# Algebra II : Polynomials

## Example Questions

### Example Question #74 : Factoring Polynomials

Factor:

Explanation:

To factor this, use trial and error to see what works. Since we have a  as the leading coefficient, it's helpful to remember that there's only one way to get  . Same with  as our third term--there's only one way to get  . Use these facts as you try to factor. Remember that signs matter. Therefore, your answer is: .

### Example Question #75 : Factoring Polynomials

Factor the polynomial:

Explanation:

The best method for factoring this polynomial is by grouping:

### Example Question #76 : Factoring Polynomials

Factor the following expression completely:

Explanation:

The expression is in quadratic form, so it can be factored as though it is quadratic.

The resulting expression can be factored further, as there are two difference of squares quadratic expressions.

### Example Question #77 : Factoring Polynomials

Factor the following polynomial:

Explanation:

In order to factor a polynomial, you have to first find the greatest common factor. In this case, each term has an x in it, so you can easily factor that out. Also, the greatest common factor for the three terms is 3. After you take both the 3 and the x out you are left with

The next step is to figure out which two numbers multiply together to equal , but also add together to equal . The factors of 52 are . The only ones that add to equal are , and in order to equal a positive 52, and , both of the numbers need to be negative.

### Example Question #78 : Factoring Polynomials

Factor .

Explanation:

We need two numbers that add to get , and multiply to get .  It looks like  and  would do the trick.  To double check our answer, we can use the quadratic formula.

Knowing our solutions are  and , we can insert them into an equation:

and set each equal to .

so:

### Example Question #79 : Factoring Polynomials

Factor .

Explanation:

Because of the leading  in the quadratic term of the function, it's hard to think of a clever combination of solutions, but we can just rely on the quadratic formula to help us out:

Setting each equation equal to  yields:

If we were to expand this function we would find that it's wrong by a factor of . To correct this, we multiply one of the terms by .  In this case we can multiply the  term by 2, which also serves to clean up the fraction.  This gives us an end result of:

### Example Question #80 : Factoring Polynomials

Factor .

Explanation:

We need to find two numbers that multiply to get , but we don't have to worry about them adding at all because there's a quadratic term and a linear term. If we check, we can see that the quadratic and linear terms multiply to get , which is very convenient for us.  Because the quadratic term is negative, we pair that sign with the linear term when factored:

### Example Question #81 : Factoring Polynomials

Factor .

Explanation:

We can start by factoring the numerator and the denominator.  Starting with the numerator, we need to find two numbers that multiply to get  and add to get .  It happens that 3 and 3 work, so:

For the denominator, we need two numbers that multiply to get , and add to get .  Here,  and  work, so:

If we needed to, we always could have used the quadratic formula to factor each of the above equations.  We can now put the factored equations back in to our original function:

Then we can cancel the  terms.

### Example Question #82 : Factoring Polynomials

Factor .

Explanation:

In the beginning, we can treat the polynomials in the numerator and denominator as separate functions. We can use the quadratic formula for each, if we can't come up with a clever guess at the solution:

Returning the solutions back into the original function gives:

We cancel the  terms and have:

Factor .