The Lumenary · Grades 9–12

Polynomials

Name: ______________________

Date: ______________________

You will learn to

Add, multiply and factorise polynomials, using special products such as $a^2 - b^2 = (a - b)(a + b)$
Divide polynomials by long and synthetic division, and prove the remainder theorem
Prove the factor theorem and use it, with the rational root test, to find roots and factors
Prove that a non-zero polynomial of degree $n$ has at most $n$ roots, and explain why it may have fewer
Explain which hypothesis the root bound needs by finding a quadratic with four roots in clock arithmetic

1.
What is the degree of $(x^3 + 2x)(4x^5 - x + 1)$ ?
2.
Find the remainder when $p(x) = x^3 - 4x^2 + 2x + 7$ is divided by $x - 3$ .
3.
For which number $k$ is $x - 2$ a factor of $x^3 + kx - 6$ ?
4.
Find all real roots of $x^3 - 2x^2 - 5x + 6$ . Enter them as a set, for example {1, 2, 3}.
5.
For $x \ne 0$ , which expression equals $\dfrac{x^2 + 1}{x}$ ?
- A. $x + 1$
- B. $x + \dfrac{1}{x}$
- C. $x^2$
- D. $2x$
6.
A polynomial with real coefficients has degree $4$ . How many different real roots could it have? Choose all that are possible.
- A.0
- B.1
- C.2
- D.3
- E.4
- F.5
7.
In the proof of the root bound we wrote: $0 = p(b) = (b - a)\,q(b)$ and $b - a \ne 0$ , so $q(b) = 0$ . Modulo $8$ this step fails and $x^2 - 1$ has four roots. Which property of the real numbers does the step use?
- A.If $uv = 0$ then $u = 0$ or $v = 0$ (the zero-product property).
- B.Multiplication is commutative.
- C.Every number has a negative.
- D.The real numbers are ordered.
8.
Which Lean statement formalises the factor theorem for a polynomial p : R[X] and a number a : R? (Recall: C a is the constant polynomial $a$ and p.eval a is $p(a)$ .)
- A.X - C a ∣ p ↔ p.eval a = 0
- B.X + C a ∣ p ↔ p.eval a = 0
- C.p.eval a = 0 → p = X - C a
- D.∀ a, X - C a ∣ p
9.
Write divide_by_linear(coeffs, a) that divides $p(x)$ by $x - a$ using synthetic division. coeffs lists the coefficients of $p$ from the highest power down (so $2x^3 - 3x^2 - 11x + 6$ is [2, -3, -11, 6]). Return a pair (quotient, remainder), where quotient is the list of the quotient's coefficients.
Computational — Practice

Answer key — Polynomials

1.
The leading terms multiply to $x^3 \cdot 4x^5 = 4x^8$ , and no other product of terms reaches $x^8$ . Since $4 \ne 0$ , the degree is $3 + 5 = 8$ .
2.
By the remainder theorem the remainder is $p(3) = 27 - 36 + 6 + 7 = 4$ . Synthetic division gives the same: the rows are $1, -1, -1, 4$ , so $p(x) = (x - 3)(x^2 - x - 1) + 4$ .
3.
We need $p(2) = 8 + 2k - 6 = 2 + 2k = 0$ , so $k = -1$ . Check: $x^3 - x - 6 = (x - 2)(x^2 + 2x + 3)$ .
4.
$p(1) = 1 - 2 - 5 + 6 = 0$ , so $x - 1$ is a factor. Synthetic division gives $x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6) = (x - 1)(x - 3)(x + 2)$ . The roots are $\{1, 3, -2\}$ . A cubic has at most three roots, so there are no others.
5.
$\dfrac{x^2 + 1}{x} = \dfrac{x^2}{x} + \dfrac{1}{x} = x + \dfrac{1}{x}$ . The numerator $x^2 + 1$ has no factor $x$ , so nothing cancels. (In fact this is the division algorithm in disguise: $x^2 + 1 = x \cdot x + 1$ .)
6.
Any number from $0$ to $4$ is possible: $x^4 + 1$ , $x^4$ , $x^4 - 1$ , $x^2(x^2 - 1)$ and $(x^2 - 1)(x^2 - 4)$ . Five is impossible by the root bound. (For odd degree, $0$ would be impossible: every odd-degree real polynomial has a real root.)
7.
The step uses the zero-product property: a product of two numbers is zero only if one factor is zero. Real and complex numbers have it; the integers modulo $8$ do not ( $2 \cdot 4 \equiv 0$ ). In Lean this hypothesis is written IsDomain R.
8.
The factor theorem says $x - a \mid p(x) \iff p(a) = 0$ , which is X - C a ∣ p ↔ p.eval a = 0. The symbol ∣ (a vertical bar, typed \mid) means “divides”.
9.
```
def divide_by_linear(coeffs, a):
    running = []
    value = 0
    for c in coeffs:
        value = value * a + c
        running.append(value)
    return running[:-1], running[-1]
```
The running values are exactly the bottom row of the synthetic division table. The last one is $(\cdots(a_n a + a_{n-1})a + \cdots)a + a_0 = p(a)$ , which is the remainder theorem in action.