Math Vale

⚔️ Algebra Kingdom · Algebra

Polynomial Functions

Divide one polynomial by another, use the remainder and factor theorems to find zeros, and read the end behaviour and shape of a polynomial graph from its degree and leading coefficient.

In short

  • Dividing p(x) by (x - k) gives p(x) = (x - k)·quotient + remainder, and the quotient is always one degree lower than p(x).
  • Synthetic division needs the coefficients of every power in order, including a 0 for each power that is missing, and the k that makes the divisor zero — so (x + 3) means k = -3.
  • The remainder theorem: the remainder on dividing by (x - k) is p(k), so a substitution replaces the whole division.
  • The factor theorem: (x - k) is a factor of p(x) exactly when p(k) = 0, which is how a cubic is broken open by hand.
  • The multiplicity of a zero is the exponent on its bracket: odd multiplicity means the graph crosses the axis there, even multiplicity means it touches and turns back.
  • End behaviour comes from the degree and the sign of the leading coefficient alone — even degree sends both arms the same way, a positive leading coefficient lifts the right-hand arm.
  • A polynomial of degree n has at most n - 1 turning points, and its distinct zeros are counted by brackets, not by exponents.

Cutting a plot along a line

Surveyor Idris fences land for a living, and on Polynomial Plains a polynomial is a plot: p(x) = 2x3 - 5x2 - 4x + 3, its degree the highest power, its leading coefficient the number in front of that power. Dividing p(x) by a line such as x - 3 asks how many whole strips of that width fit in the plot, and what narrow piece is left over.

Long division works exactly as it does with numbers. You match the leading term, multiply the whole divisor by what you wrote, subtract, and bring the next term down.

2x2 + x - 1 x - 3 ) 2x3 - 5x2 - 4x + 3 2x3 - 6x2 x2 - 4x x2 - 3x -x + 3 -x + 3 0

Synthetic division is the same arithmetic with the letters stripped out, and it works whenever the divisor is x - k. Write k outside, the coefficients of p(x) along the top, bring the first one down, then multiply-and-add all the way across.

k = 3 | 2 -5 -4 3 | 6 3 -3 ----------------------- 2 1 -1 | 0

The bottom row reads 2, 1, -1 with a 0 at the end: the quotient is 2x2 + x - 1 — one degree lower than p(x) — and the remainder is 0.

Two details cause almost every wrong answer here.

  • The k is the number that makes the divisor zero. Dividing by (x - 3) means k = 3; dividing by (x + 3) means k = -3, because x + 3 is x - (-3).
  • Every missing power needs a 0. To divide x3 - 7x + 6 the top row is 1, 0, -7, 6. Leaving that 0 out shifts every later coefficient into the wrong column and the whole answer comes out wrong.

The result is always written p(x) = (divisor)(quotient) + remainder, so the quotient and the remainder are two different things and never share one box.

The remainder theorem and the factor theorem

Once the divisor is x - k there is a shortcut so short it feels like cheating.

The remainder theorem. The remainder when p(x) is divided by x - k is exactly p(k). No division needed — substitute and stop.

Why? Write the division out: p(x) = (x - k)·q(x) + r. Put x = k into both sides. The bracket (k - k) is 0, so the whole first term vanishes and p(k) = r.

For p(x) = 2x3 - 5x2 - 4x + 3 divided by x - 3:

p(3) = 2(27) - 5(9) - 4(3) + 3 = 54 - 45 - 12 + 3 = 0

which agrees with the 0 the synthetic division produced.

The factor theorem is the same statement read the other way. If p(k) = 0 then the remainder is 0, so the division came out exactly, so x - k is a factor of p(x) — and if x - k is a factor then p(k) must be 0. One test, both directions:

x - k is a factor <=> p(k) = 0

So "is x + 2 a factor of p(x)?" is really "is p(-2) equal to 0?". Read the k out of the bracket with its sign flipped, substitute, and look at the answer. A remainder of anything other than 0 settles it: not a factor.

This is what makes a cubic solvable by hand. Test small values — the ones that divide the constant term are the sensible candidates — until one gives 0. That value is a zero, its bracket is a factor, and dividing it out leaves a quadratic you already know how to handle.

Zeros, and how many times each one counts

A zero of p(x) is an x-value with p(x) = 0, and on the graph it is a point where the curve meets the x-axis. Fully factored, a polynomial hands you every zero at once:

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

Set each bracket to 0. Solve, do not copy: (x - 1) gives x = 1, and (x + 3) gives x = -3, the opposite of the number you can see.

The exponent on a bracket is that zero's multiplicity — how many copies of the factor the product holds. Above, x = 1 has multiplicity 2 and x = -3 has multiplicity 1. Multiplicity decides what the curve *does* at that point.

  • Odd multiplicity: the graph crosses. It goes through the axis and comes out the other side. Multiplicity 3 crosses too, but flattens as it goes.
  • Even multiplicity: the graph touches and turns back. It comes down to the axis, kisses it, and returns the way it came.

Count carefully, because there are two different counts. p(x) = 2(x - 1)2(x + 3) has two distinct zeros — two places on the x-axis — but degree 3, because the exponents add to 3. Counted with multiplicity there are three zeros; counted as places on the axis there are two.

One trap worth naming: a polynomial has to be fully factored before any multiplicity can be read.

p(x) = (x - 2)(x2 - 4) = (x - 2)(x - 2)(x + 2) = (x - 2)2(x + 2)

The exponent you could see was 1, but the quadratic was hiding another (x - 2), so the multiplicity of x = 2 is really 2 and the graph touches there rather than crossing.

Where the arms go, and how often the curve turns

Far from the origin the leading term drowns out every other one — at x = 100 the term 3x4 is larger than anything below it by a mile. So the end behaviour, what the two arms of the graph do, depends only on the degree and the sign of the leading coefficient.

  • Even degree: the arms go the same way, like a parabola.
  • Odd degree: the arms go opposite ways, like a cubic.
  • Positive leading coefficient: the right-hand arm rises.
  • Negative leading coefficient: the right-hand arm falls.

Put the two together and there are exactly four possibilities.

  • Odd degree, positive: falls to the left and rises to the right.
  • Odd degree, negative: rises to the left and falls to the right.
  • Even degree, positive: rises to the left and rises to the right.
  • Even degree, negative: falls to the left and falls to the right.

A factored form does not need multiplying out to be read this way: the degree is the sum of the exponents on the brackets, and the number in front is the leading coefficient. So -2(x - 1)2(x + 3) has degree 3 and leading coefficient -2, and therefore rises to the left and falls to the right.

Between the ends the curve wanders, and a turning point is a place where it changes from going up to going down or back again. A polynomial of degree n has at most n - 1 turning points: a parabola turns once, a cubic at most twice, a quartic at most three times. It is a ceiling and not a promise — y = x3 has degree 3 and never turns at all.

That is enough to sketch a polynomial without plotting a single table of values: mark the zeros, decide at each one whether the curve crosses or bounces, use the y-intercept p(0) for height, and let the end behaviour pull the two arms into place.

How to type your answer

Every question here asks for exactly one kind of thing, and says which.

  • A quotient is an expression in x. Type it the way you write it: 2x2 + x - 1, with powers as ^. When the prompt says to ignore the remainder, do not fold it in — 2x2 + x - 1 + 5 is a different polynomial, and the remainder belongs over the divisor as its own fraction.
  • A remainder is a whole number, and so is the p(k) of the remainder theorem. One number, nothing else.
  • "Which of these is a factor?" is multiple choice. Pick the option; there is nothing to type.
  • Two zeros are a pair, written (-3, 1) with the smaller one first. Negative zeros are common here, so check the order on a number line rather than by eye.
  • A multiplicity is a whole number, usually 1, 2 or 3 — how many times the bracket appears, not the zero itself and not the degree.
  • End behaviour is multiple choice, written in words: "falls to the left and rises to the right".

Idris has one last piece of advice, from a lifetime of pegging boundaries. Check the division by multiplying back: (divisor)(quotient) + remainder should rebuild p(x) exactly. A fence that does not close was measured wrong somewhere, and it is quicker to find out now than at the far corner.

Worked examples

Example 1

Divide p(x) = x3 - 7x + 6 by (x - 2), giving the quotient and the remainder.

  1. Write the coefficients in order, filling in every missing power with a 0. There is no x2 term, so the row is 1, 0, -7, 6.
  2. The divisor is x - 2, so k = 2. Bring the 1 down.
  3. 1 × 2 = 2, and 0 + 2 = 2. Write 2.
  4. 2 × 2 = 4, and -7 + 4 = -3. Write -3.
  5. -3 × 2 = -6, and 6 + (-6) = 0. Write 0.
  6. The bottom row is 1, 2, -3 with a final 0, so the quotient is x2 + 2x - 3 and the remainder is 0. The division was exact, which also says (x - 2) is a factor.

Example 2

Use the remainder theorem to find the remainder when p(x) = 2x3 + x2 - 13x + 6 is divided by (x + 3).

  1. Read k off the divisor. x + 3 is x - (-3), so k = -3 — not 3.
  2. The remainder theorem says the remainder is p(-3), so substitute: p(-3) = 2(-3)3 + (-3)2 - 13(-3) + 6.
  3. Work the powers first, minding the signs: (-3)3 = -27 and (-3)2 = 9.
  4. p(-3) = 2(-27) + 9 + 39 + 6 = -54 + 9 + 39 + 6 = 0.
  5. The remainder is 0, so by the factor theorem (x + 3) is a factor of p(x).

Example 3

One zero of p(x) = x3 - 2x2 - 5x + 6 is x = 1. Find the other two zeros.

  1. Since x = 1 is a zero, (x - 1) is a factor, so divide it out. Synthetic division with k = 1 on the row 1, -2, -5, 6.
  2. Bring down 1; 1 × 1 = 1 and -2 + 1 = -1; -1 × 1 = -1 and -5 + (-1) = -6; -6 × 1 = -6 and 6 + (-6) = 0.
  3. The remainder is 0 (as it had to be) and the quotient is x2 - x - 6.
  4. Factor the quotient: two numbers with product -6 and sum -1 are -3 and 2, so x2 - x - 6 = (x - 3)(x + 2).
  5. Solve each bracket: x - 3 = 0 gives x = 3, and x + 2 = 0 gives x = -2.
  6. The other two zeros, smaller first, are (-2, 3).

Example 4

For p(x) = -2(x - 4)2(x + 1), state the multiplicity of each zero, say what the graph does at each, and describe the end behaviour.

  1. The brackets give the zeros: (x - 4) is 0 at x = 4, and (x + 1) is 0 at x = -1. Two distinct zeros.
  2. The exponents give the multiplicities: x = 4 has multiplicity 2, and x = -1 has multiplicity 1 (an exponent that is not written is 1).
  3. Multiplicity 2 is even, so at x = 4 the graph touches the x-axis and turns back. Multiplicity 1 is odd, so at x = -1 it crosses.
  4. Degree: the exponents add to 2 + 1 = 3, which is odd, so the two arms go opposite ways. There is no need to multiply the brackets out.
  5. Leading coefficient: the -2 out in front is negative, so the right-hand arm falls.
  6. The graph rises to the left and falls to the right, with at most 2 turning points since the degree is 3.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Surveyor Idris cuts the plot p(x) = x2 - 4 along the fence line (x - 2), and it divides exactly with nothing left over. What is the quotient? Write your answer as an expression in x, such as x2 + 2x - 5.

Answer: x + 2

  1. Coefficients of p(x), missing powers filled with 0: 1, 0, -4
  2. Dividing by (x - 2) means k = 2.
  3. Bring the leading coefficient down: 1.
  4. 1 × 2 = 2, and 0 + 2 = 2.
  5. 2 × 2 = 4, and -4 + 4 = 0.
  6. The bottom row ends in 0, so the division is exact.
  7. Quotient: x + 2

Problem 2

Difficulty 3 of 5

p(x) = x3 + 2x2 - 85 Divide p(x) by (x - 4). What is the remainder?

Answer: 11

  1. Coefficients, with a 0 for every missing power: 1, 2, 0, -85
  2. Dividing by (x - 4) means k = 4.
  3. Bring the leading coefficient down: 1.
  4. 1 × 4 = 4, and 2 + 4 = 6.
  5. 6 × 4 = 24, and 0 + 24 = 24.
  6. 24 × 4 = 96, and -85 + 96 = 11.
  7. The last number of the bottom row is the remainder: 11.
  8. Check with the remainder theorem: p(4) = 11.

Problem 3

Difficulty 4 of 5

p(x) = 2x3 + 8x2 - 14x - 20 Which of these binomials is a factor of p(x)?

  1. x + 5
  2. x - 3
  3. x - 5
  4. x + 4

Answer: A. x + 5

  1. Factor theorem: (x + 5) is a factor exactly when p(-5) = 0.
  2. p(-5) = 0, so (x + 5) divides p(x) exactly.
  3. p(5) = 360 and p(-4) = 36, neither of which is 0.
  4. The factor is x + 5.

Common mistakes

  • Using the wrong sign for k: dividing by (x + 3) and putting 3 outside the synthetic division box instead of -3, so every number in the bottom row is wrong.
  • Forgetting the 0 placeholder for a missing power, so x3 - 7x + 6 goes in as 1, -7, 6 and every coefficient after that lands in the wrong column.
  • Tacking the remainder onto the quotient: writing the answer to "divide and give the quotient" as x2 + 2x + 5 when the 5 was the remainder and belongs over the divisor.
  • Evaluating p(-k) instead of p(k) for the remainder theorem, which is the same sign slip wearing a different hat.
  • Copying zeros straight out of the brackets: reading (x + 3)(x - 5) as zeros 3 and -5 rather than -3 and 5. Each bracket has to be solved, not read.
  • Swapping cross and touch: saying the graph passes through the axis at a zero of multiplicity 2, when an even multiplicity means it comes down, touches and turns back.
  • Reading the end behaviour of -x4 as if it were x4, so both arms are drawn rising when a negative leading coefficient sends them both down.
  • Confusing the counts: giving the degree when asked for the number of distinct zeros, or giving the exponent on the wrong bracket when asked for a multiplicity.
  • Reading a multiplicity off a form that is not fully factored: (x - 2)(x2 - 4) looks like multiplicity 1 at x = 2, but the quadratic hides a second (x - 2).

What you should be able to do

  • Divide a polynomial by a linear factor using long or synthetic division, and write the quotient and remainder.
  • Use the remainder theorem to evaluate p(k) and the factor theorem to decide whether x - k is a factor.
  • Find all the zeros of a factorable polynomial, state each multiplicity, and say whether the graph crosses or touches at each.
  • Describe the end behaviour of a polynomial from its degree and leading coefficient, and match a polynomial to its graph.

Where this fits in the curriculum

Common Core

  • HSA-APR.B.2

    High school — Know and apply the Remainder Theorem: for a polynomial p(x) and a number a, the remainder on division by x − a is p(a), so p(a) = 0 if and only if (x − a) is a factor of p(x).

  • HSA-APR.B.3

    High school — Identify zeros of polynomials when suitable factorisations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.

  • HSA-APR.D.6

    High school — Rewrite simple rational expressions in different forms: write a(x)/b(x) in the form q(x) + r(x)/b(x), where the degree of r(x) is less than the degree of b(x), using inspection or long division.

  • HSF-IF.C.7.C

    High school — Graph polynomial functions, identifying zeros when suitable factorisations are available, and showing end behaviour.

SAT

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