Polynomials

1. What This Chapter Covers

Polynomials were introduced in class IX. This chapter does more with them, and it opens with two situations rather than a definition.

A flower bed. A triangular flower bed has its longest side 3 times the smallest, and the smallest side 2 units shorter than the intermediate side. Let P be the smallest side. Then the intermediate side is P + 2 and the longest is 3P, so the perimeter is P + 3P + P + 2 = 5P + 2.

A dining hall. A rectangular hall is twice as long as it is broad. If the breadth is x, the length is 2x, and the area is (2x)(x) = 2x².

The book's point is that the two answers are polynomials of different degrees, which is what the rest of the chapter is about.

The chapter is allotted 8 periods in July and runs from textbook page 51 to page 76.

2. What Counts as a Polynomial, and its Degree

A polynomial in x is an algebraic expression containing the sum of a finite number of terms of the form axⁿ for a real number a, where a ≠ 0 and n is a whole number.

The whole-number condition on the exponent is what does the excluding. The book puts examples on each side of the line:

A square root is the exponent 1/2, and 1/x is the exponent −1; neither is a whole number. Note that 1/3 as a coefficient is perfectly fine — the restriction is on the powers, not on the numbers in front of them.

The degree of p(x) is the highest power of x in it, and the first few degrees have names:

Degrees do not stop at 3. The book gives 7u⁶ − (3/2)u⁴ + 4u² − 8 as degree 6 and x¹⁰ − 3x⁸ + 4x⁵ + 2x² − 1 as degree 10, and then states the general form:

p(x) = a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² + … + aₙ₋₁x + aₙ is a polynomial of nth degree in x, where a₀, a₁, …, aₙ are real coefficients and a₀ ≠ 0.

Note the indexing: in this book a₀ is the leading coefficient, the one attached to the highest power. Many other books number the other way round, with aₙ on xⁿ. Read the subscripts off whichever book is in front of you rather than from memory.

3. The Value of a Polynomial, and its Zeroes

Substituting a number for x gives a number back.

If p(x) is a polynomial in x and k is a real number, the value obtained by substituting x = k is called the value of p(x) at x = k, denoted p(k).

For p(x) = x² − 2x − 3: p(1) = 1 − 2 − 3 = −4, and p(0) = −3.

Some substitutions give zero, and those are the ones that matter:

• p(3) = 9 − 6 − 3 = 0
• p(−1) = 1 + 2 − 3 = 0
• p(2) = 4 − 4 − 3 = −3, which is not zero

A real number k is said to be a zero of a polynomial p(x) if p(k) = 0.

So 3 and −1 are zeroes of x² − 2x − 3, and 2 is not. Note the definition says a real number — this chapter never leaves the real numbers.

For a linear polynomial the zero can be found once and for all. If k is a zero of p(x) = ax + b with a ≠ 0, then ak + b = 0, so

the zero of the linear polynomial ax + b is k = −b/a.

The zero of 2x + 5, for instance, is −5/2. Already the zero is tied to the coefficients, which is the thread the whole chapter follows.

4. The Zero of a Linear Polynomial, Geometrically

The graph of y = ax + b is a straight line. Take y = 2x + 3 and tabulate it:

The zero of the polynomial and the x-coordinate of the crossing point are the same number, read two different ways.

The line meets the X-axis between x = −1 and x = −2, at (−3/2, 0). And −3/2 is exactly −b/a with a = 2 and b = 3.

The graph of y = ax + b (a ≠ 0) is a straight line intersecting the X-axis at exactly one point, namely (−b/a, 0). So a linear polynomial has exactly one zero.

5. Zeroes of a Quadratic: the Three Cases

Now take x² − 3x − 4 and tabulate it:

Plot the table's eight points, join them with a smooth curve, and the shape is a parabola, not a straight line.

The zeroes of a quadratic polynomial ax² + bx + c (a ≠ 0) are precisely the x-coordinates of the points where the parabola y = ax² + bx + c meets the X-axis.

Whether the parabola opens upwards or downwards depends on the sign of a. Either way, only three things can happen.

In the middle panel the two crossing points of the first have merged into one, which is why a repeated zero is counted as two coincident zeroes.

• Case (i). The curve cuts the X-axis at two distinct points A and A′. Their x-coordinates are the two zeroes.
• Case (ii). The curve touches the X-axis at exactly one point — two coincident points. That x-coordinate is the only zero.
• Case (iii). The curve lies entirely above or entirely below the X-axis. There is no zero.

A quadratic polynomial has two distinct zeroes, or two equal zeroes, or no zero. So a polynomial of degree 2 has at most two zeroes.

6. Zeroes of a Cubic

The same question for degree 3. Take x³ − 4x:

A cubic need not use all three crossings: y = x³ meets the axis only at the origin, and y = x³ − x² only at 0 and 1.

The book checks two more cubics to show the count can fall short. For y = x³ the only crossing is at 0, so there is one zero; for y = x³ − x² the crossings are at 0 and 1, so there are two distinct zeroes. In every case the number is three or fewer.

Note. Given a polynomial p(x) of degree n, the graph of y = p(x) meets the X-axis at at most n points, so p(x) has at most n zeroes. The book calls this the fundamental theorem of Algebra.

That attribution is looser than it looks. The Fundamental Theorem of Algebra proper is a statement about complex roots — a degree-n polynomial has exactly n of them, counted with multiplicity. The "at most n real zeroes" used here is a consequence of it, not the theorem itself.

7. Zeroes and Coefficients: the Quadratic

For a linear polynomial the zero was −b/a. Is there a similar tie for a quadratic?

Take p(x) = 2x² − 8x + 6 and split the middle term. The product needed is 6 × 2x² = 12x², so −8x splits as −6x − 2x:

2x² − 8x + 6 = 2x² − 6x − 2x + 6 = 2x(x − 3) − 2(x − 3) = (2x − 2)(x − 3) = 2(x − 1)(x − 3)

So the zeroes are 1 and 3. Now compare them with the coefficients 2, −8 and 6:

• Sum = 1 + 3 = 4 = −(−8)/2 = −(coefficient of x) / (coefficient of x²)
• Product = 1 × 3 = 3 = 6/2 = (constant term) / (coefficient of x²)

The book repeats the check on 3x² + 5x − 2 = (3x − 1)(x + 2), whose zeroes are 1/3 and −2: the sum is −5/3 and the product is −2/3, matching again.

Then it proves it. If α and β are the zeroes of ax² + bx + c, then (x − α) and (x − β) are factors, so for some constant k

ax² + bx + c = k(x − α)(x − β) = k[x² − (α + β)x + αβ] = kx² − k(α + β)x + kαβ

Comparing coefficients gives a = k, b = −k(α + β) and c = kαβ, and dividing away the k:

α + β = −b/a and αβ = c/a

Two examples run it in both directions. Forwards, x² + 7x + 10 = (x + 2)(x + 5) has zeroes −2 and −5, sum −7 = −7/1 and product 10 = 10/1. Backwards, to build a quadratic with sum −3 and product 2, take a = 1, so b = 3 and c = 2, giving x² + 3x + 2.

The backwards direction has one subtlety worth holding onto. Any a = k works, producing kx² + 3kx + 2k, so the answer is not unique — every non-zero k gives a polynomial with the same zeroes. With zeroes 2 and −1/3 the general answer is k[x² − (5/3)x − 2/3], which for k = 3 is 3x² − 5x − 2 and for k = 6 is 6x² − 10x − 4.

8. Zeroes and Coefficients: the Cubic

For p(x) = 2x³ − 5x² − 14x + 8 the zeroes are 4, −2 and 1/2, and there are now three relationships rather than two.

• Sum = 4 + (−2) + 1/2 = 5/2 = −(−5)/2 = −b/a
• Sum of products two at a time = 4(−2) + (−2)(1/2) + (1/2)(4) = −8 − 1 + 2 = −7 = −14/2 = c/a
• Product = 4 × (−2) × 1/2 = −4 = −8/2 = −d/a

The proof is the same move as before. Writing the cubic as (x − α)(x − β)(x − γ) and expanding gives x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ; multiplying by a and comparing coefficients gives b = −a(α + β + γ), c = a(αβ + βγ + γα) and d = −aαβγ. Hence:

For ax³ + bx² + cx + d with zeroes α, β, γ:

α + β + γ = −b/a , αβ + βγ + γα = c/a , αβγ = −d/a

Watch the middle one: it is a sum of products taken two at a time, not a product. The book's own working page labels its value "constant of x / coefficient of x³", where "coefficient of x" is meant; the arithmetic underneath, −14/2, is right.

Example 7 checks all three on 3x³ − 5x² − 11x − 3 with zeroes 3, −1 and −1/3. Substitution confirms each is a zero, and then 3 − 1 − 1/3 = 5/3 = −b/a; (3)(−1) + (−1)(−1/3) + (−1/3)(3) = −3 + 1/3 − 1 = −11/3 = c/a; and (3)(−1)(−1/3) = 1 = −(−3)/3 = −d/a.

9. The Division Algorithm for Polynomials

A cubic has at most three zeroes — but if you are handed only one of them, how do you get the others? By dividing the known factor out.

Suppose 1 is a zero of x³ − 3x² − x + 3. Then x − 1 is a factor, and dividing gives the quotient x² − 2x − 3, which splits as (x + 1)(x − 3). So

x³ − 3x² − x + 3 = (x − 1)(x + 1)(x − 3)

and the three zeroes are 1, −1 and 3.

Careful with this example in the printed book. Page 71 introduces the polynomial as x³ + 3x² − x − 3 and then, two lines later, factorises x³ − 3x² − x + 3. The two are different polynomials, and only the second is consistent with the quotient, the factors and the stated zeroes 1, −1, 3 — as Exercise 3.3 question 4 on the same page confirms. The signs in the opening line are the misprint. (For the record, x³ + 3x² − x − 3 factorises as (x − 1)(x + 1)(x + 3), with zeroes 1, −1 and −3.)

Dividing polynomials works like long division of numbers. Example 8: dividing 2x² + 3x + 1 by x + 2 gives quotient 2x − 1 and remainder 3, and checking, (2x − 1)(x + 2) + 3 = 2x² + 3x − 2 + 3 = 2x² + 3x + 1.

Example 9 divides by a quadratic. To divide 3x³ + x² + 2x + 5 by 1 + 2x + x², first write both in standard form — decreasing order of exponents — so the divisor becomes x² + 2x + 1. Then:

1. Divide the highest term of the dividend by the highest term of the divisor: 3x³ ÷ x² = 3x. Subtracting leaves −5x² − x + 5.
2. Repeat: −5x² ÷ x² = −5. Subtracting leaves 9x + 10.
3. Stop, because the degree of 9x + 10 is less than the degree of x² + 2x + 1.

Checking: (x² + 2x + 1)(3x − 5) + (9x + 10) = 3x³ + x² + 2x + 5.

The remainder condition is what makes the quotient and remainder unique, exactly as 0 ≤ r < b did for whole numbers in chapter 1.

Division Algorithm for polynomials. If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where either r(x) = 0 or degree of r(x) < degree of g(x).

Four consequences follow, and the third is the most useful:

1. If g(x) is linear, then r(x) = r is a constant.
2. If the degree of g(x) is 1, then degree of p(x) = 1 + degree of q(x).
3. If p(x) is divided by (x − a), the remainder is p(a).
4. If r = 0, then g(x) divides p(x) exactly, and g(x) is a factor of p(x).

Example 11 puts it all together. To find the remaining zeroes of 2x⁴ − 3x³ − 3x² + 6x − 2 given that √2 and −√2 are two of them: those two zeroes mean (x − √2)(x + √2) = x² − 2 is a factor. Dividing gives the quotient 2x² − 3x + 1 with remainder 0, and splitting the middle term gives (2x − 1)(x − 1). So the four zeroes are √2, −√2, 1 and 1/2.

10. What the Exercises Ask, and What the Book Answers

Four numbered exercises plus an Optional Exercise, all except the last with printed answers at textbook pages 374 and 375.

• Exercise 3.1 (5 questions) is vocabulary: coefficient, degree and constant term; five true-or-false statements with reasons; evaluating p(t) = t³ − 1 at five points; and checking given numbers against two polynomials.
• Exercise 3.2 (4 questions) is the graphical meaning — counting zeroes from six given graphs, finding zeroes by factorising, drawing five parabolas, and a verification.
• Exercise 3.3 (4 questions) is the zero-coefficient relationships, in both directions: find the zeroes and verify, or build a polynomial from a given sum and product.
• Exercise 3.4 (5 questions) is division: quotient and remainder, factor checks, finding remaining zeroes from two known ones, recovering a divisor from the quotient and remainder, and constructing examples to order.
• Optional Exercise (5 questions, marked "For extensive learning") includes the cubic built from all three symmetric sums and two harder divisor problems.

The chapter-3 key is in far better shape than chapter 2's, but two answers are worth checking against the definitions:

• Exercise 3.1 question 4 asks whether −2 and 2 are zeroes of x⁴ − 16. The key prints "Yes, −2 and −2 are zeroes"; the second should read 2. Both are in fact zeroes, since (±2)⁴ = 16.
• Exercise 3.2 question 2(iv) asks for the zeroes of x⁴ − 16, and the key gives −2, 2, ±√−4. The last pair are imaginary. This chapter defines a zero as a real number k with p(k) = 0 and never introduces complex numbers, so within its own terms the answer is −2 and 2.

11. Summary

A polynomial is a finite sum of terms axⁿ with n a whole number; fractional and negative exponents disqualify an expression, though fractional coefficients are fine. Its degree is the highest power present, and degrees 1, 2 and 3 are called linear, quadratic and cubic.

The value p(k) comes from substituting x = k, and k is a zero when p(k) = 0. Geometrically the zeroes of p(x) are the x-coordinates of the points where y = p(x) meets the X-axis — one crossing for a line, at most two for a parabola, at most three for a cubic, and at most n for degree n.

A quadratic meets the axis in exactly one of three ways: cutting it twice, touching it once, or missing it entirely, which is why it has two zeroes, one, or none.

The zeroes are tied to the coefficients. For ax² + bx + c, α + β = −b/a and αβ = c/a. For ax³ + bx² + cx + d, α + β + γ = −b/a, αβ + βγ + γα = c/a and αβγ = −d/a. Running these backwards builds a polynomial from its zeroes — but only up to a constant multiple, so the answer is a family, not a single polynomial.

Finally the division algorithm: p(x) = g(x)q(x) + r(x) with r(x) = 0 or deg r < deg g. Its practical use is that a known zero gives a known factor, and dividing it out reduces the degree until what is left can be factorised by hand.