepts as shown:

(i) ; ],[ ba bxa ≤≤ (ii) ; ),( ba bxa << (iii) ; ),[ ba bxa <≤ (iv) ; ],( ba bxa ≤< (v) ; ),0[ ∞ 0≥x (vi) ; etc. ),0( ∞ 0>x

We can classify intervals as closed (i) above, open (ii), (vi) and half closed or half open intervals (iii), (iv) and (v).

Manipulating Inequalities

The following rules generalize the properties of inequalities in arithmetic operations: I: Adding, subtracting, multiplying or dividing a positive number both sides

of inequality, does not alter the sign of inequality II: Multiplying or dividing both sides by a negative number reverses the

inequality sign i.e. If and k are real numbers, and then, ba, ba >

• kb for all values of k . ka +>+• bkak > for positive values of k . • bkak < for negative values of k .

2.5.2 Solving Quadratic inequalities

When an inequality contains an unknown quantity, rules given in the previous section can be used to ‘solve’ it. A quadratic inequality is one in which the variable appears to the power of

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exponent 2, e.g. . The solution is a range or ranges of values of the variable with two boundaries.

xx 232 >−

A quadratic inequality is one that can be written in one of the following standard forms:

or or or In other words, a quadratic inequality is in standard form when the inequality is set to 0.

• Methods for solving quadratic inequalities

1. Using a sign graph of factors. 2. Using the test-point method.

Using a Sign Graph of the Factors

• This method of solving quadratic inequalities only works if there are quadratic factors.

Step 1: Write the quadratic inequality in standard form

• It is VERY important that one side of the inequality is 0. • With this technique we will be looking at the sign of a number to determine if it is a solution or not.

Step 2: Solve the quadratic equation, , by factoring to get the boundary point(s).

The boundary point(s) on the number line will create test intervals

Step 4: Find the sign of every factor in every interval

You can choose ANY value in an interval to plug into each factor. Whatever the sign of the factor is with that value gives you the sign you need for that factor in that interval. Step 5: Using the signs found in Step 4, determine the sign of the overall quadratic function in each interval.

When you look at the signs of your factors in each interval, keep in mind that they represent a product of the factors that make up your overall quadratic function.

You determine the sign of the overall quadratic function by using basic multiplication sign rules:

• The product of two factors that have the same sign is positive.

The product of two factors that have the opposite signs is negative.

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, , -5 and 3 are boundary points. 01522 =−+ xx 0)3)(5( =−+ xxStep 3: Graph of boundary points Below is a graph that marks off the boundary points -5 and 3 and shows the three sections that those points have created on the graph

Note that the two boundary points create three sections on the graph: , , and . Step 4: To find the sign of every factor

• You can choose ANY point in an interval to represent that interval.

• If we chose a number in the first interval, , like -6

For the two factors: -6 + 5 = -1 and -6 - 3 = -9

• If we chose a number in the second interval, , like 0 (I could have used -4, -1, or 2 as long as it is in the interval), it would make x + 5 positive and x - 3 negative: i.e. 0 + 5 = 5 and 0 - 3 = -3

• If we chose a number in the third interval, , like 4 (I could have used 10, 25, or 10000 as long as it is in the interval), it would make both factors positive:

4 + 5 = 9 and 4 - 3 = 1

Interval notation: Graph:

Example 2.5.2: Solve using a sign graph of factors, write your answer in interval notation and graph the solution set: Solution: 051766517 22 ≥++⇒−≥+ xxxx

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( 0)13)(52 + + =xx , so 5/2 and -1/3 are boundary points

Note that the two boundary points create three sections on the graph: , , and . The sign of every factor in each interval • If we chose a number in the first interval, , like -4 , It make both factors

negative: 2(-4) + 5 = -3 and 3(-4) + 1 = -11

• If we chose a number in the second interval, , like -1. It make 2x + 5 positive and 3x + 1 negative: 2(-1) + 5 = 3 and 3(-1) + 1 = -2

• If we chose a number in the third interval, , like 0 , both factors positive:

2(0) + 5 = 5 and 3(0) + 1 = 1

To determining the sign of the overall function in each interval In the first interval, , we have a negative times a negative, so the sign of the quadratic in that interval is positive. In the second interval, , we have a positive times a negative, so the sign of the quadratic in that interval is negative. In the third interval, , we have two positives, so the sign of the quadratic in that interval is positive. Keep in mind that our inequality is . Since we are looking for the quadratic expression to be greater than or equal to 0,that means we need our sign to be positive (or o).

Using the Test-Point Method

The test-point method for solving quadratic inequalities works for any quadratic that has a real number solution, whether it factors or not

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Generally the same procedures as in the first method will apply except when there is no factors: see the example.

Example 2.5.3 : Solve using the test-point method, write your answer in interval notation and graph the solution set: .

Solution: Since cannot be solved by factoring, how can we find the solution?

By using the quadratic formula : a

acbbx2

42 −±−=

and are boundary points.

Note that the two boundary points create three sections on the graph:

, , and .

Keep in mind that our original problem is . Since we are looking for the quadratic expression to be GREATER THAN 0, that means we need our sign to be POSITIVE

From the interval , I choose to use 0 to test this interval: Note that is approximately .35. Since 2 is positive and we are looking for values that cause our quadratic expression to be

greater than 0, would be part of the solution

From the interval , I choose to use 1 to test this interval. (I could have used 2, 3, or 5 as long as it is in the interval) Note that is approximately .35 and is approximately 5.65 Since -3 is negative and we are looking for values that cause our expression to be greater

than 0, would not be part of the solution

From the interval , I choose 6 to use to test this interval. (I could have used 10, 25, or 10000 as long as it is in the interval) Note that is approximately 5.65. Since 2 is positive and we are looking for values that cause our quadratic expression to be

greater than 0, would be part of the solution

nterval notation:

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Forms of Quadratic Inequality graphs

Graphs will look like a parabola with a solid (complete line) or dotted line and a shaded section.

The graph could be shaded inside the parabola or outside.

Steps for graphing 1. Sketch the parabola y=ax2+bx+c (dotted line for < or >, solid line for ≤ or ≥) 2. Choose a test point and see whether it is a solution of the inequality. 3. Shade the appropriate region. (if the point is a solution, shade where the point is, if it’s not a solution, shade the other region) Example: Graph 462 −+≤ xxy

x -1 -2 -3 -4 -5 y -9 -12 -13 -12 -9

Vertex: (-3,-13)

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Test point

• Test Point: (0,0)

0≤02+6(0)-4 0≤-4 (Not a solution) - So shade where the point is NOT

2.6 Polynomials

Introduction

In this part we will be looking at the different components of polynomials. Then we will move on to adding, subtracting, dividing and multiplying them.

Basic terminologies Let’s start with defining some words before we get to our polynomial. Term: A term is a number, variable or the product of a number and variable(s).

Examples of terms are , z. Coefficient: A coefficient is the numeric factor of the term.

Here are the coefficients of the terms listed above: Term Coefficient

3

5

2

z 1

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Constant Term : A constant term is a term that contains only a number/numeric value. In other words, there is no variable in a constant term. Examples of constant terms are 4, 100, and -5. Standard Form of a Polynomial

• Every polynomial in one variable is equivalent to a polynomial with the form

012

21

1 ...)( axaxaxaxaxf nn

nn

nn +++++= −

−−

− or

012

21

1 ...)( axaxaxaxaxP nn

nn

nnn +++++= −

−−

−

Where - are constants (coefficients) and . 011,1, ,,..., aaaaa nnn −− 0≠na

- is a non-negative integer. n

In other words, a polynomial is a finite sum of terms where the exponents on the variables are non-negative integers. Note that the terms are separated by +'s and -'s.

An example of a polynomial expression is .

This form is sometimes taken as the definition of a polynomial in one variable.

Degree of a Term: The degree of a term is the sum of the exponents on the variables contained in the term.

For example, the degree of the term would be 1 + 1 = 2. The exponent on a is 1 and on b is 1 and the sum of the exponents is 2.

The degree of the term would be 3 since the only variable exponent that we have is 3. Degree of the Polynomial: The degree of the polynomial is the largest degree of all its terms. Descending Order: Note that the standard form of a polynomial that is shown above is written in descending order. This means that the term that has the highest degree is written first, the term with the next highest degree is written next, and so forth.

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Also note that a polynomial can be “missing” terms. For example, the polynomial written above starts with a degree of 5, but notice there is not a term that has an exponent of 4. That means the coefficient on it is 0, so we do not write it. Some Types of Polynomials

Type Definition Example Monomial A polynomial with one term 5x Binomial A polynomial with two terms 5x - 10 Trinomial A polynomial with three terms

Combining Like Terms Recall that like terms are terms that have the exact same variables raised to the exact same

exponents. One example of like terms is . Another example is .

You can only combine terms that are like terms. You can think of it as the reverse of the distributive property.

It is like counting apples and oranges. You just count up how many variables you have the same and write the number in front of the common variable part

Adding and Subtracting Polynomials Remove the brackets, ( ).

If there is only a + sign in front of brackets, then the terms inside of brackets remain the same when you remove the brackets. If there is a - in front of the brackets, then distribute it by multiplying every term in the brackets by a -1 (or you can think of it as negating every term in the brackets). Combine like terms

Example 2.6.1: Perform the indicated operation and simplify: .

[Remove the brackets and *Add like terms together] Example 2.6.2: Perform the indicated operation and

simplify: .

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Solution:

Multiplying Polynomials In general, when multiplying two polynomials together, use the distributive property until every term of one polynomial is multiplied times every term of the other polynomial. Make sure that you simplify your answer by combining any like terms. On this page we will look at some of the more common types of polynomials to illustrate this idea (Monomial)(Monomial): In this case, there is only one term in each polynomial. You simply multiply the two terms together.

Example 2.6.3: Find the product . Solution:

(Monomial)(Polynomial): In this case, there is only one term in one polynomial and more than one term in the other. You need to distribute the monomial to EVERY term of the other polynomial.

Example 2.6.4: Find the product .

Solution:

(Binomial)(Binomial): In this case, both polynomials have two terms. You need to distribute both terms of one polynomial times both terms of the other polynomial. One way to keep track of your distributive property is to use the FOIL method. Note that this method only works on (Binomial)(Binomial). F-First terms, O-Outside terms, I-Inside terms, L-Last terms: This is a fancy way of saying to take every term of the first binomial times every term of the second binomial. In other words, do

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the distributive property for every term in the first binomial.

Example 2.6.5: Find the product . Solution:

Special product rule for a binomial squared:

and

Product of the sum and difference of two terms

Example 2.6.6: Find the product

Solution:

(Polynomial)(Polynomial): As mentioned above, use the distributive property until every term of one polynomial is multiplied by every term of the other polynomial. Make sure that you simplify your answer by combining any like terms.

Example 2.6.7: Find the product .

Solution:

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Special product rule for binomial cubed:

,

Example 2.6.8: Find the product . Solution:

2.7 Factoring Polynomials

Introduction

Factoring is to write an expression as a product of factors. For example, we can write 10 as (5)(2), where 5 and 2 are called factors of 10. We can also do this with polynomial expressions. In this section we are going to look at several ways to factor polynomial expressions. By the time I'm through with you, you will be a factoring machine. Basically, when we factor, we reverse the process of multiplying the polynomial. Greatest Common Factor (GCF): The GCF for a polynomial is the largest monomial that divides (is a factor of) each term of the polynomial.

Example 2.6.9: Factor out the GCF:

Solution: The largest monomial that we can factor out of each term is 2y.

Divide the GCF out of every term of the polynomial.

Factoring a Polynomial with Four Terms by Grouping

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In some cases there is no a GCF for ALL the terms in a polynomial. If you have four terms with no GCF, then try factoring by grouping.

Step 1: Group the first two terms together and then the last two terms together. Step 2: Factor out a GCF from each separate binomial. Step 3: Factor out the common binomial.

Example 2.6.10: Factor by grouping:

Solution:

Note how there is not a GCF for ALL the terms. So let’s go ahead and factor this by grouping.

then finally

Example 2.6.11: Factor by grouping: .

Solution: NOTE: Group the first two terms together and then the last two terms together.

and then Note that if we multiply our answer out that we do get the original polynomial. Factoring a Perfect Square Trinomial

OR

Examples 2.6.12: Factor the expressions: (a) (b) .

Solution: (a) Since it is a trinomial, you can try factoring this by trial and error shown above But if you can recognize that it fits the form of a perfect square trinomial, you can save yourself some time.

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Factoring a Difference of Two Squares

, Factor the difference of two squares:

Note that the sum of two squares DOES NOT factor. Factoring a Sum of Two Cubes

Example2.6.13: : Factor the sum of cubes: .

First note that there is no GCF to factor out of this polynomial.

This fits the form of the sum of cubes. So we will factor using that rule

Factoring a Difference of Two Cubes

The difference of two cubes has to be exactly in this form to use this rule. When you have the difference of two cubes, you have a product of a binomial and a trinomial. The binomial is the difference of the bases that are being cubed. The trinomial is the first base squared, the second term is the opposite of the product of the two bases found, and the third term is the second base squared.

Example 2.6.14: Factor the difference of cubes: .

Solution:

Now that you have a list of different factoring rules, let’s put it all together. The following is a checklist of the factoring rules that we have covered in our study.

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Factoring Strategy

I. GCF: Always check for the GCF first, no matter what.

II. Binomials: (a) (b)

(c)

III. Trinomials: (a). (b). Trial and error:

(c.) Perfect square trinomial: ,

IV. Polynomials with four terms: Factor by grouping

Example 2.6.15: Factor completely Solution: The first thing that we always check when we are factoring is WHAT?

The GCF. In this case, there is one.

Factoring out the GCF of 3 we get:

Next, we assess to see if there is anything else that we can factor. We have a trinomial inside the ( ). It fits the form of a perfect square trinomial, so we will factor it accordingly:

Example 2.6.16: Factor completely.

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Next we assess to see if there is anything else that we can factor. Note how the second binomial is another difference of two squares. That means we have to continue factoring this problem.

There is no more factoring that we can do in this problem. Note that if we would multiply this out, we would get the original polynomial. Division methods: Long division and Synthetic division The synthetic division

A method of dividing polynomials when the divisor is a polynomial of the first degree, by using only the coefficients of the terms.

To test the possible zeros and find an actual zero:

Recall that if you apply synthetic division and the remainder is 0, then c (from first degree divisor cx − ) is a zero or root of the polynomial function.

Dividend / Divisor = Quotient + Remainder / Divisor

General Background of Synthetic Division

Step1: An easy way to do this is to first set it up as if you are doing long division and then set up your synthetic division.

When you write out the dividend make sure that you write it in descending powers and you insert 0's for any missing terms.

• For example, if you had the problem , the polynomial , starts out with degree 4, then the next highest degree is 1.

• It is missing degrees 3 and 2. So if we were to put it inside a division box we would write it like this:

.

Step 2: Bring down the leading coefficient to the bottom row.

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Step 3: Multiply c by the value just written on the bottom row. Place this value right beneath the next coefficient in the dividend:

Step 4: Add the column created in step 3. Write the sum in the bottom row:

Step 5: Repeat until done.

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