Algebra 2 · A.APR.B.2 · A.APR.B.3

The remainder and factor theorems

Dividing a polynomial by x minus a number takes a page of work. Can you know the remainder without dividing, and tell at a glance when there is none?

You need

Synthetic divisionAlgebra 2

Unlocks

The fundamental theorem of algebraAlgebra 2 The rational root theoremAlgebra 2

Where you are 3 in Remainder and factor theorems

Algebra9

Polynomials10

Remainder and factor theorems3

The remainder and factor theorems

The fundamental theorem of algebra The rational root theorem

Why this exists

Polynomial long division and Synthetic division answer “what is p(x) divided by x - c?” with a quotient and a remainder. Often only the remainder matters, and most of all whether it is 0: a zero remainder means x - c is a factor, and factors are what turn a cubic into pieces the Solving quadratics by factoring can solve.

Two theorems cut the work. The remainder is simply p(c), one evaluation. And x - c is a factor exactly when p(c) = 0. So one known zero of a polynomial splits off a factor and lowers the degree by one, which is how most cubics and quartics are factored by hand.

Definition

Dividing a polynomial p(x) by x - c produces a quotient q(x) and a remainder r with

p(x) = (x - c)\,q(x) + r

for every value of x. The remainder always has lower degree than the divisor, and x - c has degree 1, so r is a number.

When r = 0, p(x) = (x - c)\,q(x), and x - c is a factor of p(x); we also say p(x) is divisible by x - c.

A zero of p is a number c with p(c) = 0. The divisor fixes c with its sign: x + 2 is x - (-2), so c = -2.

Intuition

Take p(x) = x^3 - x^2 - 4x + 4 and divide by x + 1. Synthetic division by -1 on the coefficients 1, -1, -4, 4 gives the bottom row 1, -2, -2, 6, so

p(x) = (x + 1)(x^2 - 2x - 2) + 6

Now simply evaluate: p(-1) = -1 - 1 + 4 + 4 = 6. The same number, and no accident. Put x = -1 into the identity and the factor x + 1 becomes 0, wiping out the whole product and leaving only the remainder.

So the remainder on dividing by x - c is the height of the graph of p above c. Where the graph meets the x-axis the height is 0, the remainder is 0, and x - c is a factor. Here p(x) = (x + 2)(x - 1)(x - 2), and the graph crosses at exactly -2, 1 and 2.

Figure 1The remainder on dividing p(x) by x - c is the height of the graph above c. At c = -1 it is 6. At the three crossings it is 0, and there x - c is a factor.

Results

The remainder theorem. When a polynomial p(x) is divided by x - c, the remainder is p(c).

The factor theorem. x - c is a factor of p(x) if and only if p(c) = 0.

Four views of one fact. For a real number c, these all say the same thing.

View	Statement
zero of the function	p(c) = 0
root of the equation	x = c solves p(x) = 0
x-intercept of the graph	(c, 0) lies on y = p(x)
factor of the polynomial	x - c divides p(x) with remainder 0

The first three are one fact by Roots, zeros and x-intercepts; the factor theorem adds the fourth.

Splitting off a zero. If p(c) = 0, then p(x) = (x - c)\,q(x), where q has degree one less than p, and the zeros of p are c together with the zeros of q.

Why it is true

The remainder theorem. Division gives p(x) = (x - c)\,q(x) + r with r a number. This is an identity, true for every value of x, so it is true at x = c:

p(c) = (c - c)\,q(c) + r = 0 \cdot q(c) + r = r

The factor theorem, one way. If p(c) = 0, the remainder theorem says r = 0, so p(x) = (x - c)\,q(x) and x - c is a factor.

The other way. If p(x) = (x - c)\,q(x) for some polynomial q, then p(c) = 0 \cdot q(c) = 0. This half needs no division at all.

Splitting off a zero. Degrees add when polynomials are multiplied, so q has degree one less than p. By the zero-product property, (x - c)\,q(x) = 0 exactly when x = c or q(x) = 0.

Worked examples

Example 1

A remainder without dividing

Find the remainder when 2x^5 - x^3 + 4 is divided by x + 2.

Solution

Write the divisor as x - (-2), so c = -2. By the remainder theorem the remainder is p(-2), where p(x) = 2x^5 - x^3 + 4:

\begin{aligned} p(-2) &= 2(-32) - (-8) + 4 \\ &= -64 + 8 + 4 = -52 \end{aligned}

The remainder is -52.

Check with synthetic division by -2. The missing powers x^4, x^2 and x need 0 as a placeholder, so the coefficients are 2, 0, -1, 0, 0, 4, and the bottom row is 2, -4, 7, -14, 28, -52. The last entry agrees, and the row also gives the quotient, so p(x) = (x + 2)\,q(x) - 52 with

q(x) = 2x^4 - 4x^3 + 7x^2 - 14x + 28

Evaluating took one line; dividing took six columns. When only the remainder is wanted, evaluate.

Example 2

Testing for a factor

Let p(x) = 2x^3 + 5x^2 - 11x + 4. Is (a) x - 1, (b) x + 2, (c) 2x - 1 a factor of p(x)?

Solution

(a) p(1) = 2 + 5 - 11 + 4 = 0. By the factor theorem, x - 1 is a factor.

(b) p(-2) = -16 + 20 + 22 + 4 = 30. This is not 0, so x + 2 is not a factor; dividing by it leaves remainder 30.

(c) 2x - 1 is not of the form x - c, but 2x - 1 = 2\left(x - \tfrac{1}{2}\right), and a nonzero constant multiple does not change whether something divides p(x). So test \displaystyle c = \frac{1}{2}:

\begin{aligned} p\left(\tfrac{1}{2}\right) &= 2 \cdot \tfrac{1}{8} + 5 \cdot \tfrac{1}{4} - \tfrac{11}{2} + 4 \\ &= \tfrac{3}{2} - \tfrac{11}{2} + 4 = 0 \end{aligned}

So 2x - 1 is a factor. In fact p(x) = (x - 1)(2x - 1)(x + 4).

Example 3

Factoring from one known zero

Given that x = -1 is a zero of p(x) = x^3 - 2x^2 - 13x - 10, factor p(x) completely and find all its zeros.

Solution

Since p(-1) = 0, the factor theorem says x + 1 is a factor. Synthetic division by -1 on 1, -2, -13, -10 finds the other factor: the bottom row is 1, -3, -10, 0. The remainder 0 confirms the given zero, and the rest is the quotient:

p(x) = (x + 1)(x^2 - 3x - 10)

The quadratic factors as in Factored form: the numbers with product -10 and sum -3 are -5 and 2. So

p(x) = (x + 1)(x - 5)(x + 2)

and the zeros are -2, -1 and 5. Check by expanding: (x + 1)(x^2 - 3x - 10) is x^3 - 3x^2 - 10x + x^2 - 3x - 10, which collects to p(x).

Each zero now shows all four views. For 5: p(5) = 0, x = 5 solves p(x) = 0, the graph crosses at (5, 0), and x - 5 is a factor.

Example 4

Two zeros given, complex zeros left over

The numbers 2 and -3 are zeros of p(x) = x^4 - 5x^3 - 2x^2 + 46x - 60. Find all four zeros.

Solution

Divide by x - 2: synthetic division by 2 on 1, -5, -2, 46, -60 gives 1, -3, -8, 30, 0, so p(x) = (x - 2)(x^3 - 3x^2 - 8x + 30). Call the cubic q(x).

The number -3 is a zero of q too, since 0 = p(-3) = (-3 - 2)\,q(-3) = -5\,q(-3) forces q(-3) = 0. Dividing q rather than p keeps the work short: synthetic division by -3 on 1, -3, -8, 30 gives 1, -6, 10, 0. So

p(x) = (x - 2)(x + 3)(x^2 - 6x + 10)

The quadratic has discriminant 36 - 40 = -4, so by the quadratic formula

x = \frac{6 \pm \sqrt{-4}}{2} = \frac{6 \pm 2i}{2} = 3 \pm i

The zeros are 2, -3, 3 + i and 3 - i. Only 2 and -3 are x-intercepts of the graph. The other two arrive as a conjugate pair, as they must for real coefficients.

Example 5

Choosing a coefficient

Find k so that x - 3 is a factor of p(x) = x^3 + kx^2 - 4x + 3.

Solution

By the factor theorem, x - 3 is a factor exactly when p(3) = 0. Here

p(3) = 27 + 9k - 12 + 3 = 18 + 9k

and 18 + 9k = 0 gives k = -2.

Check: with k = -2, synthetic division of 1, -2, -4, 3 by 3 gives 1, 1, -1, 0, a remainder of 0, and p(x) = (x - 3)(x^2 + x - 1).

Example 6

A factor for every power

Show that x - 1 is a factor of x^n - 1 for every whole number n \geq 1, and use this to show that 7^n - 1 is always a multiple of 6.

Solution

Let p(x) = x^n - 1. Then p(1) = 1^n - 1 = 0 whatever n is, so by the factor theorem x - 1 is a factor. The quotient can be named. Multiplying out

(x - 1)(x^{n-1} + x^{n-2} + \dots + x + 1)

gives x^n + x^{n-1} + \dots + x minus x^{n-1} + \dots + x + 1, and everything cancels except x^n - 1. For example, x^4 - 1 = (x - 1)(x^3 + x^2 + x + 1).

Now put x = 7:

7^n - 1 = (7 - 1)(7^{n-1} + \dots + 7 + 1)

The second bracket is a sum of whole numbers, so 7^n - 1 is 6 times a whole number. Check: 7^2 - 1 = 48 and 7^3 - 1 = 342, both multiples of 6. The same argument shows that a^n - 1 is a multiple of a - 1 for every whole number a \geq 2.

Boundaries

The divisor must be linear. The theorems are about x - c. For ax - b with a \neq 0, the same proof at \displaystyle x = \frac{b}{a} shows the remainder is \displaystyle p\left(\frac{b}{a}\right), and ax - b is a factor exactly when \displaystyle p\left(\frac{b}{a}\right) = 0, as in Example 2(c). For a divisor of degree 2 or more, such as x^2 + 1, the remainder can be a polynomial like 3x - 2 rather than one number, and no single evaluation finds it.

The theorems test a value; they do not find one. Factoring this way needs a first zero from somewhere: given, read off a graph, or guessed and then tested. The The rational root theorem narrows the guesses.

A zero can repeat. After dividing out x - c, the number c may still be a zero of the quotient, as with (x - 4)^2. What a repeated factor does to the graph is the subject of Zeros and multiplicity.

Complex zeros have no intercept. The factor theorem holds for complex c too, but only a real zero is an x-intercept (Example 4).

Where this goes next

Splitting off one zero at a time factors a polynomial into linear pieces once enough zeros are known. The The fundamental theorem of algebra guarantees that a polynomial of degree n has n complex zeros, counting repeats, and the factor theorem turns them into n linear factors; the same reasoning gives the limit of n real zeros used in Degree, end behavior and the shape of a polynomial graph. The The rational root theorem supplies candidates for the first zero, Graphing polynomial functions uses the factors to sketch, and in calculus a ratio of polynomials that gives \displaystyle \frac{0}{0} at c has the factor x - c on top and bottom, which is why Limits by algebra: factoring and rationalizing works.

Exercises

Remainders

Find the remainder by evaluating.

(x^4 - 3x^2 + 5x - 1) \div (x - 2)
(3x^3 + 2x^2 - 7) \div (x + 1)
(4x^3 - 6x^2 + x + 2) \div \left(x - \tfrac{1}{2}\right)
Use synthetic division to find p(4) for p(x) = x^4 - 5x^3 + 2x^2 + 7x - 3, then check by substituting.

Testing factors

Is x - 3 a factor of x^3 - 4x^2 + x + 6?
Is x + 2 a factor of 2x^3 + x^2 - 5x + 4?
Is 3x + 1 a factor of 3x^3 + 7x^2 - 4x - 2? If so, find the other factor.
Which of x - 1, x + 1, x - 2 and x + 2 are factors of x^4 + x^3 - 7x^2 - x + 6?

Factoring from a known zero

Factor completely and give every zero.

x^3 + 2x^2 - 5x - 6, given that 2 is a zero
2x^3 - 3x^2 - 11x + 6, given that 3 is a zero
x^3 - 6x^2 + 7x + 2, given that 2 is a zero
x^3 - 5x^2 + 23x - 51, given that 3 is a zero
x^4 - 2x^3 - 13x^2 + 14x + 24, given that -1 and 4 are zeros
x^3 - 3x - 2, given that -1 is a zero

Finding a coefficient

Find k so that x + 2 is a factor of x^3 + kx^2 + 5x + 6.
Find k so that dividing 2x^3 - x^2 + kx - 4 by x - 2 leaves remainder 10.
Find a and b so that x^3 + ax^2 + bx + 6 is divisible by x - 2 and leaves remainder 6 when divided by x + 1. Then factor it.
Find the cubic p(x) with zeros 1, -2 and 4 and with p(0) = 16.

In context

The bell chamber at the top of Beata’s bell tower is a box x meters wide, x + 1 meters deep and x + 4 meters tall, with a volume of 84 cubic meters, so x^3 + 5x^2 + 4x - 84 = 0. Corin guesses x = 3. Show that he is right, that no other real x works, and give the chamber’s dimensions.

Explaining

Show that x + 1 is a factor of x^n + 1 when n is odd, and is not when n is even.
Explain why x - 1 is a factor of every polynomial whose coefficients add up to 0. Use this to factor 5x^3 - 2x^2 - 7x + 4 into a linear and a quadratic factor.
Ferdinand divides p(x) = x^3 + 2x - 1 by x + 3 and says the remainder theorem gives a remainder of p(3) = 32. What went wrong, and what is the remainder?
Use the factor theorem to explain why a nonzero polynomial of degree n has at most n different zeros.

Answers

c = 2: 16 - 12 + 10 - 1 = 13. The remainder is 13.
c = -1: -3 + 2 - 7 = -8. The remainder is -8.
\displaystyle c = \frac{1}{2}: \displaystyle 4 \cdot \frac{1}{8} - 6 \cdot \frac{1}{4} + \frac{1}{2} + 2 = \frac{1}{2} - \frac{3}{2} + \frac{1}{2} + 2 = \frac{3}{2}. The remainder is \displaystyle \frac{3}{2}.
Synthetic division of 1, -5, 2, 7, -3 by 4 gives 1, -1, -2, -1, -7, so p(4) = -7. Check: 256 - 320 + 32 + 28 - 3 = -7.
Yes: p(3) = 27 - 36 + 3 + 6 = 0.
No: p(-2) = -16 + 4 + 10 + 4 = 2, so dividing leaves remainder 2.
Yes. \displaystyle 3x + 1 = 3\left(x + \frac{1}{3}\right), and \displaystyle p\left(-\frac{1}{3}\right) = -\frac{1}{9} + \frac{7}{9} + \frac{4}{3} - 2 = 0. Synthetic division by \displaystyle -\frac{1}{3} gives 3, 6, -6, 0, so \displaystyle p(x) = \left(x + \frac{1}{3}\right)(3x^2 + 6x - 6), which is (3x + 1)(x^2 + 2x - 2). The other factor is x^2 + 2x - 2.
p(1) = 0, p(-1) = 0, p(2) = 0 and p(-2) = -12. So x - 1, x + 1 and x - 2 are factors and x + 2 is not. (The fourth factor is x + 3.)
Division by 2 gives 1, 4, 3, 0: (x - 2)(x^2 + 4x + 3) = (x - 2)(x + 1)(x + 3). Zeros 2, -1, -3.
Division by 3 gives 2, 3, -2, 0: (x - 3)(2x^2 + 3x - 2) = (x - 3)(2x - 1)(x + 2). Zeros 3, \displaystyle \frac{1}{2}, -2.
Division by 2 gives 1, -4, -1, 0: (x - 2)(x^2 - 4x - 1). The quadratic formula gives \displaystyle x = 2 \pm \sqrt{5}, so \displaystyle p(x) = (x - 2)(x - 2 - \sqrt{5})(x - 2 + \sqrt{5}). Zeros 2, \displaystyle 2 + \sqrt{5}, \displaystyle 2 - \sqrt{5}.
Division by 3 gives 1, -2, 17, 0: (x - 3)(x^2 - 2x + 17). The discriminant is 4 - 68 = -64, so the quadratic has zeros \displaystyle \frac{2 \pm 8i}{2} = 1 \pm 4i. Zeros 3, 1 + 4i, 1 - 4i.
Division by -1 gives 1, -3, -10, 24, 0; then dividing x^3 - 3x^2 - 10x + 24 by x - 4 gives 1, 1, -6, 0. So p(x) = (x + 1)(x - 4)(x^2 + x - 6), which is (x + 1)(x - 4)(x - 2)(x + 3). Zeros -1, 4, 2, -3.
Division by -1 on 1, 0, -3, -2 gives 1, -1, -2, 0: (x + 1)(x^2 - x - 2) = (x + 1)^2(x - 2). Zeros -1 and 2; -1 is a repeated zero, the subject of Zeros and multiplicity.
p(-2) = -8 + 4k - 10 + 6 = 4k - 12, and 4k - 12 = 0 gives k = 3.
p(2) = 16 - 4 + 2k - 4 = 8 + 2k, and 8 + 2k = 10 gives k = 1.
p(2) = 8 + 4a + 2b + 6 = 0 gives 2a + b = -7, and p(-1) = -1 + a - b + 6 = 6 gives a - b = 1. Adding, 3a = -6, so a = -2 and b = -3. Then x^3 - 2x^2 - 3x + 6 = (x - 2)(x^2 - 3), which is \displaystyle (x - 2)(x - \sqrt{3})(x + \sqrt{3}).
By the factor theorem x - 1, x + 2 and x - 4 are factors; together they already make degree 3, so what is left is a number a: p(x) = a(x - 1)(x + 2)(x - 4). Then p(0) = a(-1)(2)(-4) = 8a = 16, so a = 2 and p(x) = 2(x - 1)(x + 2)(x - 4), which expands to 2x^3 - 6x^2 - 12x + 16.
27 + 45 + 12 - 84 = 0, so x = 3 works and x - 3 is a factor. Division gives 1, 8, 28, 0, so the equation is (x - 3)(x^2 + 8x + 28) = 0. The quadratic has discriminant 64 - 112 = -48, which is negative, so it has no real zeros. The chamber is 3 by 4 by 7 meters.
p(-1) = (-1)^n + 1. For odd n, (-1)^n = -1, so p(-1) = 0 and x + 1 is a factor. For even n, (-1)^n = 1, so p(-1) = 2: not a factor, and the remainder is 2.
Every power of 1 is 1, so p(1) is the sum of the coefficients. If that sum is 0, then p(1) = 0 and x - 1 is a factor. For 5x^3 - 2x^2 - 7x + 4, 5 - 2 - 7 + 4 = 0, and division by 1 gives 5, 3, -4, 0: the polynomial is (x - 1)(5x^2 + 3x - 4).
The divisor x + 3 is x - (-3), so c = -3, not 3. The remainder is p(-3) = -27 - 6 - 1 = -34.
Suppose c_1 is a zero. By the factor theorem p(x) = (x - c_1)\,q_1(x) with q_1 of degree n - 1. Any other zero c gives 0 = p(c) = (c - c_1)\,q_1(c) with c - c_1 \neq 0, so c is a zero of q_1. Each new zero splits off another factor and lowers the degree of what is left by one. After n different zeros, p(x) = a(x - c_1)(x - c_2)\cdots(x - c_n) with a a nonzero number, and this product is 0 only at c_1, \dots, c_n. So there is no room for another zero.