college algebra review

7/31/2019 College Algebra Review

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Factoring and expanding polynomials.

In algebra, students learn to factor polynomials like the quadratic equation. Factoring ismuch easier to understand once the student has learned how to expand a polynomial,which is simply multiplying two or more factors to form one polynomial. It is the exact

opposite of factoring. The general quadratic equation has the form ax^2 + bx + c = 0and its factors will usually have the form (mx+n) (jx + k), where "x" is a variable and allthe other values are constant..

Difficulty: Moderately Easy

InstructionsExpanding

1. Step 1

Write the factors in parentheses side-by-side. If one polynomial has more terms thanthe other, write the shorter one first.

(x + 3)(2x ^2 - x + 7)

2. Step 2

Multiply the first term of the first polynomial by each term in the second polynomial.

(x + )(2x ^2 - x + 7) = 2x^3 - x^2 +7x

3. Step 3

Multiply the next term of the first polynomial through the second polynomial. Repeatthis step for each additional term in the first polynomial, if necessary.

( + 3)(2x ^2 - x + 7) = 6x^2 - 3x +21

4. Step 4

Combine the solutions and then group like terms together.

2x^3 - x^2 +7x + 6x^2 - 3x + 212x^3 - x^2 +6x^2 + 7x - 3x + 21

5. Step 5

Simplify the solution by combining the like functions.

2x^3 -x^2 +6x^2 + 7x -3x + 21(x + 3)(2x ^2 - x + 7) = 2x^3 + 5x^2 + 4x + 21


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Factoring

6. Step 1

Write the polynomial with terms in rank order and then write two sets of parentheses

after the equal sign.

5x - 8 + 3x^2 = 45x - 8 + 3x^2 - 4 = 03x^2 + 5x -12 = ( )( )

7. Step 2

Factor the first term and put the resulting values in the left side of the parentheses.

3x^2 = 3x * x

3x^2 + 5x -12 = (3x )(x )

8. Step 3

Factor the last term and place the factors in the right-hand side of the parentheses.If more than one set of factors exist, choose one at random.

-12 = 4 * -3 or 3 * -43x^2 + 5x -12 = (3x + 4)(x - 3)

9. Step 4

Expand the factor to see if they match the original polynomial.

3x^2 + 5x -12 = (3x + 4)(x - 3)3x^2 + 5x -12 does not equal 3x^2 - 5x - 12

10.Step 5

Try the next set of factors for the last term if the first set did not work. Continue untilyou find the correct set.

3x^2 + 5x -12 = (3x - 4)(x + 3)3x^2 + 5x -12 = 3x^2 + 5x -12

Linear inequalities

Solving linear inequalities is something that many students struggle with. Most knowhow basic inequalities work, and they know how to solve a basic linear equation, butputting the two together can cause some confusion. With a basic understanding of howeach work, you can easily combine the two to complete a linear inequality.


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Instructions

1. Step 1

Understand the basics of inequalities and linear equations. In an inequality, we saythat that one number is greater than (>) another number or less than ( 6In this equation we will be solving for x. We want to know what x is greater than(>). We know this because the open side of the inequality is facing the number that

will be the greatest of the two.

3. Step 3

Pretend that the inequality sign--in this case the > sign--is an = sign for the timebeing. When you do this, the original equation looks just like a traditional linearequation. The new form of the equation would be written as:x + 2 = 6

4. Step 4

Solve for x by subtracting 2 from both sides of the = sign. It looks like this:x + 2 - 2 =

6 - 2Completing this equation, we find that the 2s on the left side cancel each otherout, and the subtraction of 2 from the 6 on the right side of give us 4:x = 4

5. Step 5

Convert it back into an inequality. In Step 3 we exchanged the > for the = sign tomake the equation look like a traditional linear equation, but now it must be turnedback into an inequality. Do this by dropping the = and replacing it with the >. Fromthis we now know that x does not equal for but that:x > 4 (x is greater than 4)For thisequation, x can be any number that is larger than 4, so x could be 5, 6, 7 and so on.

Tips & Warnings

More advanced equations may require an additional step. An equation with anumber before the x, like 2x, would require you to divide both sides of theequation by the 2 to get the x by itself.

This is usually the first component of linear equations. After the inequality isdetermined, you may be asked to graph the inequality. For more information onhow to do this please see the Resources listed below.


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Don't forget to change the = sign back to the inequality sign. In a linear inequality,x will never = a number but will be < or > the number.

linear equations

Solving linear equations is one of the most fundamental skills an algebra student canmaster. Most algebraic equations require the skills used when solving linear equations.This fact makes it essential that the algebra student becomes proficient in solving theseproblems. By using the same process over and over, you can solve any linear equationthat your math teacher sends your way.

Difficulty: ModerateInstructionsThings You'll Need:

Paper Pencil

1. Step 1

Start by moving all of the terms that contain a variable to the left-hand side of theequation. For example, if you are solving 5a + 16 = 3a + 22, you will move the 3a tothe left-hand side of the equation. To do this, you must add the opposite of 3a toboth sides. When you add -3a to both sides, you get 2a + 16 = 22.

2. Step 2

Move the terms that do not contain variables to the right-hand side of the equation.In this example, you will add the opposite of +16 to both sides. This is -16, so youwill have 2a + 16 - 16 = 22 - 16. This gives you 2a = 6.

3. Step 3

Look at the variable (a) and determine if there are any other operations beingperformed on it. In this example, it is being multiplied by 2. Do the oppositeoperation, which is dividing by 2. This gives you 2a/2 = 6/2, which simplifies to a = 3.

4. Step 4

Check your answer for accuracy. To do this, put the answer back in to the originalequation. 5 * 3 + 16 = 3 * 3 + 24. This gives you 15 + 16 = 9 + 22. This is true,because 31 = 31.

5. Step 5


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Use the same process, even if the equation contains negatives or fractions. Forinstance, if you are solving (5/4) x + (1/2) = 2x - (1/2), you would begin by movingthe 2x to the left-hand side of the equation. This requires you to add the opposite.Since you will be adding it to a fraction (5/4), change the 2 to a fraction with acommon denominator (8/4). Add the opposite: (5/4) x - (8/4) x + (1/2) = (8/4) x - (8/4)

x -1/2, which gives (-3/4) x + (1/2) = - 1/2.

6. Step 6

Move the + 1/2 to the right-hand side of the equation. To do this, add the opposite (-1/2). This gives (-3/4) x + (1/2) - (1/2) = (-1/2) - (1/2), which simplifies to -3/4 x = -1.

7. Step 7

Divide both sides by -3/4. To divide by a fraction, you must multiply by the reciprocal(-4/3). This gives (-4/3) * (-3/4) x = -1 * (-4/3), which simplifies to x = 4/3.

8. Step 8

Check your answer. To do this, plug 4/3 in to the original equation. (5/4) * (4/3) +(1/2) = 2 * (4/3) -(1/2). This gives (5/3) + (1/2) = (8/3) - (1-2). This is true, because13/6 = 13/6.

Tips & Warnings

Using a calculator actually makes solving linear equations longer. If possible, dothis work by hand, especially when working with fractions.

Always check your answer. Making mistakes along the way is quite easy when

solving linear equations. Checking your answers will ensure that you do not getthe problem wrong.

quadratic equations

quadratic equation is an equation that can be written in the form:ax^2 + bx + c = 0,where a, b and c are real numbers and a is not 0.Quadratic equations have twosolutions, which are not necessarily unique.Algebra introduces quadratic equations andpossible ways to solve them. This article will address four different methods for solving

quadratic equations: factoring, completing the square, using the quadratic formula andusing Microsoft Excel.The first step in each method is to write the equation in standardquadratic equation form, ax^2 + bx + c = 0.

Difficulty: Moderately ChallengingInstructionsThings You'll Need:

Calculator


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Microsoft Excel

1. Step 1

Solve by factoring:Example:x^2 = 9Write the equation in standard quadratic form by

subtracting 9 from both sides: x^2 - 9 = 0Factor to write the polynomial as a product:(x + 3)(x - 3) = 0Set each factor equal to 0: (x + 3) = 0 or (x - 3) = 0Solve eachfactor: x = -3 or x = 3

2. Step 2

Solve by completing the square:Example:x^2 = 9Write in standard quadraticequation form by subtracting 9 from both sides: x^2 - 9 = 0Apply the square rootproperty: x = +/- square root of 9Solve the square root: x = +/- 3

3. Step 3

Solve by using the quadratic formula:Example:3x^2 + 16x + 5 = 0This example isalready written in the standard quadratic equation form; therefore, we know that a =3, b = 16 and c = 5.Substitute the values for a, b and c into the quadratic formula:x =(-b +/- square root(b^2 - 4ac)) / (2a)x = (-16 +/- square root(16^2 - 4(3)(5))) / (2(3))x= (-16 +/- square root(256 - 60)) / 6x = (-16 +/- square root(196)) / 6x = (-16 +/- 14) /6x = (16 - 14) / 6 or x = (16 + 14) / 6x = -1/3 or x = -5Apply the square root property:x = +/- square root of 9Solve the square root: x = +/- 3

4. Step 4

Solve by using Microsoft Excel:Example:3x^2 + 16x + 5 = 0This example is alreadywritten in the standard quadratic equation form; therefore, we know that a=3, b=16and c=5.In ExcelColumn A = aColumn B = bColumn C = cColumn D = the firstsolution for x=((-B2)+SQRT((B2*B2)-4*A2*C2))/(2*A2)Column E = the secondsolution for x=((-B2)-SQRT((B2*B2)-4*A2*C2))/(2*A2)Substitute the values for a, band c into the quadratic formula:x = (-b +/- square root(b^2 - 4ac)) / (2a)x = (-16 +/-square root(16^2 - 4(3)(5))) / (2(3))x = (-16 +/- square root(256-60)) / 6x = (-16 +/-square root(196)) / 6x = (-16 +/- 14) / 6x = (16 - 14) / 6 or x = (16 + 14) / 6x = -1/3 orx = -5

quadratic inequalities


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A quadratic inequality

Quadratic equations are polynomial equations in which the leading term variable issquared and, to solve them, the process of factoring needs to be utilized. Factoring isthe breakdown of polynomial equations into simpler equations, which, when multipliedtogether, give the same result. The inequality aspect means that the solution is notequal to only one number. The factoring aspect of quadratics can create more than onesolution that can all hold true or be any one of of several. Factoring is the key process tosolving quadratic inequalities.


InstructionsThings You'll Need:

An inequality Paper Pencil

Inequalities With Greater Than or Less Than

1. Step 1

Write the inequality on paper.

2. Step 2

Simplify the inequality, if possible. It is not possible here.

3. Step 3
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Get all terms on one side of the inequality and 0 on the other. Subtract 10 from bothsides.

4. Step 4

Factor the terms. The factors are x + 5 and x - 2.

5. Step 5

Decide the positive and negative combinations required to satisfy the inequality. Apositive is required, so both factors need to be positive. This results in two possiblecombinations.

6. Step 6

Write each factor as an inequality equal to 0 with the signs determined in Step 5.

These are the critical values.

7. Step 7
http://i.ehow.com/images/a04/7i/tj/solve-quadratic-inequalities-2.6-800X800.jpghttp://i.ehow.com/images/a04/7i/tj/solve-quadratic-inequalities-2.4-800X800.jpghttp://i.ehow.com/images/a04/7i/tj/solve-quadratic-inequalities-2.3-800X800.jpg


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Solve each inequality for x.

8. Step 8

Determine which critical value satisfy the inequality and this is the answer.Sometimes both sets of critical values can be part of the solution. Both sets arepossible.

9. Step 9

Simplify the answer, if possible. X is greater than 2 or less than -5.

Inequalities With Greater Than or Equal To or Less Than or Equal To

10.Step 1

Write the inequality on paper.

11.Step 2
http://i.ehow.com/images/a04/7i/tj/solve-quadratic-inequalities-2.8-800X800.jpghttp://i.ehow.com/images/a04/7i/tj/solve-quadratic-inequalities-2.7-800X800.jpg


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Simplify the inequality, if possible. Distribute the -x and divide both sides by -2,which means the inequality sign is reversed.

12.Step 3

Get all terms on one side of the inequality and 0 on the other. Add 12 to both sides

13.Step 4

Factor the terms. The factors are x - 3 and x - 4.

14.Step 5

Decide the positive and negative combinations required to satisfy the inequality. Anegative is required, so one factor needs to be negative and the other a positive.This results in two possible combinations.

15.Step 6


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Write each factor as an inequality equal to 0 with the signs determined in Step 5.These are the critical values.

16.Step 7

Solve each inequality for x.

17.Step 8

Determine which critical value(s) satisfy the inequality and this is the answer.Sometimes both sets of critical values can be part of the solution. The first set is notpossible, so it cannot be a solution. The second set works and that is the answer.

18.Step 9

Simplify the answer, if possible. X is greater than or equal to 3 AND less than orequal to 4.


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absolute inequalities

Don't let absolute value symbols in an inequality have you throwing your hands up in theair in frustration. This article will guide you through how to solve absolute value

inequalities without ripping your hair out.


InstructionsThings You'll Need:

Paper Pencil

Solving Absolute Value Inequalities When the Equation is Greater Than

1. Step 1

Look at the absolute value inequality that you are given, and make note of theprocesses that are in the equation.Is there subtraction or addition? Are you going tohave to multiply or divide? Are there any exponents? Knowing what processes are inthe absolute value inequality will let you know what processes you will need to doand it will also let you begin to decide what processes will need to be done beforeothers. Remember that when you are solving any equation for a variable you willneed to do the reverse order of operations to both sides of the equation in order toget your solution. This means that you will begin with addition/subtraction, thenmove to multiplication/division, followed by exponents, and then finish up with any

parentheses. An easy way to remember this is the normal order of operations isPEMDAS, so the reverse order is SADMEP, with each letter standing for a process.

2. Step 2

Acknowledge what the absolute value symbols mean for the inequality that you areabout to solve. When you are dealing with a "great than" inequality such as | 3x 7 |> 10 you have to not only solve the inequality 3x-7>10, you also have to solve theinequality 3x-710 and 3x-717 and 3x


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4. Step 4

Continue using the reverse order of operations to solve the two inequalities to getyour solutions to the original absolute value inequality.Divide both sides of bothinequalities by 3, which will leave you with the following solutions:x>17/3 and x


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Step 2

Break up the equation, more than once if necessary, if there is more than one set ofabsolute value brackets.

For example, for ||x-3| + |x+4|| = 11, it must be considered whether |x-3| + |x+4| is

positive or negative (fortunately it is always positive, so we can just remove its absolutevalue brackets). Then consideration is made whether x-3 is positive or not and whetherx+4 is positive or not.

Step 3

Make a line diagram of critical values if that would be helpful.

The critical values for ||x-3| + |x+4|| = 11 are at x=3 and x=-4, since that's when thesigns of the arguments of the absolute value brackets change.

Step 4

Fill out the line with the signs of the arguments of the absolute value brackets.

Below -4, both x-3 and x+4 are negative. So the equation becomes (3-x) + (-x-4) = 11,or x = -6.

Between -4 and 3, x-3 is negative and x+4 is positive. So the equation becomes (3-x) +(x+4) = 11, for which x has no solution.

Above 3, the equation merely needs all absolute value brackets removed, sinceeverything is positive. So the equation becomes (x-3) + (x+4) = 11, or x = 5.

Systems of equationsStep 1

Eliminate one variable to get the value of the other variable by adding or subtractingscalar products of the equations from each other.

For example, for the system

x + y = 12x - y = 2

add the two equations to cancel out the y's, to get

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

or

3x = 3.

So x = 1.


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Step 2

Plug the value of the known variable back into one of the equations, to solve for theother variable's value.

Continuing from the example in Step 1:

(1) + y = 1

Therefore, y = 0.

Step 3

Solve for three variables in three equations by performing more steps, adding scalarproducts of equations, to achieve the desired isolation of each variable one-by-one.

Exponential equations

algebra an exponential equation has variables in the exponent of a power, rather thanthe base. For example, 5 = 3^x is an exponential equation because 3 is raised to thepower x.

These math problems are encountered in high school algebra and can be solved withlogarithms. In practical use, exponential equations occur in financial math problemsinvolving compound interest.


Instructions

1. Step 1

First, review the basic rules of logarithms. For any two positive numbers a and b, thefollowing rules can be applied:

Log(ab) = Log(a) + Log(b)Log(a/b) = Log(a) - Log(b)Log(a^b) = (b)Log(a)

In the examples below, it will not matter if you use the "LOG" or "LN" buttons on yourcalculator. The difference between LOG and LN is that LOG is base 10, and LN isbase e=2.71828

2. Step 2

First, practice solving a simple exponential equation. Let's use the one given in theintro 5 = 3^x. The first step is to take the logarithm of both sides, so


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Log(5) = Log(3^x)

Using the third rule of logs, we can simplify it to

Log(5) = (x)Log(3)

And if we divide both sides by Log(3) we can isolate x, so

Log(5)/Log(3) = x

On the calculator, Log(5)/Log(3) = .68987/.47712 = 1.46497.

3. Step 3

Now try a more complicated one, 5^x = 4^(x+3). The first step is always to take the

natural log of both sides.

5^x = 4^(x+3)(x)Log(5) = (x+3)Log(4)(x)Log(5) = (x)Log(4) + (3)Log(4)

Put all the terms with x on one side of the equation and factor out the x.

(x)Log(5) - (x)Log(4) = (3)Log(4)(x)[Log(5) - Log(4)] = (3)Log(4)(x)Log(5/4) = (3)Log(4)

x = (3)Log(4)/Log(5/4)

So, x = (3)(.60206)/(.09691) = 18.638

4. Step 4

Now try using logs to solve an exponential equation in a compound interest problem.Meg invests $200 in a savings account that earns 4% interest per year, and $75 inanother account that earns 8% interest per year. At what point will the two accountshave the same amount of money?

5. Step 5

Set up the exponential equation. We are solving for x which represents a length oftime at which point the tow accounts have the same amount of money. Using theformula for compound interest, we have

200(1.04)^x = 75(1.08)^x


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If we take log of both sides and use the rules, we get

Log(200) + (x)Log(1.04) = Log(75) + (x)Log(1.08)

Put all the x terms on side and factor out the x, so

Log(200) - Log(75) = (x)[Log(1.08) - Log(1.04)]Log(200/75) = (x)Log(1.08/1.04)Log(2.66667) = (x)Log(1.03846)

And thus x = Log(2.66667)/Log(1.03846) = 26. So in 26 years the accounts will havethe same amount of money.

logarithmic equations

In math, a logarithm is the inverse function of an exponential function. For example, if

b^x = y, then x = Log(y) in base b. Logarithms are used to solve algebra equations thatinvolve exponents, and conversely, exponents are used to solve equations that involvelogarithms. If you encounter a logarithm equation in an algebra or precalculus class,here are the steps to solve the equation.


Instructions

1. Step 1

First, apply the 4 basic rules of logs to simplify the equation. The 4 rules hold trueregardless of the base of the logarithm.

(1) log(a) + log(b) = log(ab)(2) log(a) - log(b) = log(a/b)(3) log(a^n) = n[log(a)](4) log(1) = 0

For example, if you equation is log(x) + log(x+3) = 1, it can be rewritten as log(x+3x)= 1 by using the first rule.

2. Step 2

Perform the same operations to both sides of the equation so that all of the termswith variables are on one side. For example, the equation2log(x) + log(5) = log(x) + 3can be simplified to2log(x) - log(x) = 3 - log(5)log(x) = 3 - log(5).

3. Step 3


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Use the appropriate base to cancel out the logarithms. Conventionally, theabbreviation "log" denotes logarithm base 10, and "ln" denotes logarithm base e,where e 2.71828. If the logarithm is with another base, that number will be writtenas a subscript below the "g" in the abbreviation "log."

4. Step 4

To illustrate steps 1 through 3, consider the following example:2log(x) - log(x-2.5) = 1.

First, apply rule three to make the left side intolog(x) - log(x-2.5) = 1.

Then, apply rule two to obtainlog(x/(x-2.5)) = 1.

Now make both sides into exponents base 10:10^[log(x/(x-2.5))] = 10^1x/(x-2.5) = 10.

And now use regular algebra to solve.x/(x-2.5) = 10x = 10x - 25x - 10x + 25 = 0(x-5) = 0x = 5.

college algebra review

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