Polynomial Factors and Roots

Task

Limited factoring of polynomials; rewriting to reduce order or eliminate terms; using discriminant to count roots of a quadratic.

Basic methods

Quadratic formula; division (when given a root); monomial or linear change of variables (complete the square).

Students are expected to be able to routinely factor out common terms (e.g. write as ) and should know the correspondence between roots and linear factors.

Teaching notes

Diagnostic Aids, Symbols.

The quadratic formula

Quadratics are used very heavily in later work because they are one of the few families of equations that can be solved easily. Accordingly, students should know the quadratic formula by heart—like a poem—and be able to use it accurately without hesitation.

• Complex roots are discussed in the next section.

Problem 1

1. Find the roots of .
2. Factor .
3. Find the roots of

(alternate: write as products of linear factors)

These are quick applications of the quadratic formula.

Problem 2

Find two real roots of .

Notes: Students should recognize this as a quadratic in . The quadratic has roots . Roots of the original are obtained by formally applying the cube root function: .

Problem 3

For which values of does have (1) one root; (2) two real roots; (3) no real roots?

This is a straightforward application of the quadratic formula, complicated by having an answer involving inequalities ( , , and respectively).

Problem 4

Find so that has one (repeated) root as a polynomial in .

A word problem

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Word problems that genuinely use this material tend to be long and perhaps better considered as “projects”.

A sphere of radius is dropped into a cup with cross-section given by (in usual coordinates). Find the height of the center of the sphere above the bottom of the cup.

Notes: The height is determined by the fact that the sphere is tangent to the cup surface. Use symmetry to see the center of the sphere is on the axis, and then tangency happens when the common solutions of the equations and have a single value of . Use the first equation to eliminate from the second, then the problem is to find so the resulting quadratic in has a single solution.

Problem 5

is a root of . Write the polynomial as a product of linear factors.
(alternate: find all roots)

Since is a root must be a factor. Divide and use the quadratic formula on the quotient.

Complex roots

Essentially all of elementary algebra works over the complex numbers, and the extension is completely routine except that the arithmetic takes a little longer. It is helpful for students to understand that the usual restriction to real numbers is a matter of choice and not because the complexes are scary or weird.

Complex numbers cannot be avoided when studying roots of polynomials. Exposure can be kept to a minimum, namely

• add and multiply complex numbers;
• form conjugates and know the product of a number and it's conjugate is real;

There is no immediate need to multiply or divide complex polynomials so this is limited to an illustration in Problem 1.

Key Fact: If a polynomial with real coefficients has a root of the form with real but not real (i.e then the conjugate must also be a root. Consequently the polynomial is divisible by a real quadratic with these two roots. See “Irrational roots” below for an analogous situation.

Problem 1

Write as a (complex) polynomial in .

Problem 2

1. Write as a product of complex linear factors.
2. Write as a product of complex linear factors.

Problem 3

is a root of . Write the polynomial as a product of linear factors.

Divide by and use the quadratic formula on the quotient, just as in Problem 5 in the previous section.

Problem 4

1. Find a quadratic with real coefficients and with as a root.
2. Find a quadratic with real coefficients and with as a root.

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There are two ways to solve this: multiplying conjugate linear terms, and pattern matching using the quadratic formula:

• If is a root then is also a root and the product has real coefficients. Rewrite as and use the familiar expansion of this as . Expand further to get an explicit polynomial. Sometimes arithmetic can be simplified by multiplying by a number to clear denominators.
• The roots of are . Since the roots are complex must be negative and matching terms gives and . Solve for .

Pattern-matching is a powerful tool and worth exploring. It is not, however, a blank check: some qualitative information is needed to justify its use. The justification here is that real and purely imaginary numbers are linearly independent (heuristically “don't mix”) over the reals.

Irrational Roots

Exactly analogous to the complex-root fact, and true for essentially the same reason, we have:

Fact: If a polynomial with rational coefficients has a root of the form with rational but irrational then the “conjugate” must also be a root. Consequently the polynomial is divisible by a rational quadratic with these two roots.

Problem 1

Find a quadratic with rational coefficients and with as a root.

Note: This can be done by multiplying “conjugate” linear factors or by pattern matching; see the Notes following Problem 4 in the previous section.

Problem 2

The polynomial has root . Find the other roots (alternate: write as a product of linear factors).

Note: In principle we could divide by and then use the quadratic formula. The arithmetic would be awful. Instead note the fact implies the “conjugate” is also a root. Find a rational quadratic with these roots and divide to find the remaining root.

Problem 3

is a root of . Express the polynomial as a product of linear factors.

Note: Find a rational quadratic with roots the given one and it's conjugate, divide, and use the quadratic formula on the quotient.