Quadratic Equations - Algebra

To factor a quadratic equation in the form

, where , find two integers that have a sum of and a product of .

For this equation, that would be 9 and 5.

Therefore, the solutions to this equation are and .

1. After adding like terms and setting the equation equal to zero, the immediate next step in solving any quadratic is to simplify. If the coefficients of all three terms have a common factor, pull it out. So go ahead and divide both sides (and therefore ALL terms on BOTH sides) by 4.

Since zero divided by four is still zero, only the left side of the equation changes.

1. Either factor by grouping or use the square trick.

Grouping:

1 + 1 = 2

(The "1" was pulled out only to make the next factoring step clear.)

x + 1 = 0, x = –1

OR

Perfect Square:

x = –1

To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

So (x + 2)(x + 3) = 0

x = –2 or x = –3

To find the domain, find all areas of the number line where the fraction is defined.

because the denominator of a fraction must be nonzero.

Factor by finding two numbers that sum to -2 and multiply to 1. These numbers are -1 and -1.

Given the equation

Realizing that the domain is restricted by the denominator, meaning that the denominator can not be equal to 0.

Set the denominator = to 0 and solve.

First factoring ,

Zero-product Property, setting both quantities equal to 0 and solve.

So when x is 6 or -1, our denominator will be 0. Meaning those would be our domain restrictions.

To factor the polynomial, we need the numbers that multiply to –12 and add to +1. This leads us to –3 and +4. We solve the polynomial by setting it equal to 0.

So either x = 3 or x = –4 will make the product equal to 0.

This is a factoring problem, so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract 128 from both sides to get .

We then notice that all four numbers are divisible by four, so we can simplify the expression to .

Think of the equation in this format to help with the following explanation.

We must then factor to find the solutions for x. To do this we must make a factor tree of c (which is 32 in this case) to find the possible solutions. The possible numbers are 1 * 32, 2 * 16, and 4 * 8.

Since c is negative, we know that our factoring will produce a positive and negative number.

We then look at b to see if the greater number will be positive or negative. Since b is positive, we know that the greater number from our factoring tree will be positive.

We then use addition and subtraction with the factoring tree to find the numbers that add together to equal b. Remember that the greater number is positive and the lesser number is negative in this example.

Positive 8 and negative 4 equal b. We then plug our numbers into the factored form of .

We know that anything multiplied by 0 is equal to 0, so we plug in the numbers for x which make each equation equal to 0. In this case .

The quadratic can be solved as . Setting each factor to zero yields the answers.

Begin by setting the equation equal to 0 by subtracting 88 from both sides.

Now that the equation is in the form , find two integers that sum to and have a product of .

For this equation, those integers are and .

Therefore, the solutions to this equation are

Begin by setting the equation equal to zero by adding 105 to each side.

For an equation in the form , where , find two integers that have a sum of and a product of .

For this equation, that would be 8 and 9.

Therefore, the solutions to this equation are

When we attempt to factor a quadratic, we must first look for the factored numbers. When quadratics are expressed as the factored numbers are and . Since , we know the factors for 1 are 1 and 1. So we know the terms will be

Looking at our constant, , we see a positive 6. So 6 factors into either 2 and 3 or 1 and 6 (since and ). Since our constant is a positive number, we know that our factors are either both positive, or both negative. (Note: you should know that 2 negative numbers multiplied becomes a positive number).

So to figure out what we must use we look at the part of the quadratic. We are looking for 2 numbers which add up to our . So, 1 and 6 do not work, since . But, 2 and 3 are perfect since .

But, since our is a negative 5, we know we must use negative numbers in our factored expression. Thus, our factoring must become

or

Determine the signs of the binomials. Because the quadratic has a negative middle term and a positive end term, the signs will be both negative.

To factor, find the roots of that will either add or subtract to get a coefficient.

The only possibility is the set .

Substitute the roots to the factored form .

The expression can be factored by finding terms that multiply back to the original expression. The easiest way is to find two numbers that add to the middle term as well as multiply to the last term The numbers that satisfy both of these conditions are and , so the answer is .

Factoring is basically removing quantities from an equation. For instance has an x in each term. Therefor the fully factored form would have the most amount of x's removed from each term and would be . You cannot take out more than is present in the smallest term (smallest in the sense of what you are trying to factor out). Here we could not remove more than one x since 2x was limiting. For Quadratic equations such as we have a little easier time of factoring because we know that it will break down or factor into two binomials or two equations with two terms each. To get started recognize that since the "a" term is 1 , our first term in each binomial will just be x. Next you have to come up with two numbers that will multiply to equal the c term (the last one, 8), but add or subtract to give the middle "b" term. This comes from the fact the if you foiled your choices of your two factors the numbers must add or subtract to equal the middle term and multiply to equal the last term. Let's begin:

Start by writing two parentheses ( ) ( )

Since our "a" term is 1, the first term in each binomial will be x!

Now what numbers multiply to give the product of 8? 8 and 1. 2 and 4. Good. Which of these (if any because not every quadratic is factorable) are able to add to give the middle term of 6? Of course it is 2 and 4. So start by assuming the two factors are:

Choose an addition sign because there is no way that 2 and 4 can subtract to equal a positive 6. Now us the FOIL method to check to see if this does become when combined.

multply the first x by everything in the second parenthesis:

now mulitply 4 by every term in the second parenthesis and combine all like terms:

So are the factors of . This step by step method makes the trial and error nature of this type of factoring a little more precise. Remember, if your "a" term is one then the first term in each of your factors are x. Then just think.. "what two numbers multiply to give me the last term of my quadratic I am factoring, but also add or subtract to give the middle term.

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into quadratic form, but this is already done.

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of are and , we know that

Next, factor out our constant, ignoring the sign for now. The factors of are either and or and . We must select those factors which have either a difference or a sum equal to the value of in our trinomial. In this case, and cannot sum or difference to , but and can. Now, we can add in our missing values:

One last step remains. We must check our signs. Since is negative in our trinomial, one and only one of our two binomials must have a negative sign. To figure out which, check the sign of in our trinomial. Since is negative, the larger of the two numbers in our binomial must be negative.

Thus, our two binomial factors are and .

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into quadratic form, but this is already done.

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of are and , we know that

Next, factor out our constant, ignoring the sign for now. The factors of are either and , and , or and , We must select those factors which have either a difference or a sum equal to the value of in our trinomial. In this case, neither and nor and can sum to , but and can. Now, we can add in our missing values:

One last step remains. We must check our signs. Since is positive in our trinomial, Either both signs are negative or both are positive. To figure out which, check the sign of in our trinomial. Since is negative, both signs in our binomials must be negative.

Thus, our two binomial factors are and .

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into quadratic form.

--->

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of are and , we know that

Next, factor out our constant, ignoring the sign for now. The factors of are either and , and , or and , We must select those factors which have either a difference or a sum equal to the value of in our trinomial. In this case, neither and nor and can sum to , but and can. Now, we can add in our missing values:

One last step remains. We must check our signs. Since is positive in our trinomial, Either both signs are negative or both are positive. To figure out which, check the sign of in our trinomial. Since is negative, both signs in our binomials must be negative.

Thus, our two binomial factors are and . Note that this can also be written as , and if you graph this, you get a result identical to .

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into quadratic form, but this is already done. Note that this is a special trinomial: , and thus only and are present to be manipulated.

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of are and , we know that

Next, factor out our constant, ignoring the sign for now. The only factors of are either and , or and , We must select those factors which have either a difference or a sum equal to the value of in our trinomial. In this case, only and sum to , which we use since is not present.

One last step remains. We must check our signs. Since is negative in our trinomial, one sign is positive and one is negative.

Thus, our two binomial factors are and .

This is called a difference of squares. If you see a trinomial in the form , the roots are always and .

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into quadratic form. It isn't ideal to work with , but in this equation cannot be reduced.

--->

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of are and , we know that

Next, factor out our constant, ignoring the sign for now. The factors of are and . Note that these do not normally sum or difference to , but the presence of as a term means one term will be doubled in value when creating . This means we either get and (no good), or and (good). Remember to add the number we want to double inside the binomial opposite , so it multiplies correctly. Now, we can add in our missing values:

One last step remains. We must check our signs. Since is negative in our trinomial, one sign is positive and one is negative. To figure out which, check the sign of in our trinomial. Since is positive, the bigger of our two terms after any multipliers are completed is the positive value. Since our terms after mutiplying are and , the must be the positive term.

Thus, our two binomial factors are and .