Solving by Factoring - GED Math

This problem involves the difference of two cubic terms. We need to use a special factoring formula that will allow us to factor this equation.

But before we can use this formula, we need to manipulate to make it more similar to the left hand side of the special formula. We do this by making the coefficients (343 and 64) part of the cubic power.

Comparing this with , and .

Plug these into the formula.

Begin by factoring out a 2:

Then, we recognize that the trinomial can be factored into two terms, each beginning with :

Since the last term is negative, the signs of the two terms are going to be opposite (i.e. one positive and one negative):

Finally, we need two numbers whose product is negative thirty-five and whose sum is positive two. The numbers and fit this description. So, the factored trinomial is:

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 from both sides to get

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

We must then factor to find the solutions for . To do this we must make a factor tree of which is 28 in this case to find the possible solutions. The possible numbers are , , .

Since is positive we know that our factoring will produce two positive numbers.

We then use addition with the factoring tree to find the numbers that add together to equal . So , , and

Success! 14 plus 2 equals . 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 which make each equation equal to 0 so in this case .

The center is located at (3,4) which means the standard equation of a circle which is:

becomes

which equals to

Change division into multiplication by the reciprocal which gives us the following

Now

this results in the following:

Simplification gives us

which equals

This expression involves the difference of two cubic terms. To factor an expression in this format, we can use a special formula.

Before we can use this formula, we need to manipulate our original expression to identify and .

Comparing this with the formula, and . Now we can use the formula to factor.

You can factor this trinomial by breaking it up into two binomials that lead with :

You will fill in the binomials by finding two factors of 36 that add up to 5. This is achieved with positive 9 and negative 4:

You can then set each of the two binomials equal to 0 and solve for :

Therefore:

The area of a right triangle with legs of length and is

.

Substitute and for and and 600 for , then solve for :

We can now factor the quadratic expression:

Set each linear binomial to 0 and solve to get possible solutions:

Since must be positive, we throw out the negative solution.

.

This is a quadratic equation in standard form, so first we need to factor .

This can be factored out as

where .

By trial and error we find that , so

can be expressed as

.

Set each linear binomial equal to 0 and solve separately:

The solution set is .

To factor a polynomial of the form , we want to look at the factors of and the factors of . We want to find the combination of factors which when multiplied and added together give the value of .

In our case, , , and .

The factors for are .

The factors for are.

Since is we will want to use the factors because .

Therefore when we put these factors into the binomal form we get,

.

Also see that

will foil out into the original polynomial, as , the coefficient for our term, and , the constant.

First, factor out the greatest common factor (GCF), which here is . If you don't see the whole GCF at once, factor out what you do see (here, either the or the ), and then check the result to see if any more factors can be pulled out.

Then, to factor a quadratic trinomial, list factors of the quadratic term and the constant (no variables) term, then combine them into binomials that when multiplied back out will give the original trinomial.

Here, the quadratic term has only one factorization: .

The constant term has factorizations of , , , and .

We know the constant term is positive, so the binomials both have the same operation in them (adding or subtracting), since a positive times a positive OR a negative times a negative will both give a positive result.

But since the middle term is negative, both binomial factors must contain subtraction. And .

For quadratic equations, you need to factor in order to solve for your variable. You do this after the equation is set equal to zero. Luckily, this is already done for you! Thus, start by factoring:

into

Then, you set each factor equal to . Solve each "small" equation:

or

or

BOTH of these are answers to the equation.

For quadratic equations, you need to factor in order to solve for your variable. You do this after the equation is set equal to zero. Thus, you get:

Next do your factoring:

into

Then, you set each factor equal to . Solve each "small" equation:

or or

BOTH of these are answers to the equation.

For quadratic equations, you need to factor in order to solve for your variable. You do this after the equation is set equal to zero. Thus, you get:

Next do your factoring. You know that both groups will be positive. Also, given that the middle term is , you only have one possible choice for your factors of :

Then, you set each factor equal to . Solve each "small" equation:

or

or

BOTH of these are answers to the equation.

For quadratic equations, you need to factor in order to solve for your variable. You do this after the equation is set equal to zero. Thus, you get:

Next do your factoring. You know that both groups will be negative. This will give you a positive last factor but a negative middle term. Given the value of the middle term, the factors of needed will be and

Then, you set each factor equal to . Solve each "small" equation:

or

or

BOTH of these are answers to the equation.

This is a quadratic equation, so first, move all terms to the same side by subtracting :

The quadratic polynomial can be factored using the (or grouping) method. We want to split the middle term by finding two integers with sum and product ; through some trial and error, we find and . The equation becomes

Regroup:

Distribute out common factors as follows:

Since the product of these two binomial expressions is equal to 0, one of them is equal to 0; set both to 0 and solve:

Subtract 1 from both sides:

Divide both sides by 3:

or

Add 3 to both sides:

The solution set of the equation is .

Start by rearranging the given equation:

Next, factor the equation.

Finally, set each factor equal to zero and solve.

and

Since can equal to either , we know that must then equal to either .

Start by factoring the equation.

For this equation, you want two numbers that multiply up to and add to . The only numbers that fit this criterion are and .

Thus,

Now, set each of these factors equal to zero and solve.