Systems of Equations - Algebra 1

First pull out 3u from both terms.

3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember, but pat yourself on the back for getting to such hard questions! The difference of cubes formula is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v:

3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]

= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept .

The given expression is a special binomial, known as the "difference of squares". A difference of squares binomial has the given factorization: . Thus, we can rewrite as and it follows that

The product of is .

For the equation ,

must equal and must equal .

Thus and must be and , making the answer .

This is a quadratic equation. We can solve for either by factoring or using the quadratic formula. Since this equation is factorable, I will present the factoring method here.

The factored form of our equation should be in the format .

To yield the first value in our original equation (), and .

To yield the final term in our original equation (), we can set and .

Now that the equation has been factored, we can evaluate . We set each factored term equal to zero and solve.

First, find the factors of 90 and 315. The greatest common factor is the largest factor shared by both of the numbers: 45.

Then, find the least common multiple of 5 and 15. This will be the smallest number that can be divided by both 5 and 15: 15.

Finally, the greatest common factor (45) divided by the least common multiple (15) = 45 / 15 = 3.

First factor the numerator. We need two numbers with a sum of 3 and a product of 2. The numbers 1 and 2 satisfy these conditions:

Now, look to see if there are any common factors that will cancel:

The in the numerator and denominator cancel, leaving .

The general form for a factored expression of order 2 is

, which, when FOILED, gives .

Comparing this generic expression to the one given in the probem, we can see that the term should equal , and the term should equal 2.

The values of and that satisfy the two equations are and ,

so your factored expression is

The first step in solving the equation is to add to both sides so that we can set the new equation equal to so that we can factor it.

We now need to factor it. We need to find two numbers, and , such that and .

Those two numbers are

Our new factored expression becomes

Now we can easily identify that the values that satisfy this equation are .

In order to set this up for completing the square, we need to move the 135 to the other side:

Now the equation is in the form:

To complete the square we need to add to both sides the folowing value:

So we need to add 9 to both sides of the equation:

Now we can factor the left side and simplify the left side:

Now we need to take the square root of both sides:

**NOTE: Don't foget to add the plus or minus symbol. We add this becase there are two values we can square to get 144:

and

**End note

Now we can split into two equations and solve for x:

and

So our solution is:

There is a common variable in each term. Pull this out as a common factor.

Factor .

The common factors that will achieve the middle term and will have a product of 36 is four and nine. Write the binomials.

The cubics here cannot be simplified any further.

The answer is:

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.