Polynomials - Algebra 2

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept .

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept .

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept .

By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

Set each factor to zero and solve:

Since no real number squared is equal to a negative number, no real solution presents itself here.

The solution set is .

By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

Set each factor to zero and solve:

Since no real number squared is equal to a negative number, no real solution presents itself here.

The solution set is .

By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

Set each factor to zero and solve:

Since no real number squared is equal to a negative number, no real solution presents itself here.

The solution set is .

By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

Set each factor to zero and solve:

Since no real number squared is equal to a negative number, no real solution presents itself here.

The solution set is .

You need to factor by grouping but the important step is to remember the difference of squares.

You need to use the sum of two cubes equation

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

LCM of

LCM of

and since

The LCM

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.

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)

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: