Factoring Polynomials - Algebra

The polynomial factors into (x + 3) (3x - 2).

3x² + 7x – 6 = (a + b)(c + d)

There must be a 3x term to get a 3x² term.

3x² + 7x – 6 = (3x + b)(x + d)

The other two numbers must multiply to –6 and add to +7 when one is multiplied by 3.

b * d = –6 and 3d + b = 7

b = –2 and d = 3

3x² + 7x – 6 = (3x – 2)(x + 3)

(x + 3) is the correct answer.

The product of the last two numbers should be 6, while the sum of the products of the inner and outer numbers should be 5x. Factors of six include 1 and 6, and 2 and 3. In this case, our sum is five so the correct choices are 2 and 3. Then, our factored expression is (x + 2)(x + 3). You can check your answer by using FOIL.

y = x2 + 5x + 6

2 * 3 = 6 and 2 + 3 = 5

(x + 2)(x + 3) = x2 + 5x + 6

If we convert the given quadratic equation from standard form to vertex form, we get:

Hence the axis of symmetry is

and the minimum value at is ( this function is concave down)

When working with a rational expression, you want to first put your monomials in standard format.

Re-order the bottom expression, so it is now reads .

Then factor a out of the expression, giving you .

The new fraction is .

Divide out the like term, , leaving , or .

First factor the equation. Find two numbers that multiply to 24 and sum to -10. These numbers are -6 and -4:

Set both expressions equal to 0 and solve for x:

Looking at the first two terms, the greatest common factor is 3b, hus we can factor out 3b.

Looking at the last two terms, the greatest common factor is –1. Factor out the –1.

Notice that you can factor out the .

This is your final answer.

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.

Because both terms are perfect squares, this is a difference of squares:

The difference of squares formula is .

Here, a = x and b = 5. Therefore the answer is .

You can double check the answer using the FOIL method:

The solutions indicate that the answer is:

and we need to insert the correct addition or subtraction signs. Because the last term in the problem is positive (+4), both signs have to be plus signs or both signs have to be minus signs. Because the second term (-5x) is negative, we can conclude that both have to be minus signs leaving us with:

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:

Using the FOIL rule, only yields the same polynomial as given in the question.

When asked to factor a difference of squares, the solution will always be the square roots of the coefficients with opposite signs in each pair of parentheses.

We will discuss coefficients in the general equation:

In this case, is positive and is negative, and , so we know our answer involves two negative numbers that are factors of and add to . The answer 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 .

By factoring one gets

Now setting each of the two factors to 0 (using the zero property), one gets

or

Here . Multiply and and you get which can be factored as and and when one adds and you get . Hence the quadratic equation can be rewritten as

Now you factor by grouping the first two terms and the last two terms giving us

which can be further factored resulting in

By setting each of the two factors to 0 we get

The vertex form of a quadratic equation is given by

Where the vertex is located at

giving us .

Factor 2 as GCF from the first two terms giving us:

Now we complete the square by adding 4 to the expression inside the parenthesis and subtracting 8 ( because ) resulting in the following equation:

which is equal to

Hence the vertex is located at

For the quadeatic equation . Applying these to the quadratic formula

we get

resulting in

and