Factoring Polynomials - Algebra II

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

1. The first step would be to simplify, but since 2, 15, and 25 have no common factors greater than 1, simplification is impossible.

Now we factor. Multiply the first coefficient by the final term and list off the factors.

2 * 25 = 50

Factors of 50 include:

1 + 50 = 51

2 + 25 = 27

5 + 10 = 15

1. Split up the middle term to make factoring by grouping possible.

Note that the "2" and the "10," and the "5" and the "25," have to go together for factoring to come out with integers. Always make sure the groups actually have a common factor to pull.

1. Pull out the common factors from both groups, "2x" from the first and "5" from the second.

1. Factor out the "(x+5)" from both terms.

1. Set each parenthetical expression equal to zero and solve.

2x + 5 = 0, x = –5/2

x + 5 = 0, x = –5

Factor out the 7:

Take the 8 from the x-term, cut it in half to get 4, then square it to get 16. Make this 16 equal to C/7:

Solve for C:

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:

This is a difference of perfect cubes so it factors to . Only the first expression will yield an answer when set equal to 0, which is 1. The second expression will never cross the -axis. Therefore, your answer is only 1.

Factor the equation to . Set and get one of your 's to be . Then factor the second expression to . Set them equal to zero and you get .

The quadratic can be solved as . Setting each factor to zero yields the answers.

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.

To factor trinomials like this one, we need to do a reverse FOIL. In other words, we need to find two binomials that multiply together to yield .

Finding the "first" terms is relatively easy; they need to multiply together to give us , and since only has two factors, we know the terms must be and . We now have  , and this is where it gets tricky.

The second terms must multiply together to give us , and they must also multiply with the first terms to give us a total result of . Many terms fit the first criterion. , , and all multiply to yield . But the only way to also get the "" terms to sum to is to use . It's just like a puzzle!

The given expression is a special binomial, known as the "difference of squares". A difference of squares binomial has the given factorization: . Thus, we can rewrite as and it follows that

For the trinomial to be factorable, we would have to be able to find two integers with product 36 and sum ; that is, would have to be the sum of two integers whose product is 36.

Below are the five factor pairs of 36, with their sum listed next to them. must be one of those five sums to make the trinomial factorable.

1, 36: 37

2, 18: 20

3, 12: 15

4, 9: 13

6, 6: 12

Of the five choices, only 16 is not listed, so if , then the polynomial is prime.

LCM of

LCM of

and since

The LCM

Use the -method to split the middle term into the sum of two terms whose coefficients have sum and product . These two numbers can be found, using trial and error, to be and .

and

Now we know that is equal to .

Factor by grouping.