Pull out an  term

Two numbers are needed that add to  and multiply to be . Guess and check results in  and .

Each term must be set equal to 0 to find the roots.

The polynomial is degree 4 so there are 4 roots. To make the roots easier to find the expression can be written as

The roots are 

In order to find the roots, factorize the quadratic.

The multiple to the integer 52 are:

The last set can produce the middle term.

Write the binomials.

Setting the equation equal to zero, we have two equations:

Solving for the equations, we will have  as the possible roots.

The answer is:  

Use the quadratic equation to solve for the roots.

Substitute the values of the polynomial  into the equation.

Simplify the quadratic formula.

Since we have a negative discriminant, we will have complex roots even though there are no real roots.

The roots are:  

One of the possible roots given is:  

There are multiple methods to solve quadratics. Use whichever is easiest for the problem.

Solve by factoring:

What numbers will multiply to get  and what two numbers will multiply to get -14?

To solve for the roots, set the factors equal to 0 and solve:

To find the roots, or solutions, of this quadratic equation, first factor it.

Recall that when a function is in the  form, the factors of a and c when multiplied and added together must equal b.

First, identify factors of 6 that could equal -1. Y

ou have to think of your positive and negative signs here. Remember that  and you need to have one positive and one negative number to get -6.

Therefore, after factoring, you should get: 

.

Then, set each of those expressions equal to 0.

Therefore, your roots are:

.

Since this function is already factored and equal to 0, you can just set each expression equal to 0 to get your roots.

and

.

The equation will need to be simplified to its standard form.

Simplify this equation by using the FOIL method to expand .

Simplify the terms.

The equation becomes:

Distribute the negative two through the parentheses.

Combine like terms.

The equation in standard for becomes:  

The standard form for a polynomial is:  

Write the quadratic equation.

Substitute the coefficients corresponding to the equation in standard form.

Simplify the radical.  The roots will be imaginary.

Simplify the fractions and replace  with .

The answer is:  

You can solve this quadratic in many different ways, including by graphing or using the quadratic formula. This particular quadratic can be factored:

we're looking for 2 numbers that add to 5 and multiply to

The numbers that work are 8 and -3:

continue factoring

The factors are and . Set each factor equal to zero:

You can solve this quadratic equation in many different ways, including by graphing or factoring. You can also use the quadratic formula:

This will give us two answers:

This quadratic equation can be solved in several ways, including the quadratic equation or by graphing. It can also be solved by factoring:

For this quadratic, we are looking for 2 numbers that add to -19 and multiply to 6x10=60. The numbers satisfying these conditions are -4 and -15:

continue factoring

The factors are and . Set each equal to zero: