Factoring and expanding polynomials.

In algebra, students learn to factor polynomials like the quadratic equation. Factoring is much 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 = 0 and its factors will usually have the form (mx+n) (jx + k), where "x" is a variable and all the other values are constant..

Difficulty: Moderately EasyInstructions

Expanding1. Step 1

Write the factors in parentheses side-by-side. If one polynomial has more terms than the 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. Repeat this 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

Factoring6. 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 * x3x^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 * -4 3x^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 until you 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 know how basic inequalities work, and they know how to solve a basic linear equation, but putting the two together can cause some confusion. With a basic understanding of how each work, you can easily combine the two to complete a linear inequality.

Difficulty: Moderately EasyInstructions1. Step 1

Understand the basics of inequalities and linear equations. In an inequality, we say that that one number is greater than (>) another number or less than (<) another. On occasion we say that something is greater than or equal to, or less than or equal to. In a basic linear equation, we solve for the variable of x by subtracting the value from the opposite side or the = sign.

2. Step 2

Start by writing your equation on a sheet of paper. It will look something like this:x + 2 > 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 time being. When you do this, the original equation looks just like a traditional linear equation. 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 other out, 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 to make the equation look like a traditional linear equation, but now it must be turned back into an inequality. Do this by dropping the = and replacing it with the >. From this we now know that x does not equal for but that:x > 4 (x is greater than 4)For this equation, 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 a

number before the x, like 2x, would require you to divide both sides of the equation by the 2 to get the x by itself.

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

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 can master. Most algebraic equations require the skills used when solving linear equations. This fact makes it essential that the algebra student becomes proficient in solving these problems. By using the same process over and over, you can solve any linear equation that 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 the equation. For example, if you are solving 5a + 16 = 3a + 22, you will move the 3a to the left-hand side of the equation. To do this, you must add the opposite of 3a to both 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 you will 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 being performed on it. In this example, it is being multiplied by 2. Do the opposite operation, 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 original equation. 5 * 3 + 16 = 3 * 3 + 24. This gives you 15 + 16 = 9 + 22. This is true, because 31 = 31.

5. Step 5

Use the same process, even if the equation contains negatives or fractions. For instance, if you are solving (5/4) x + (1/2) = 2x - (1/2), you would begin by moving the 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 a common 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, because 13/6 = 13/6.

Tips & Warnings Using a calculator actually makes solving linear equations longer. If possible, do

this 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 get the 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 two solutions, which are not necessarily unique.Algebra introduces quadratic equations and possible ways to solve them. This article will address four different methods for solving quadratic equations: factoring, completing the square, using the quadratic formula and using Microsoft Excel.The first step in each method is to write the equation in standard quadratic equation form, ax^2 + bx + c = 0.

Difficulty: Moderately ChallengingInstructionsThings You'll Need: Calculator 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 each factor: x = -3 or x = 3

2. Step 2

Solve by completing the square:Example:x^2 = 9Write in standard quadratic equation form by subtracting 9 from both sides: x^2 - 9 = 0Apply the square root property: 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 is already 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 already written in the standard quadratic equation form; therefore, we know that a=3, b=16 and c=5.In ExcelColumn A = aColumn B = bColumn C = cColumn D = the first solution for x=((-B2)+SQRT((B2*B2)-4*A2*C2))/(2*A2)Column E = the second solution for x=((-B2)-SQRT((B2*B2)-4*A2*C2))/(2*A2)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 = -5

quadratic inequalities

A quadratic inequality

Quadratic equations are polynomial equations in which the leading term variable is squared and, to solve them, the process of factoring needs to be utilized. Factoring is the breakdown of polynomial equations into simpler equations, which, when multiplied together, give the same result. The inequality aspect means that the solution is not equal to only one number. The factoring aspect of quadratics can create more than one solution that can all hold true or be any one of of several. Factoring is the key process to solving quadratic inequalities.

Difficulty: Moderately EasyInstructionsThings You'll Need:

An inequality Paper

Pencil

Inequalities With Greater Than or Less Than1. Step 1

Write the inequality on paper.

2. Step 2

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

3. Step 3

Get all terms on one side of the inequality and 0 on the other. Subtract 10 from both sides.

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. A positive is required, so both factors need to be positive. This results in two possible combinations.

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

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 are possible.

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 To10. Step 1

Write the inequality on paper.

11. Step 2

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

15. Step 6

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 not possible, 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 or equal to 4.

absolute inequalities

Don't let absolute value symbols in an inequality have you throwing your hands up in the air in frustration. This article will guide you through how to solve absolute value inequalities without ripping your hair out.

Difficulty: Moderately EasyInstructionsThings You'll Need: Paper Pencil

Solving Absolute Value Inequalities When the Equation is Greater Than1. Step 1

Look at the absolute value inequality that you are given, and make note of the processes that are in the equation.Is there subtraction or addition? Are you going to have to multiply or divide? Are there any exponents? Knowing what processes are in the absolute value inequality will let you know what processes you will need to do and it will also let you begin to decide what processes will need to be done before others. Remember that when you are solving any equation for a variable you will need to do the reverse order of operations to both sides of the equation in order to get your solution. This means that you will begin with addition/subtraction, then move 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 is PEMDAS, 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 are about 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 the inequality 3x-7<-10 because the absolute value symbols will make any negative solutions positive.

3. Step 3

Begin to solve the two inequalities using the reverse order of operations.We have the two equations that we came up with in the previous step, 3x-7>10 and 3x-7<-10.Now we will use the reverse order of operations on each inequality, we begin by adding seven to both sides of both inequalities which leaves us with the following two inequalities:3x>17 and 3x<-3

4. Step 4

Continue using the reverse order of operations to solve the two inequalities to get your solutions to the original absolute value inequality.Divide both sides of both inequalities by 3, which will leave you with the following solutions:x>17/3 and x<-1.

5. Step 5

Remember, if you have to divide or multiply by a negative number you need to flip the inequality symbol.

Solving Absolute Value Inequalities When the Equation is Less Than6. Step 1

Follow Step 1 from above.

7. Step 2

Acknowledge what the absolute value symbols mean for the inequality.If we were solving the problem from above, only the problem was a "less than" equation we would need to set up the inequality as follows to account for the absolute value symbols,We would begin with the absolute value inequality, | 3x –7 | < 10, which would then become -10<3x-7<10 because the absolute value symbols turn negative solutions into positive solutions.

8. Step 3

Use the reverse order of operations like we did with the "greater than" equation to solve for the solutions.Add 7 to all three parts of the equation:-3<3x<17Divide all three parts of the equation by 3 to get our solution:-1<x<17/3.

9. Step 4

Remember, if you have to multiply or divide by a negative number to get your solution you also need to flip the inequality symbols.

absolute value equations. Step 1

Try to isolate x, but as with inequalities, split the equation into two when both positive and negative values are possible.

For example, for |x-3| = 9, account must be taken of x-3 being negative and x-3 being positive. If the former, then x-3 = -9, and the solution is x = -6. If x-3 is positive, just solve x-3 = 9.

Step 2

Break up the equation, more than once if necessary, if there is more than one set of absolute 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 absolute value brackets). Then consideration is made whether x-3 is positive or not and whether x+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 the signs 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, since everything is positive. So the equation becomes (x-3) + (x+4) = 11, or x = 5.

Systems of equations Step 1

Eliminate one variable to get the value of the other variable by adding or subtracting scalar 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.

Step 2

Plug the value of the known variable back into one of the equations, to solve for the other 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 scalar products 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 than the base. For example, 5 = 3^x is an exponential equation because 3 is raised to the power x.

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

Difficulty: Moderately EasyInstructions1. Step 1

First, review the basic rules of logarithms. For any two positive numbers a and b, the following 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 your calculator. The difference between LOG and LN is that LOG is base 10, and LN is base e=2.71828

2. Step 2

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

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 in another account that earns 8% interest per year. At what point will the two accounts have the same amount of money?

5. Step 5

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

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

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 have the 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 that involve exponents, and conversely, exponents are used to solve equations that involve logarithms. If you encounter a logarithm equation in an algebra or precalculus class, here are the steps to solve the equation.

Difficulty: Moderately EasyInstructions1. Step 1

First, apply the 4 basic rules of logs to simplify the equation. The 4 rules hold true regardless 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(x²+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 terms with variables are on one side. For example, the equation 2log(x) + log(5) = log(x) + 3can be simplified to 2log(x) - log(x) = 3 - log(5)log(x) = 3 - log(5).

3. Step 3

Use the appropriate base to cancel out the logarithms. Conventionally, the abbreviation "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 written as 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(x²) - log(x-2.5) = 1.

Then, apply rule two to obtainlog(x²/(x-2.5)) = 1.

Now make both sides into exponents base 10:10^[log(x²/(x-2.5))] = 10^1x²/(x-2.5) = 10.

And now use regular algebra to solve.x²/(x-2.5) = 10x² = 10x - 25x² - 10x + 25 = 0(x-5)² = 0x = 5.