Factoring Polynomials - Math

We can factor this trinomial using the FOIL method backwards. This method allows us to immediately infer that our answer will be two binomials, one of which begins with and the other of which begins with . This is the only way the binomials will multiply to give us .

The next part, however, is slightly more difficult. The last part of the trinomial is , which could only happen through the multiplication of 1 and 2; since the 2 is negative, the binomials must also have opposite signs.

Finally, we look at the trinomial's middle term. For the final product to be , the 1 must be multiplied with the and be negative, and the 2 must be multiplied with the and be positive. This would give us , or the that we are looking for.

In other words, our answer must be

to properly multiply out to the trinomial given in this question.

Here you have an expression with three variables. To factor, you will need to pull out the greatest common factor that each term has in common.

Only the last two terms have so it will not be factored out. Each term has at least and so both of those can be factored out, outside of the parentheses. You'll fill in each term inside the parentheses with what the greatest common factor needs to be multiplied by to get the original term from the original polynomial:

To find the greatest common factor, we must break each term into its prime factors:

The terms have , , and in common; thus, the GCF is .

Pull this out of the expression to find the answer: .

First, begin by factoring out a common term, in this case :

Then, factor the terms in parentheses by finding two integers that sum to and multiply to :

To factor, we are looking for two terms that multiply to give and add together to get .

Possible factors of :

Based on these options, it is clear our factors are and .

Our final answer will be:

To factor, we are looking for two terms that multiply to give and add together to get . There are numerous factors of , so we will only list a few.

Possible factors of :

Based on these options, it is clear our factors are and .

Our final answer will be:

Factor out the largest quantity common to all terms:

Factor the simplified quadratic:

can be looked as . When A=1, as it does in this case, we can ignore it. So now we need to look at factors of "C" that add up to "B."

Factors of 20 are:

1 20

2 10

4 5

Of these three options, the 4 & 5 will add to 9, so we write

needs to be seen as . Then we need to ask what factors of "C" will equal "B" if one of them is multiplied by "A"

So first thing is to find factors of "C," luckily 5 only has 2:

1 and 5

We need to write an equation that uses either 5, 1 and a "*2" to equal -9.

Now we can write out our factors knowing that we will be using -5,2,and 1.

(__X+      )(      X+        )

from the work above, we know we have to multiply the 2 and the -5, so they need to be in opposite factors

(2X+__)(X-5)

that only leaves one space for the 1

needs to be seen as . Then you need to ask what factors of "C" will equal "B" when added together.

C=-18, factors of -18 are:

-1,18

-2,9

-3,6

1,-18

2,-9

3,-6

Of those factors, only -3,6 will give us "B", which in this case, is "3"

so

becomes

Set , and, consequently, . Substitute to form a quadratic polynomial in :

Factor this trinomial by finding two numbers whose product is 8 and whose sum is . Through trial and error, these numbers can be found to be and , so

Substitute back for :

Both factors are the difference of perfect cubes and can be factored further as such using the appropriate pattern:

Set , and, consequently, . Substitute to form a quadratic polynomial in :

Factor this trinomial by finding two numbers whose product is 8 and whose sum is . Through trial and error, these numbers can be found to be and , so

Substitute back for :

The first binomial is the difference of squares; the second is prime since 8 is not a perfect square. Thus, the final factorization is

Step 1: Break down into factors...

Step 2: Find two numbers that add or subtract to .

We will choose and .

Step 3: Look at the equation and see which number needs to change sign..

According to the middle term, must be negative.

So, the factors are and .

Step 4: Factor by re-writing the solutions in the form:

So...

This is a difference of squares. The difference of squares formula is _a_2 – _b_2 = (a + b)(a – b).

In this problem, a = 6_x_ and b = 7_y_:

36_x_2 – 49_y_2 = (6_x_ + 7_y_)(6_x_ – 7_y_)

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)

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 .

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