# Review division factoring and Root Finding

(This is a brush-up from last semester.)

The division algorithm allows us to write any two

polynomials f(x) and p(x) ≠ 0 uniquely in the form

f(x) = p(x) q(x)+r(x), where q(x) is called the quo-

tient and r(x) is called the remainder.

In lecture we used the division algorithm to show that

factoring is the same as root- nding. In particular, the

remainder theorem tells you that f(c) is the remainder

when you divide f(x) by (x-c). (To see this, write

f(x) = (x-c) q(x) + r. Then f(c) = r.)

**Problems.**

**1. Constructing a polynomial with given roots.**

Construct a polynomial with real coefficients that

has exactly the given zeros and degree.

(a) 3-4i, degree 2

(b) 3-4i, 7, degree 3

(c) 3-4i, 7, degree 4

**2. Rational roots.**

Completely factor the following polynomials.

HINT: The rational roots theorem and synthetic

division can be a big help.

**3. Complex arithmetic.**

Let z_{1} = 2 cis (150°) and z_{2} = 3 cis (-60°). Find:

a) z_{1} and z_{2}
in a + bi form

(b) z_{2}/z_{1}

(c) z_{2}-z_{1}

(e) All solutions z to z^2 = z_{1}.

**4. Derivative of a polynomial with a root of
multiplicity greater than 1.**

Have each person in your group invent a polynomial p that
has a root of multiplicity 2. Now

differentiate it to get the polynomial p'. Factor p'. Do you notice a common
pattern? Do you think

that this happens in general? Can you prove it?

**5. Factoring a quadratic by depressing it.**

Suppose you want to factor a quadratic polynomial, i.e. a
polynomial of the form Ax^2+Bx+C.

If you divide by A, you will get an equation of the form p = x^2 +βx + γ. We
wish to put this

equation in the "depressed" form y^2 + c. To accomplish this, substitute x = y-α
and expand

out to get an equation of the form y^2 + by + c. What should α be in order to
make the y-term

zero? Choose this α, solve for x, and expand your solution in terms of the
original parameters of the

quadratic polynomial (A,B, and C). Do you get a familiar formula?

**6. Factoring cubics part I: depressing a cubic.**

Ever wonder how to get a general formula for the factors
of a cubic polynomial? Begin with a general cubic,

Divide by A to get a
cubic of the form We wish to put this
equation in the "depressed" form The trick
is to make a substitution of the form x = y-α .What should α be in order to make
the y^2 term zero?

**7. Factoring cubics part II: solving a depressed cubic.**

Suppose you are trying to factor a depressed cubic,

p = y^3 + cy + d. So you are trying to solve

the equation y^3 +cy +d = 0. The trick is to look

for a solution of the form y = s-t. Plug this in.

Show that we've got a solution if:

So if we can solve this system for s and t in terms

of c and d, we're done. Do you see how to solve

it?

**8. Factoring cubics part III: try an example!**

Now let's do an example. To have a nice example

to test out our method, let's make up a cubic

where we already know the roots. I don't know,

how about p(x) = (x-3)(x^2 +1)? Or maybe you

would prefer real roots, e.g. p(x) = (x-5)(x +

1)(x + 4)? Multiply out the factored polynomial

that you pick and then use the methods in the

previous two sections to factor it. Does it work?

You also might try starting with a random depressed

cubic, such as p(x) = x^3-3x + 1. Compute the roots.

Then find the zeros by graphing or using Newton's method.

Do your answers agree?

**9. Factoring a quartic.**

To factor a quartic equation, first figure out a

substitution that \depresses" it (i.e. gets rid of

the cube term). Then assume that you can write

it as a product of quadratic polynomials with

generic coefficients. Multiply the quadratics to

get a quartic and equate the coefficients with the

coefficients of the depressed quartic. This gives

a system of equations which, with some strategic

elimination of variables, you can reduce to the

problem of finding the roots of a cubic.

**10. Factoring a quintic.**

Show that it is impossible to represent the roots

of the polynomial f(x) = x^5-x-1 using addition,

subtraction, multiplication, division, or extraction

of roots. Note that this result shows that

it is impossible to find a general formula for the

roots of a polynomial of degree 5 or higher. (Hint:

read up on Galois theory! :)