1) Begin the problem by factoring the final term. Include the negative when factoring.

–2 + 2 = 0

–4 + 1 = –3

–1 + 4 = 3

All options are exhausted, therefore the problem cannot be solved by factoring, which means that the roots either do not exist or are not rational numbers. We must use the quadratic formula.

Put the quadratic in standard form:

Factor:

An easy way to factor (and do so with less trial and error) is to think of what two numbers could multiply to equal "c", but add to equal "b". These letters come from the designations in the standard form of a quadratic equation: . As you can see the product of -6 and 2 is -12 and they both add to 4.

Use the quadratic formula to solve the equation.

Plug in these values and solve.

We start by subtracting  on both sides in order to get a quadratic expression on one side of the equation:

We can factor this into:

Thus, either  or .

So,  or 

To solve for x, we must first simplify the trinomial into two binomials.

To simplify the trinomial, its general form given by , we must find factors of  that when added give us . For our trinomial,  and the two factors that add together to get  are  and .

Now, using the two factors, we can rewrite  as the sum of the two factors each multiplied by x:

Next, we group the first two and last two terms together and factor each of the groups:

Now, simplify further:

Finally, set each of these binomials equal to zero and solve for x:

Write the quadratic equation.

The quadratic is in the form:  

Substitute the known coefficients.

Simplify the numerator and denominator.

Because we have a negative value inside the square root, we will have complex roots instead of real roots.  The roots are imaginary.

The answer is:  

Reverse Foil Method

Factors of 

Factors that add to 

Expand Quadratic Equation

Regroup 

Roots occur at 

In order to find the perimeter, start by defining the variables. It is typically easier to define one of the variables in terms of the other; therefore, only one unknown will need to be calculated to find the perimeter. The problem states that the length is eleven more than twice the width; thus, we can define our variables in the following way:

The farmer knows that the field's area is thirty square meters. Area is found using the following formula:

Substitute in the known value for the area and the defined variables for the length and width. 

Notice that this equation possesses all the components of a quadratic. Use the information in the equation to construct a quadratic equation that can be factored to obtain an answer. Start by multiplying the first term by the variable on the right side of the equation.

In order to make the quadratic equal to zero, subtract 30 from both sides of the equation. 

Now, factor the quadratic and solve for the variable. We can use the ac method to solve for the variable. Quadratics can be written in the following format:

We need to find two numbers whose product equals a multiplied by c and whose sum equals b; therefore, the product of the factors must be -60 and their sum must equal 11. Write out the prime factorization of 60.

There is one factor of -60 that when added together sum to equal 11: 15 and -4.

Use the factors and split the middle term in the quadratic in order to make factoring by grouping possible.

Pull the greatest common factor from each pair of terms:  from the first and 15 from the second.

Factor out the quantity  from both terms.

Set each factor equal to zero and solve for w.

We can cross out the this negative option because the width of a dimension cannot be a negative value. Solve for w in the second factor.

The width of the field is 2 meters. Substitute 2 in for the variable w and solve for the perimeter.

Perimeter is found using the formula:

The perimeter of the field is 34 meters.  

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 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