### All Algebra II Resources

## Example Questions

### Example Question #11 : Non Quadratic Polynomials

Factor .

**Possible Answers:**

**Correct answer:**

First, we can factor a from both terms:

Now we can make a clever substitution. If we make the function now looks like:

This makes it much easier to see how we can factor (difference of squares):

The last thing we need to do is substitute back in for , but we first need to solve for by taking the square root of each side of our substitution:

Substituting back in gives us a result of:

### Example Question #171 : Polynomials

Simplify:

**Possible Answers:**

**Correct answer:**

Use the difference of squares to factor out the numerator.

The term is prime, but can still be factored by another difference of squares.

Replace the fraction.

Simplify the top and bottom.

The answer is:

### Example Question #171 : Polynomials

Find a polynomial function of the lowest order possible such that two of the roots of the function are:

**Possible Answers:**

**Correct answer:**

Find a polynomial function of the lowest order possible such that two of the roots of the function are:

Recall that by roots of a polynomial we are referring to values of such that .

Because one of the roots given is a complex number, we know there must be a second root that is the complex conjugate of the given root. This is .

Because is a root, the unknown function must have a factor

The other roots are complex numbers, so there must be a quadratic factor.

To find the quadratic factor start with the value for one of the complex roots:

Isolate the imaginary term onto one side and square,

Expand the left side, and note on the right side the factor reduces as follows:

So now we have,

The quadratic factor for is therefore . Combining this with the factor give a factored expression for the desired function:

Now we carry out the multiplication to write the final form of ,

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