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Thursday 21st of October Home Square Roots and Radical Expressions Solving Radical Equations Simplifying Radical Expressions Irrational Numbers in General and Square Roots in Particular Roots of Polynomials Simplifying Radical Expressions Exponents and Radicals Products and Quotients Involving Radicals Roots of Quadratic Equations Radical Expressions Radicals and Rational Exponents Find Square Roots and Compare Real Numbers Radicals Radicals and Rational Exponents Theorems on the Roots of Polynomial Equations SYNTHETIC DIVISION AND BOUNDS ON ROOTS Simplifying Radical Expressions Exponents and Radicals Properties of Exponents and Square Roots Solving Radical Equations Rational Exponents and Radicals,Rationalizing Denominators Rational Exponents and Radicals,Rationalizing Denominators Quadratic Roots Exponents and Roots Multiplying Radical Expressions Exponents and Radicals Solving Radical Equations Solving Quadratic Equations by Factoring and Extracting Roots Newton's Method for Finding Roots Roots of Quadratic Equations Studio Roots, Radicals, and Root Functions Review division factoring and Root Finding Radicals Simplifying Radical Expressions Multiplying and Simplifying Radical Expressions LIKE RADICALS Multiplication and Division of Radicals Radical Equations BOUNDING ROOTS OF POLYNOMIALS
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# 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 z1 = 2 cis (150°) and z2 = 3 cis (-60°). Find:

a) z1 and z2 in a + bi form
(b) z
2/z1
(c) z
2-z1 (e) All solutions z to z^2 = z1.

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.

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! :)