Solving Polynomials - Pre-Algebra

Add 45 to both sides:

Find two numbers that add to -6 and multiply to 8. These numbers are -4 and -2.

Then set each term equal to 0:

Solve:

and

We can replace each instance of in the original expression with 3.

Now, solve using order of operations:

We can replace each instance of in the original expression with 3.

Start by simplifying our numerators and denominators.

In order to subtract our fractions, a common denominator must be found. The least common multiple of 15 and 20 is 60.

Multiply both the numerator and denominator of the first fraction by 4, and the second by 3.

Now, the fractions can be combined using subtraction:

Simplify the fraction to find the final answer:

First add 2 to both sides of the inequality.

Now, divide by 3 on both sides.

This gives us our final answer:

Factor the polynomial to reveal the quadratic equation inside the polynomial. Then, based on your preference, find the solutions to the quadratic equation with one of the various methods (we will use factoring to determine the roots).

The roots or solutions to this polynomial are x=-4, x=-1, and x=0. We find these roots by setting each factor equal to zero and solving.

An easy way to factor quadratics when the "a" term is 1 is to recognize the first term of each factor will just be x. Then, what two numbers can multiply to equal the "c" term, but add or subtract to equal the "b" term. All letter designations come from , the standard form of a quadratic equation.

Set the polynomial to zero by subtracting four on both sides.

Factor.

To factor this polynomial find the two factors of the integer term, , that when added together results in the middle term of the polynomial, .

Separate the two solutions.

The answers are .

In order to solve the equation, we have to isolate the variable. We do this by performing the same operation to either side of the equation.

Subtract 56 from both sides of the equation.

Factorize the left side.

These can be split into 2 separate equations.

The solutions are:

Subtract one from both sides of the equation.

Simplify the equation.

Divide by three on both sides.

Simplify the fractions.

Square root both sides to eliminate the square.

When taking the square root, there are two roots to the solution. These two solutions must satisfy the original equation.

Check to see if the roots will satisfy the original equation.

Both solutions satisfy the original equation, so the answers are:

This is a polynomial with an order of two. We should expect 2 solutions.

Pull out a common factor. The common factor is .

Do not divide by on both sides or this will eliminate a root to this solution!

The above equation can be separated into two equations.

Subtract eight from both sides.

The solutions are:

Substitute these back to the original equation to ensure that the roots are not invalid.

The correct answer is:

Find a solution to the following equation:

This looks intimidating, but we can make it a lot easier by trying one trick.

What happens when x is equal to 0?

So, all our x terms become 0 and we are left with

So, the point (0,134) is on the equation, and is a solution.

Add 45 to both sides:

Find two numbers that add to -6 and multiply to 8. These numbers are -4 and -2.

Then set each term equal to 0:

Solve:

and

We can replace each instance of in the original expression with 3.

Now, solve using order of operations:

We can replace each instance of in the original expression with 3.

Start by simplifying our numerators and denominators.

In order to subtract our fractions, a common denominator must be found. The least common multiple of 15 and 20 is 60.

Multiply both the numerator and denominator of the first fraction by 4, and the second by 3.

Now, the fractions can be combined using subtraction:

Simplify the fraction to find the final answer:

First add 2 to both sides of the inequality.

Now, divide by 3 on both sides.

This gives us our final answer:

Factor the polynomial to reveal the quadratic equation inside the polynomial. Then, based on your preference, find the solutions to the quadratic equation with one of the various methods (we will use factoring to determine the roots).

The roots or solutions to this polynomial are x=-4, x=-1, and x=0. We find these roots by setting each factor equal to zero and solving.

An easy way to factor quadratics when the "a" term is 1 is to recognize the first term of each factor will just be x. Then, what two numbers can multiply to equal the "c" term, but add or subtract to equal the "b" term. All letter designations come from , the standard form of a quadratic equation.

Set the polynomial to zero by subtracting four on both sides.

Factor.

To factor this polynomial find the two factors of the integer term, , that when added together results in the middle term of the polynomial, .

Separate the two solutions.

The answers are .

In order to solve the equation, we have to isolate the variable. We do this by performing the same operation to either side of the equation.

Subtract 56 from both sides of the equation.

Factorize the left side.

These can be split into 2 separate equations.

The solutions are:

Subtract one from both sides of the equation.

Simplify the equation.

Divide by three on both sides.

Simplify the fractions.

Square root both sides to eliminate the square.

When taking the square root, there are two roots to the solution. These two solutions must satisfy the original equation.

Check to see if the roots will satisfy the original equation.

Both solutions satisfy the original equation, so the answers are: