Dividing polynomials and factoring uniquely

Mathematics · Abstract Algebra · ages 22-24

Name ______________________ Date ____________

Divide x squared plus 7x plus 10 by x plus 2. The quotient is ax plus b. What is a plus b?
Answer: ______________
Let f equal x cubed minus 2x squared plus 3x minus 6. What is the remainder on division by x minus 2?
- f(2) equals 0, so x minus 2 is a factor
- f(2) equals 4, so x minus 2 is not a factor
- f(0) equals minus 6, so x minus 2 is not a factor
Divide x squared plus 5x plus 3 by x plus 1. Which statement is correct?
- Quotient x plus 6 with remainder 3
- Quotient x plus 4 with remainder minus 1, so x plus 1 is not a factor
- Quotient x plus 4 with remainder 0, so x plus 1 is a factor
Divide x^2 + 7x + 10 by x + 2 using polynomial long division. What is the quotient? Write your answer in the form ax + b, then give the value of a + b.
Answer: ______________
Divide 2x^2 + 7x + 3 by x + 3. The quotient has the form ax + b and the remainder is 0. What is a + b?
Answer: ______________
Factor x^3 - 3x^2 + 7x - 21 by grouping. One factor is x - 3. The other factor is x^2 + c. What is c?
Answer: ______________
Divide x^2 + 5x + 3 by x + 1. Which statement is correct?
- Quotient x + 4 with remainder -1, so x + 1 is not a factor
- Quotient x + 4 with remainder 0, so x + 1 is a factor
- Quotient x + 6 with remainder 3, so x + 1 is a factor
- Quotient x - 4 with remainder 7, so x + 1 is not a factor
Factor x^3 + 4x^2 + 5x + 20 completely by grouping.
- (x^2 + 5)(x + 4)
- (x^2 + 4)(x + 5)
- (x + 4)(x + 5)
- (x^2 - 5)(x - 4)
Which chain of ideas explains why polynomials over a field factor uniquely into primes (irreducibles), just like whole numbers factor uniquely into prime numbers?
- Division with remainder lets us build a greatest common divisor process, which forces every polynomial to factor into irreducibles in one unique way
- Every polynomial has a root, so we can always pull out linear factors until nothing is left
- Polynomials of degree 2 always factor, and higher degrees follow by graphing
- Unique factorisation holds because coefficients are always whole numbers
Factor x cubed minus 3x squared plus 7x minus 21 by grouping. One factor is x minus 3 and the other is x squared plus c. What is c?
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

Dividing polynomials and factoring uniquely W1-mt_m2G9PIHj6y-s1

6 · The quotient is x plus 5, so 1 plus 5 is 6.
f(2) equals 0, so x minus 2 is a factor · The Remainder Theorem gives f(2), which is 0 here.
Quotient x plus 4 with remainder minus 1, so x plus 1 is not a factor · Long division gives quotient x plus 4 and nonzero remainder minus 1.
6 · The quotient is x + 5, so a = 1 and b = 5, and a + b = 6. The remainder is 0, which tells us x + 2 is a factor.
3 · The quotient is 2x + 1 with remainder 0, so a = 2, b = 1, and a + b = 3. The zero remainder confirms x + 3 is a factor.
7 · Grouping gives x^2(x - 3) + 7(x - 3) = (x^2 + 7)(x - 3), so c = 7.
Quotient x + 4 with remainder -1, so x + 1 is not a factor · Long division gives quotient x + 4 and remainder -1. Because the remainder is not zero, x + 1 is not a factor of the polynomial, and x = -1 is not a root.
(x^2 + 5)(x + 4) · Grouping splits the four terms into two pairs that share the common factor x + 4, giving (x^2 + 5)(x + 4). Over the reals x^2 + 5 cannot be factored further.
Division with remainder lets us build a greatest common divisor process, which forces every polynomial to factor into irreducibles in one unique way · Division with remainder powers the Euclidean algorithm for greatest common divisors, and that machinery is exactly what proves every polynomial factors into irreducibles in essentially one way.
7 · Grouping gives (x squared plus 7) times (x minus 3).