# Algebra II : Polynomials

## Example Questions

### Example Question #11 : How To Factor A Trinomial

For what value of  allows one to factor a perfect square trinomial out of the following equation:

Explanation:

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:

### Example Question #1 : Factoring Polynomials

Factor the trinomial .

Explanation:

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.

### Example Question #81 : Systems Of Equations

Solve for x.

x = –2/5, –5

x = –5

x = –2/3, –3

x = –5/2, –5

x = –5, 5

x = –5/2, –5

Explanation:

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

2) 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.

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

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

5) Set each parenthetical expression equal to zero and solve.

2x + 5 = 0,  x = –5/2

x + 5 = 0, x = –5

Explanation:

### Example Question #101 : Systems Of Equations

Factor the following quadratic equation.

Explanation:

When we attempt to factor a quadratic, we must first look for the factored numbers. When quadratics are expressed as  the factored numbers are  and . Since , we know the factors for 1 are 1 and 1. So we know the terms will be

Looking at our constant, , we see a positive 6. So 6 factors into either 2 and 3  or 1 and 6 (since  and ). Since our constant is a positive number, we know that our factors are either both positive, or both negative. (Note: you should know that 2 negative numbers multiplied becomes a positive number).

So to figure out what we must use we look at the  part of the quadratic. We are looking for 2 numbers which add up to our . So, 1 and 6 do not work, since . But, 2 and 3 are perfect since .

But, since our  is a negative 5, we know we must use negative numbers in our factored expression. Thus, our factoring must become

or

### Example Question #51 : Factoring Polynomials

Factor the following:

Explanation:

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

### Example Question #52 : How To Factor A Polynomial

Factor the following polynomial:

Can't be factored

Explanation:

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.

### Example Question #11 : How To Factor A Variable

Solve for  when

Explanation:

First, factor the numerator: .

Now your expression looks like

Second, cancel the "like" terms -  - which leaves us with .

Third, solve for , which leaves you with

### Example Question #12 : How To Factor A Variable

Factor the following polynomial: .

Explanation:

Because the  term has a coefficient, you begin by multiplying the  and the  terms () together:

Find the factors of  that when added together equal the second coefficient (the  term) of the polynomial:

There are four factors of , and only two of those factors, , can be manipulated to equal  when added together and manipulated to equal  when multiplied together:

### Example Question #53 : Factoring Polynomials

Solve for :

None of the other answers.

Explanation:

Step 1: Subtract 30 from both sides of the equation in order to make the equation equal to 0.

Step 2: Factor out a 3.

Step 3: Factor the trinomial.

At this step you set both factors equal to 0 and solve for .