Polynomial Functions

Study Sheet

Polynomial Functions

Vocabulary, end behavior, division, zeros, and graphs

Polynomial Vocabulary

Concept
The language of polynomials

A polynomial in one variable is a sum of terms of the form anxna_n x^n, where each exponent is a whole number.

f(x)=anxn+an−1xn−1+⋯+a1x+a0f(x)=a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0
  • Degree: the greatest exponent on the variable.
  • Leading coefficient: the coefficient ana_n of the highest-degree term.
  • Standard form: terms written from highest degree to lowest.
  • Constant term: a0a_0, the term with no variable.

Classify by degree: 00 constant, 11 linear, 22 quadratic, 33 cubic, 44 quartic, 55 quintic.

A degree-nn polynomial has at most nn real roots and its ends point the way its leading term points. The remainder theorem (P(a)P(a) is the remainder on dividing by x−ax - a) and the factor theorem (P(a)=0P(a) = 0 iff x−ax - a is a factor) connect roots to factors; the rational root theorem lists the only possible rational roots so you can test them.

Degree is the most informative single number about a polynomial: it caps the number of zeros (at most nn real ones), caps the number of turns (at most n−1n - 1), and, with the sign of the leading coefficient, fixes the end behavior. That is why the first step with any polynomial is to put it in standard form and read off those two numbers.

Example
Reading off the parts

Consider g(x)=7−2x3+4xg(x)=7-2x^3+4x. Write in standard form: g(x)=−2x3+4x+7g(x)=-2x^3+4x+7. Degree =3=3 (cubic), leading coefficient =−2=-2, constant term =7=7.

Side note
Where it goes wrong

Concluding a polynomial has exactly nn roots because it has degree nn. It has AT MOST nn real roots; counting complex roots with multiplicity gives exactly nn.

Tip

Tip: A polynomial must have only whole-number exponents. Expressions like x−1x^{-1}, x=x1/2\sqrt{x}=x^{1/2}, or 1x\dfrac{1}{x} are not polynomial terms.

Try it
Try it: classify

For h(x)=3x2−x4+6−2xh(x) = 3x^2 - x^4 + 6 - 2x: write it in standard form and give the degree, leading coefficient, constant term and number of terms.

Show the answer

Standard form: −x4+3x2−2x+6-x^4 + 3x^2 - 2x + 6. Degree 44 (quartic), leading coefficient −1-1, constant term 66, four terms.

End Behavior

Concept
What the “arms” do

The end behavior of a graph describes what happens to f(x)f(x) as x→+∞x\to+\infty and x→−∞x\to-\infty. It depends only on the degree parity and the sign of the leading coefficient.

degreeleading coeff.end behavioreven+up-left, up-righteven−down-left, down-rightodd+down-left, up-rightodd−up-left, down-right\begin{array}{c|c|c} \text{degree} & \text{leading coeff.} & \text{end behavior}\\\hline \text{even} & + & \text{up-left, up-right}\\ \text{even} & - & \text{down-left, down-right}\\ \text{odd} & + & \text{down-left, up-right}\\ \text{odd} & - & \text{up-left, down-right} \end{array}

For large ∣x∣|x| the leading term outgrows everything else: in −2x3+5x−1-2x^3 + 5x - 1 at x=10x = 10, the cube contributes −2000-2000 while the rest adds 4949. So the ends of the graph copy the ends of −2x3-2x^3: an odd power keeps the sign of xx, so the ends go opposite ways, and the negative coefficient flips that to up-left, down-right.

The four cases of end behavior.

Example
Predicting the ends
xy−3−2−112−8−6−4−2246
f(x)=−2x3+5x−1f(x) = -2x^3 + 5x - 1: up on the left, down on the right.

For f(x)=−2x3+5x−1f(x)=-2x^3+5x-1: degree 33 is odd and the leading coefficient −2-2 is negative, so the graph goes up on the left and down on the right. In symbols: as x→−∞, f(x)→+∞x\to-\infty,\ f(x)\to+\infty, and as x→+∞, f(x)→−∞x\to+\infty,\ f(x)\to-\infty.

Side note
Where it goes wrong

Judging end behavior from the first term as written: f(x)=5x−2x3−1f(x) = 5x - 2x^3 - 1 "starts with 5x5x, so it rises to the right." Only the highest-degree term controls the ends; here it is −2x3-2x^3, so the graph falls to the right.

Tip

Remember: Only the leading term controls end behavior. Every other term becomes negligible for very large ∣x∣|x|.

Try it
Try it: predict the ends

Describe the end behavior of g(x)=3x4−100x3g(x) = 3x^4 - 100x^3 and h(x)=7−x5h(x) = 7 - x^5.

Show the answer

gg: degree 44 (even), leading coefficient 3>03 > 0: both ends up. (The −100x3-100x^3 matters near the middle but loses once xx is large.) hh: leading term −x5-x^5, odd degree, negative: up on the left, down on the right.

Adding, Subtracting, and Multiplying Polynomials

Concept
Combining polynomials

Add / subtract: combine like terms. When subtracting, distribute the −- sign to every term first. Multiply: distribute each term of one factor across the other, then combine like terms. The degrees add: (deg⁡m)⋅(deg⁡n)(\deg m)\cdot(\deg n) product has degree m+nm+n.

Polynomials combine like multi-digit numbers written in base xx: like terms play the role of place values. That is why adding lines up powers, and why a product's degree is the sum of the degrees: the highest-degree terms multiply to xm⋅xn=xm+nx^m \cdot x^n = x^{m + n} and nothing else can reach that power.

Example
Three operations
x²−2x+5x+3x³−2x²5x3x²−6x15
Six products; collect −2x2+3x2-2x^2 + 3x^2 and 5x−6x5x - 6x.

Sum: (3x2−2x+1)+(x2+5x−4)=4x2+3x−3(3x^2-2x+1)+(x^2+5x-4)=4x^2+3x-3. Difference: (5x3−2x+4)−(2x3+x−1)=3x3−3x+5(5x^3-2x+4)-(2x^3+x-1)=3x^3-3x+5. Product: (x+3)(x2−2x+5)(x+3)(x^2-2x+5)

=x3−2x2+5x+3x2−6x+15=x3+x2−x+15.= x^3-2x^2+5x+3x^2-6x+15 = x^3+x^2-x+15.
Side note
Where it goes wrong

Subtracting only the first term: (5x3−2x+4)−(2x3+x−1)=3x3−2x+4+x−1(5x^3 - 2x + 4) - (2x^3 + x - 1) = 3x^3 - 2x + 4 + x - 1. Every term of the second polynomial changes sign: 3x3−3x+53x^3 - 3x + 5.

Tip

Tip: Line up like terms in columns for addition and subtraction, just like whole numbers. It prevents dropped terms.

Try it
Try it: multiply

Multiply (2x−1)(x2+4x−3)(2x - 1)(x^2 + 4x - 3) and state the degree.

Show the answer

2x3+8x2−6x−x2−4x+3=2x3+7x2−10x+32x^3 + 8x^2 - 6x - x^2 - 4x + 3 = 2x^3 + 7x^2 - 10x + 3. Degree 1+2=31 + 2 = 3.

Long Division and Synthetic Division

Concept
Two ways to divide

Long division works for any divisor. Synthetic division is a fast shortcut that works only when dividing by a linear factor x−cx-c. The Division Algorithm says

f(x)=(divisor)(quotient)+remainder.f(x) = (\text{divisor})(\text{quotient}) + \text{remainder}.

Polynomial long division is the same algorithm as number long division: at each step, match the leading term, subtract, and bring down, until what is left has lower degree than the divisor. Synthetic division is that process for a divisor x−cx - c with the variables deleted and the subtraction turned into addition (multiplying by cc instead of −c-c), which is why it only works for linear divisors.

Example
Long division: (2x3+3x2−x+5)÷(x+2)(2x^3+3x^2-x+5)\div(x+2)
2x2− x+1x+2 ) 2x3+3x2− x+5‾\begin{array}{r} 2x^2-\ x+1\\[-1pt] x+2\,\overline{)\,2x^3+3x^2-\ x+5}\\ \end{array}

Step by step the quotient is 2x2−x+12x^2-x+1 with remainder 33, so

2x3+3x2−x+5x+2=2x2−x+1+3x+2.\frac{2x^3+3x^2-x+5}{x+2}=2x^2-x+1+\frac{3}{x+2}.
Side note
Where it goes wrong

Using c=−1c = -1 for division by x−1x - 1 (or c=2c = 2 for x+2x + 2). Synthetic division uses the zero of the divisor: x−1=0x - 1 = 0 at c=1c = 1, and x+2=0x + 2 = 0 at c=−2c = -2.

Example
Synthetic division: (x3−4x2+5x−2)÷(x−1)(x^3-4x^2+5x-2)\div(x-1)

Use c=1c=1 with the coefficients 1, −4, 5, −21,\ -4,\ 5,\ -2:

11−45−21−321−320\begin{array}{c|rrrr} 1 & 1 & -4 & 5 & -2\\ & & 1 & -3 & 2\\\hline & 1 & -3 & 2 & 0 \end{array}

The bottom row gives quotient x2−3x+2x^2-3x+2 and remainder 00. Since the remainder is 00, (x−1)(x-1) is a factor:

x3−4x2+5x−2=(x−1)(x2−3x+2)=(x−1)2(x−2).x^3-4x^2+5x-2=(x-1)(x^2-3x+2)=(x-1)^2(x-2).
Tip

Watch out: Include a 00 placeholder for every missing power before you divide. For x3−7x^3-7 use coefficients 1, 0, 0, −71,\ 0,\ 0,\ -7.

Try it
Try it: synthetic division

Divide x3−7x^3 - 7 by x−2x - 2 synthetically.

Show the answer

Coefficients with placeholders: 1,0,0,−71, 0, 0, -7; c=2c = 2. Bring down 11; 1⋅2=21 \cdot 2 = 2, 0+2=20 + 2 = 2; 2⋅2=42 \cdot 2 = 4, 0+4=40 + 4 = 4; 4⋅2=84 \cdot 2 = 8, −7+8=1-7 + 8 = 1. Quotient x2+2x+4x^2 + 2x + 4, remainder 11: x3−7x−2=x2+2x+4+1x−2\tfrac{x^3 - 7}{x - 2} = x^2 + 2x + 4 + \tfrac{1}{x - 2}.

The Remainder and Factor Theorems

Concept
Evaluating by dividing

Remainder Theorem: When f(x)f(x) is divided by x−cx-c, the remainder equals f(c)f(c). Factor Theorem: (x−c)(x-c) is a factor of f(x)f(x) if and only if f(c)=0f(c)=0 (that is, cc is a zero).

Reminder — Remainder and Factor theorems:  P(x)=(x−c)Q(x)+P(c)⟹(x−c)∣P(x)  ⟺  P(c)=0P(x)=(x-c)Q(x)+P(c) \quad\Longrightarrow\quad (x-c)\mid P(x) \iff P(c)=0

The division algorithm says f(x)=(x−c)q(x)+rf(x) = (x - c)q(x) + r for a constant rr. Put x=cx = c: the first term vanishes and f(c)=rf(c) = r. That is the Remainder Theorem, and the Factor Theorem is its special case r=0r = 0: (x−c)(x - c) divides ff exactly when f(c)=0f(c) = 0, which is when the graph meets the xx-axis at cc.

Example
Testing a factor
xy−0.50.511.522.5−2−11
f(x)=x3−4x2+5x−2=(x−1)2(x−2)f(x) = x^3 - 4x^2 + 5x - 2 = (x - 1)^2(x - 2) touches at 11 and crosses at 22.

Is (x−2)(x-2) a factor of f(x)=x3−4x2+5x−2f(x)=x^3-4x^2+5x-2? Evaluate:

f(2)=8−16+10−2=0.f(2)=8-16+10-2=0.

Because f(2)=0f(2)=0, yes—(x−2)(x-2) is a factor, and x=2x=2 is a zero.

Side note
Where it goes wrong

Evaluating f(−c)f(-c) to test x−cx - c: testing (x−2)(x - 2) with f(−2)f(-2). The factor x−2x - 2 corresponds to the zero x=2x = 2, so compute f(2)f(2).

Tip

Tip: Synthetic division and the Remainder Theorem agree: the last number in a synthetic-division row is f(c)f(c).

Try it
Try it: remainder

Find the remainder when f(x)=2x4−3x2+x−5f(x) = 2x^4 - 3x^2 + x - 5 is divided by x+1x + 1. Is x+1x + 1 a factor?

Show the answer

By the Remainder Theorem the remainder is f(−1)=2−3−1−5=−7f(-1) = 2 - 3 - 1 - 5 = -7. It is not zero, so x+1x + 1 is not a factor.

Factoring Higher-Degree Polynomials

Concept
A factoring toolkit
  • GCF first: always pull out the greatest common factor.
  • Grouping: for four terms, group in pairs and factor each.
  • Sum of cubes: a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2).
  • Difference of cubes: a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2).
  • Quadratic form: treat x4−5x2+4x^4-5x^2+4 like u2−5u+4u^2-5u+4 with u=x2u=x^2.

Every factoring pattern is a multiplication rule read backward, and each factor found is a zero found (Factor Theorem). The difference of cubes works because (a−b)(a2+ab+b2)(a - b)(a^2 + ab + b^2) expands with every middle term cancelling in pairs; the quadratic-form trick works because x4−5x2+4x^4 - 5x^2 + 4 is a quadratic in the quantity x2x^2.

Example
Four factoring moves

GCF: 6x4−9x3+3x2=3x2(2x2−3x+1)=3x2(2x−1)(x−1)6x^4-9x^3+3x^2=3x^2(2x^2-3x+1)=3x^2(2x-1)(x-1). Grouping: x3+2x2−9x−18=x2(x+2)−9(x+2)=(x+2)(x2−9)=(x+2)(x−3)(x+3)x^3+2x^2-9x-18=x^2(x+2)-9(x+2)=(x+2)(x^2-9)=(x+2)(x-3)(x+3). Difference of cubes: 8x3−1=(2x−1)(4x2+2x+1)8x^3-1=(2x-1)(4x^2+2x+1). Quadratic form: x4−5x2+4=(x2−1)(x2−4)=(x−1)(x+1)(x−2)(x+2)x^4-5x^2+4=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2).

Side note
Where it goes wrong

Factoring a3−b3a^3 - b^3 as (a−b)3(a - b)^3 or (a−b)(a2+2ab+b2)(a - b)(a^2 + 2ab + b^2). The correct middle term is +ab+ab with no 22: 8x3−1=(2x−1)(4x2+2x+1)8x^3 - 1 = (2x - 1)(4x^2 + 2x + 1). Expanding checks it.

Tip

Remember: “Factor completely” means keep going until every factor is prime over the integers (or as far as the problem allows).

Try it
Try it: factor completely

Factor x4−16x^4 - 16 and 27x3+827x^3 + 8 completely over the integers.

Show the answer

x4−16=(x2−4)(x2+4)=(x−2)(x+2)(x2+4)x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4). Sum of cubes with a=3xa = 3x, b=2b = 2: (3x+2)(9x2−6x+4)(3x + 2)(9x^2 - 6x + 4).

Finding All Zeros

Concept
Rational Root Theorem, multiplicity, and depressing

Rational Root Theorem: any rational zero of a polynomial with integer coefficients has the form pq\dfrac{p}{q}, where pp divides the constant term and qq divides the leading coefficient. Strategy: test candidates until one works, use synthetic division to depress the polynomial to a lower degree, then repeat or factor the result. Multiplicity: the number of times a factor repeats. A zero of multiplicity 22 makes the graph touch (bounce off) the xx-axis; odd multiplicity makes it cross.

Why must a rational zero pq\tfrac pq (in lowest terms) have pp dividing the constant and qq dividing the leading coefficient? Substitute it and clear denominators: anpn+⋯+a0qn=0a_n p^n + \cdots + a_0 q^n = 0. Every term but the last is a multiple of pp, so pp divides a0qna_0q^n, and since pp shares no factor with qq, pp divides a0a_0. The same argument at the other end gives q∣anq \mid a_n.

Example
Find all zeros of f(x)=x3−2x2−5x+6f(x)=x^3-2x^2-5x+6
xy−3−2−1123−8−6−4−22468(0, 6)
Zeros −2-2, 11, 33, all among the candidates ±1,±2,±3,±6\pm1, \pm2, \pm3, \pm6.

Candidates: ±1,±2,±3,±6\pm1,\pm2,\pm3,\pm6. Test x=1x=1: 1−2−5+6=01-2-5+6=0, so (x−1)(x-1) is a factor. Depress:

11−2−561−1−61−1−60\begin{array}{c|rrrr} 1 & 1 & -2 & -5 & 6\\ & & 1 & -1 & -6\\\hline & 1 & -1 & -6 & 0 \end{array}

The quotient is x2−x−6=(x−3)(x+2)x^2-x-6=(x-3)(x+2). Therefore

f(x)=(x−1)(x−3)(x+2),zeros: x=1, 3, −2.f(x)=(x-1)(x-3)(x+2),\qquad \text{zeros: } x=1,\ 3,\ -2.
Side note
Where it goes wrong

Continuing to test candidates after reaching a quadratic. Once synthetic division leaves x2−x−6x^2 - x - 6, factor it (or use the formula); testing more candidates is slower and misses irrational or complex zeros.

Example
A case with complex zeros: f(x)=x3−2x2+x−2f(x)=x^3-2x^2+x-2

Test x=2x=2: 8−8+2−2=08-8+2-2=0. Depress with c=2c=2:

21−21−22021010\begin{array}{c|rrrr} 2 & 1 & -2 & 1 & -2\\ & & 2 & 0 & 2\\\hline & 1 & 0 & 1 & 0 \end{array}

The quotient is x2+1x^2+1, which has zeros x=±ix=\pm i. So the zeros are x=2, i, −ix=2,\ i,\ -i.

Tip

Tip: Once you depress down to a quadratic, stop dividing—finish with factoring or the quadratic formula.

Reminder — The quadratic formula:  x=−b±b2−4ac2a(a≠0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)
Try it
Try it: all zeros

Find all zeros of f(x)=2x3−3x2−11x+6f(x) = 2x^3 - 3x^2 - 11x + 6.

Show the answer

Candidates ±1,±2,±3,±6,±12,±32\pm1, \pm2, \pm3, \pm6, \pm\tfrac12, \pm\tfrac32. f(3)=54−27−33+6=0f(3) = 54 - 27 - 33 + 6 = 0. Divide by x−3x - 3: 2x2+3x−2=(2x−1)(x+2)2x^2 + 3x - 2 = (2x - 1)(x + 2). Zeros: 33, 12\tfrac12, −2-2.

Fundamental Theorem of Algebra

Concept
Counting zeros

Fundamental Theorem of Algebra: every polynomial of degree n≥1n\ge1 has exactly nn zeros in the complex numbers, counted with multiplicity. Complex Conjugate Theorem: if the coefficients are real and a+bia+bi is a zero, then its conjugate a−bia-bi is also a zero. Complex zeros always come in pairs, so a polynomial of odd degree with real coefficients has at least one real zero.

Over the complex numbers, every polynomial of degree n≥1n \ge 1 has at least one zero (the deep part of the theorem, proved in later courses). Factor it out with the Factor Theorem and the degree drops by 11; repeat, and you get exactly nn zeros. Real coefficients force complex zeros into conjugate pairs, because conjugating the equation f(z)=0f(z) = 0 gives f(zˉ)=0f(\bar z) = 0.

Example
Writing a polynomial from its zeros
xy−2−1123−16−12−8−44
(x−3)(x2+4)(x - 3)(x^2 + 4): one real crossing at 33; the other two zeros, ±2i\pm 2i, do not appear on the graph.

(a) Zeros 2, −3, 52,\ -3,\ 5:

(x−2)(x+3)(x−5)=(x2+x−6)(x−5)=x3−4x2−11x+30.(x-2)(x+3)(x-5)=(x^2+x-6)(x-5)=x^3-4x^2-11x+30.

(b) Zeros 3, 2i, −2i3,\ 2i,\ -2i (the pair is required for real coefficients):

(x−3)(x2+4)=x3−3x2+4x−12.(x-3)(x^2+4)=x^3-3x^2+4x-12.
Side note
Where it goes wrong

Building a real-coefficient polynomial from zeros 33 and 2i2i alone: (x−3)(x−2i)(x - 3)(x - 2i) has imaginary coefficients. With real coefficients, −2i-2i must be included: (x−3)(x2+4)(x - 3)(x^2 + 4).

Tip

Tip: A pair of conjugate zeros a±bia\pm bi multiplies to the real quadratic x2−2ax+(a2+b2)x^2-2ax+(a^2+b^2)—no imaginary numbers left behind.

Try it
Try it: build from zeros

Write a cubic with real coefficients, leading coefficient 11, and zeros −1-1 and 2+i2 + i.

Show the answer

The conjugate 2−i2 - i is also a zero. (x−(2+i))(x−(2−i))=x2−4x+5(x - (2 + i))(x - (2 - i)) = x^2 - 4x + 5. Then (x+1)(x2−4x+5)=x3−3x2+x+5(x + 1)(x^2 - 4x + 5) = x^3 - 3x^2 + x + 5.

Graphing Polynomial Functions

Concept
Putting it all together

To sketch y=f(x)y=f(x):

  • Find the end behavior from degree and leading coefficient.
  • Find the zeros (x-intercepts) and their multiplicities: cross at odd, touch at even.
  • Plot the y-intercept f(0)f(0).
  • A degree-nn polynomial has at most n−1n-1 turning points.
  • Connect smoothly—polynomial graphs have no breaks or sharp corners.

Near a zero cc of multiplicity mm, the graph behaves like k(x−c)mk(x - c)^m. An odd power changes sign as xx passes cc, so the graph crosses; an even power keeps its sign, so the graph touches and turns back. Together with the end behavior and a few values in between, that is enough for an accurate sketch.

Example
Graphing f(x)=x3−4x=x(x−2)(x+2)f(x)=x^3-4x=x(x-2)(x+2)
xy−3−2−112−5−4−3−2−11234
Three single zeros: the graph crosses at −2-2, 00 and 22.

Odd degree, positive leading coefficient: down-left, up-right. Zeros x=−2,0,2x=-2,0,2, each multiplicity 11 (the graph crosses at each). The curve below shows all three crossings.

Side note
Where it goes wrong

Crossing at a double zero. For (x+1)2(x−2)(x + 1)^2(x - 2), the factor (x+1)2(x + 1)^2 is positive on both sides of −1-1, so the sign of ff does not change there; the graph touches the axis and turns back.

Example
A repeated zero: f(x)=(x+1)2(x−2)=x3−3x−2f(x)=(x+1)^2(x-2)=x^3-3x-2
xy−3−2−112−5−4−3−2−112(0, −2)
Touches at −1-1 (double zero), crosses at 22.

Zeros x=−1x=-1 (multiplicity 22, the graph touches and turns) and x=2x=2 (multiplicity 11, the graph crosses). Odd degree, positive lead: down-left, up-right.

Example
An even-degree graph: f(x)=x4−5x2+4f(x)=x^4-5x^2+4
xy−3−2−112−3−2−112345
Both ends up; four crossings.

Even degree, positive lead: both ends up. Four real zeros at x=−2,−1,1,2x=-2,-1,1,2 (each crossing).

Tip

Remember: The number of real zeros can be less than the degree—some zeros may be complex, appearing as “missing” xx-intercepts even though the Fundamental Theorem still counts them.

Try it
Try it: sketch

Describe the graph of f(x)=−x(x−3)2f(x) = -x(x - 3)^2: end behavior, zeros with behavior, and yy-intercept.

Show the answer

Leading term −x3-x^3: up on the left, down on the right. Zeros: 00 (single, crosses) and 33 (double, touches). f(0)=0f(0) = 0. Between 00 and 33, e.g. f(1)=−4f(1) = -4, so the graph dips below the axis and comes back up to touch at 33.

Going Deeper: Advanced Polynomial Ideas

Concept
Vieta's formulas: roots to coefficients

For a monic polynomial with roots r1,…,rnr_1,\dots,r_n (repeated by multiplicity), expanding ∏(x−ri)\prod (x-r_i) shows that the elementary symmetric functions of the roots are the coefficients, with alternating signs. For a cubic x3+bx2+cx+dx^3+bx^2+cx+d with roots r1,r2,r3r_1,r_2,r_3:

r1+r2+r3=−b,r1r2+r1r3+r2r3=c,r1r2r3=−d.r_1+r_2+r_3=-b,\quad r_1r_2+r_1r_3+r_2r_3=c,\quad r_1r_2r_3=-d.

For a quartic x4+bx3+cx2+dx+ex^4+bx^3+cx^2+dx+e with roots r1,r2,r3,r4r_1,r_2,r_3,r_4:

∑ri=−b,∑i<jrirj=c,∑i<j<krirjrk=−d,r1r2r3r4=e.\sum r_i=-b,\quad \sum_{i<j} r_ir_j=c,\quad \sum_{i<j<k} r_ir_jr_k=-d,\quad r_1r_2r_3r_4=e.

Non-monic: for leading coefficient ana_n, divide through first—each symmetric function is (−1)kan−k/an(-1)^k a_{n-k}/a_n.

Example
Using Vieta without finding the roots

The roots of x3−6x2+11x−6x^3-6x^2+11x-6 satisfy r1+r2+r3=6r_1+r_2+r_3=6,  r1r2+r1r3+r2r3=11\ r_1r_2+r_1r_3+r_2r_3=11, and r1r2r3=6r_1r_2r_3=6. Find r12+r22+r32r_1^2+r_2^2+r_3^2 without solving. Use the identity

r12+r22+r32=(∑ri)2−2 ⁣∑i<jrirj=62−2(11)=14.r_1^2+r_2^2+r_3^2=\Big(\textstyle\sum r_i\Big)^2-2\!\sum_{i<j} r_ir_j = 6^2-2(11)=14.

(Check: the roots are 1,2,31,2,3, and 1+4+9=141+4+9=14.) Also 1r1+1r2+1r3=r1r2+r1r3+r2r3r1r2r3=116\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}=\dfrac{r_1r_2+r_1r_3+r_2r_3}{r_1r_2r_3}=\dfrac{11}{6}.

Concept
Power sums and Newton's identities

Let pk=r1k+⋯+rnkp_k=r_1^k+\cdots+r_n^k be the kkth power sum of the roots and let eke_k be the kkth elementary symmetric function (so e1=∑rie_1=\sum r_i, e2=∑i<jrirje_2=\sum_{i<j} r_ir_j, …). Newton's identities convert between them:

p1=e1,p2=e1p1−2e2,p3=e1p2−e2p1+3e3,p_1=e_1,\quad p_2=e_1p_1-2e_2,\quad p_3=e_1p_2-e_2p_1+3e_3,
p4=e1p3−e2p2+e3p1−4e4.p_4=e_1p_3-e_2p_2+e_3p_1-4e_4.

These let you compute ∑rik\sum r_i^k from the coefficients alone—no roots needed.

Example
Power sums via Newton's identities

For x3−6x2+11x−6x^3-6x^2+11x-6 we read off e1=6, e2=11, e3=6e_1=6,\ e_2=11,\ e_3=6. Then

p1=e1=6,p2=e1p1−2e2=6⋅6−2⋅11=14,p3=e1p2−e2p1+3e3=6⋅14−11⋅6+3⋅6=36.\begin{aligned} p_1&=e_1=6,\\ p_2&=e_1p_1-2e_2=6\cdot6-2\cdot11=14,\\ p_3&=e_1p_2-e_2p_1+3e_3=6\cdot14-11\cdot6+3\cdot6=36. \end{aligned}

So r13+r23+r33=36r_1^3+r_2^3+r_3^3=36. (Check with roots 1,2,31,2,3: 1+8+27=361+8+27=36.)

Concept
Remainder of P(x)P(x) modulo (x−a)(x−b)(x-a)(x-b)

Dividing P(x)P(x) by a quadratic leaves a remainder of degree at most 11, say R(x)=mx+kR(x)=mx+k. Since P(x)=(x−a)(x−b)Q(x)+mx+kP(x)=(x-a)(x-b)Q(x)+mx+k, substituting the two roots kills the quotient term:

P(a)=ma+k,P(b)=mb+k.P(a)=ma+k,\qquad P(b)=mb+k.

Solve this 2×22\times2 linear system for mm and kk. The remainder is then

R(x)=P(a)−P(b)a−b (x−a)+P(a),R(x)=\frac{P(a)-P(b)}{a-b}\,(x-a)+P(a),

a form of linear interpolation through the points (a,P(a))(a,P(a)) and (b,P(b))(b,P(b)).

Example
Remainder when dividing by (x−1)(x−2)(x-1)(x-2)

Find the remainder of P(x)=x50−3x+7P(x)=x^{50}-3x+7 upon division by (x−1)(x−2)=x2−3x+2(x-1)(x-2)=x^2-3x+2. Evaluate at the two roots:

P(1)=1−3+7=5,P(2)=250−6+7=250+1.P(1)=1-3+7=5,\qquad P(2)=2^{50}-6+7=2^{50}+1.

Write R(x)=mx+kR(x)=mx+k. Then m+k=5m+k=5 and 2m+k=250+12m+k=2^{50}+1. Subtracting,

m=250−4,k=5−m=9−250.m=2^{50}-4,\qquad k=5-m=9-2^{50}.

So the remainder is R(x)=(250−4)x+(9−250)R(x)=(2^{50}-4)x+(9-2^{50}).

Concept
Building a polynomial from finite differences

Tabulate a degree-nn polynomial at equally spaced inputs and take successive differences of the outputs. The nnth differences are constant and equal an⋅n!a_n\cdot n! (for step size 11), while the (n+1)(n+1)th differences are 00. This both detects the degree and reconstructs the leading coefficient.

x01234ff0f1f2f3f4Δf1−f0f2−f1⋯Δ2const if deg 2\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4\\\hline f & f_0 & f_1 & f_2 & f_3 & f_4\\ \Delta & & f_1-f_0 & f_2-f_1 & \cdots & \\ \Delta^2 & & & \text{const if deg }2 & & \end{array}

More generally, n+1n+1 data points determine a unique polynomial of degree ≤n\le n (interpolation).

Example
Recovering a polynomial from a table

A cubic ff gives the values f(0),…,f(4)=1, 2, 11, 34, 77f(0),\dots,f(4)=1,\,2,\,11,\,34,\,77. Build the difference table:

f12113477Δ192343Δ281420Δ366\begin{array}{c|ccccc} f & 1 & 2 & 11 & 34 & 77\\ \Delta & & 1 & 9 & 23 & 43\\ \Delta^2 & & & 8 & 14 & 20\\ \Delta^3 & & & & 6 & 6 \end{array}

Third differences are constant (66), confirming degree 33 with leading coefficient 6/3!=16/3!=1. Now fit f(x)=x3+bx2+cx+df(x)=x^3+bx^2+cx+d: from f(0)=1f(0)=1, d=1d=1; from f(1)=2f(1)=2 and f(2)=11f(2)=11,

1+b+c+1=2,8+4b+2c+1=11  ⇒  b+c=0, 4b+2c=2,1+b+c+1=2,\qquad 8+4b+2c+1=11 \;\Rightarrow\; b+c=0,\ 4b+2c=2,

giving b=1, c=−1b=1,\ c=-1. So f(x)=x3+x2−x+1f(x)=x^3+x^2-x+1. (Check f(4)=64+16−4+1=77f(4)=64+16-4+1=77.)

Concept
Roots in arithmetic or geometric progression

When a problem promises the roots follow a pattern, choose symmetric variables so Vieta collapses nicely.

  • Arithmetic (three roots): write them r−d, r, r+dr-d,\ r,\ r+d. Their sum is 3r3r, so rr equals −13(coeff. of x2)-\tfrac13(\text{coeff. of }x^2) immediately—the middle root is forced by the sum.
  • Geometric (three roots): write them rs, r, rs\dfrac{r}{s},\ r,\ rs. Their product is r3r^3, so rr is the cube root of the product of the roots.
Reminder — Vieta's formulas:  r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}
Example
Cubic with roots in arithmetic progression

The roots of x3−12x2+39x−28x^3-12x^2+39x-28 are in arithmetic progression. Let them be r−d, r, r+dr-d,\ r,\ r+d. By Vieta the sum is 3r=123r=12, so r=4r=4: one root is exactly 44. Check: f(4)=64−192+156−28=0f(4)=64-192+156-28=0. Depress by (x−4)(x-4) to get x2−8x+7=(x−1)(x−7)x^2-8x+7=(x-1)(x-7), so the roots are 1,4,71,4,7—indeed an arithmetic progression with common difference 33.

Concept
Rational Root Theorem with bounds and Descartes' Rule

Three tools sharply narrow the candidate list before you test anything.

  • Rational Root Theorem: every rational root is ±pq\pm\dfrac{p}{q} with p∣a0p\mid a_0, q∣anq\mid a_n.
  • Upper/lower bounds (synthetic division): if you divide by x−cx-c with c>0c>0 and the bottom row is all nonnegative, then cc is an upper bound—no root exceeds it. If you divide by x−cx-c with c<0c<0 and the bottom row alternates in sign, then cc is a lower bound.
  • Descartes' Rule of Signs: the number of positive real roots equals the number of sign changes in P(x)P(x), or less by an even number. Apply to P(−x)P(-x) for the count of negative real roots.
Example
Pruning candidates with Descartes and bounds

For P(x)=x3−4x2+x+6P(x)=x^3-4x^2+x+6: the sign pattern of the coefficients +,−,+,++,-,+,+ has 22 changes, so there are 22 or 00 positive roots. Then P(−x)=−x3−4x2−x+6P(-x)=-x^3-4x^2-x+6 has pattern −,−,−,+-,-,-,+ with 11 change, so exactly 11 negative root. Rational candidates are ±1,±2,±3,±6\pm1,\pm2,\pm3,\pm6. Testing x=−1x=-1: −1−4−1+6=0-1-4-1+6=0—the guaranteed negative root. Depressing gives x2−5x+6=(x−2)(x−3)x^2-5x+6=(x-2)(x-3), so the two positive roots are 22 and 33, matching Descartes exactly. Dividing by x−4x-4 leaves all-nonnegative entries, confirming 44 is an upper bound.

Tip

The “(a−b)∣P(a)−P(b)(a-b)\mid P(a)-P(b)” idea: for a polynomial PP with integer coefficients and any integers a,ba,b,

(a−b) ∣ P(a)−P(b).(a-b)\ \big|\ P(a)-P(b).

This is because each ak−bka^k-b^k is divisible by a−ba-b. It is a powerful screening tool: e.g. if PP has integer coefficients and P(0)=3P(0)=3, P(4)=6P(4)=6, no integer nn with P(n)=0P(n)=0 can exist between them that violates  n∣3 \,n\mid 3\, and  (n−4)∣3\,(n-4)\mid 3. It also proves that a polynomial taking the value 11 at three distinct integers can never equal 00 at an integer.

Why It Works

Remainder theorem: dividing by x−ax - a gives P(x)=(x−a)Q(x)+rP(x) = (x - a)Q(x) + r with rr constant (degree less than 11). Substituting x=ax = a kills the first term: P(a)=rP(a) = r. The factor theorem is the case r=0r = 0. The rational root theorem comes from clearing denominators in P(p/q)=0P(p/q) = 0: the constant term must be divisible by pp and the leading coefficient by qq, which leaves only finitely many candidates to try.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Remainder and Factor theorems
P(x)=(x−c)Q(x)+P(c)⟹(x−c)∣P(x)  ⟺  P(c)=0P(x)=(x-c)Q(x)+P(c) \quad\Longrightarrow\quad (x-c)\mid P(x) \iff P(c)=0

What it means. Dividing PP by x−cx-c leaves remainder P(c)P(c); so x−cx-c is a factor exactly when cc is a root.

Example. P(x)=x2−5x+6P(x)=x^2-5x+6: P(2)=0P(2)=0, so x−2x-2 is a factor.

Why it works. Division gives P(x)=(x−c)Q(x)+rP(x)=(x-c)Q(x)+r with rr constant (its degree is below 11). Substituting x=cx=c kills the first term and leaves P(c)=rP(c)=r.

Tip. Testing a possible root is one substitution — far faster than doing the division.

Tip
Vieta's formulas
r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}

What it means. The sum and product of a quadratic's roots read straight off its coefficients.

Example. x2−7x+12=0x^2-7x+12=0: the roots sum to 77 and multiply to 1212, so they are 33 and 44.

Why it works. If r1,r2r_1,r_2 are the roots then ax2+bx+c=a(x−r1)(x−r2)=a(x2−(r1+r2)x+r1r2)ax^2+bx+c=a(x-r_1)(x-r_2)=a\left(x^2-(r_1+r_2)x+r_1r_2\right). Matching coefficients gives both formulas.

Tip. Great for "find the sum of the roots" questions — you never have to solve the equation.