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NSWMaths Extension 1Syllabus dot point

What exactly is a polynomial, and how do its degree and coefficients control its behaviour?

Define a polynomial and use the language of degree, leading term, leading coefficient, monic and the zero polynomial, including the degree of sums and products and equating coefficients of identically equal polynomials

A first-contact answer to the HSC Maths Extension 1 dot point on the language of polynomials. Degree, leading term and coefficient, monic and the zero polynomial, the degree of sums and products, equating coefficients of identically equal polynomials, and even and odd polynomials, with worked examples.

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What this dot point is asking

This is your first formal contact with polynomials in Extension 1. NESA wants you to be fluent in the vocabulary before you start dividing, factoring and graphing them: what a polynomial actually is, its degree, its leading term and leading coefficient, what monic means, and the special status of the zero polynomial. On top of the vocabulary, two working skills are examined directly: predicting the degree of a sum or a product without expanding fully, and finding unknown coefficients by equating two polynomials that are identically equal.

The answer

What a polynomial is

A polynomial in xx is a sum of terms, each a number times a whole-number power of xx:

P(x)=anxn+anβˆ’1xnβˆ’1+β‹―+a1x+a0,P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0,

where the coefficients an,anβˆ’1,…,a0a_n, a_{n-1}, \dots, a_0 are real numbers and the powers are non-negative integers. The two conditions that do the work are hidden in that sentence. The powers must be whole numbers, so x=x1/2\sqrt{x} = x^{1/2}, 1x=xβˆ’1\dfrac{1}{x} = x^{-1} and 2x2^x are all banned: none is a polynomial. And the coefficients are just constants, so 3x2βˆ’7x+13x^2 - 7x + 1 qualifies but x2+xlog⁑exx^2 + x \log_e x does not.

Degree, leading term and leading coefficient

The vocabulary all keys off the term of highest power that actually appears.

  • The leading term is the term of highest power with a non-zero coefficient.
  • The degree is the power of the leading term.
  • The leading coefficient is the coefficient of the leading term.
  • The constant term is a0=P(0)a_0 = P(0), the term with no xx.
  • A monic polynomial is one whose leading coefficient is 11.

So for P(x)=3x4βˆ’5x2+2xβˆ’7P(x) = 3x^4 - 5x^2 + 2x - 7 the leading term is 3x43x^4, the degree is 44, the leading coefficient is 33 (so it is not monic), and the constant term is βˆ’7-7.

The anatomy of a polynomialThe polynomial P of x equals three x to the fourth minus five x squared plus two x minus seven, with labels: the highest power four is the degree, the coefficient three of that highest power is the leading coefficient, the whole term three x to the fourth is the leading term, and the number minus seven is the constant term.P(x) = 3x4 βˆ’ 5x2 + 2x βˆ’ 7leading coefficientdegree = 4constant termleading term

Low degrees have names you should use: degree 00 is a (non-zero) constant, degree 11 is linear, degree 22 is quadratic, degree 33 is cubic, degree 44 is quartic, degree 55 is quintic.

There is one trap NESA likes here. A constant function is a linear function, but a constant polynomial is not a linear polynomial: a linear polynomial has degree 11, while a non-zero constant has degree 00. The word linear means degree exactly one when it is applied to polynomials.

The zero polynomial

The constant polynomial a0a_0 has degree 00 as long as a0β‰ 0a_0 \ne 0. The single exception is the zero polynomial Z(x)=0Z(x) = 0. It has every coefficient equal to zero, so it has no leading term, and therefore it has no degree at all. It is not "degree 00"; it is the one polynomial with no degree. This sounds like a technicality, but it is exactly the case that breaks the degree rules below, which is why the rules are stated with care.

The degree of a sum or difference

Add or subtract two polynomials and you get another polynomial, formed by adding or subtracting matching coefficients. The degree usually behaves simply, but there is a catch when the two have the same degree.

Let P(x)P(x) and Q(x)Q(x) be non-zero polynomials of degrees nn and mm.

  • If nβ‰ mn \ne m, then deg⁑(P(x)Β±Q(x))\deg\big(P(x) \pm Q(x)\big) is the larger of nn and mm. The bigger leading term has nothing to cancel against.
  • If n=mn = m, the leading terms can cancel, so deg⁑(P(x)Β±Q(x))≀n\deg\big(P(x) \pm Q(x)\big) \le n, and the result might even be the zero polynomial (with no degree).

For example, (x2βˆ’3x+2)+(βˆ’x2+4x+9)=x+11(x^2 - 3x + 2) + (-x^2 + 4x + 9) = x + 11: two quadratics whose leading terms cancel, leaving a polynomial of degree 11. And P(x)+(βˆ’P(x))=0P(x) + \big(-P(x)\big) = 0 for every PP, where the opposite polynomial βˆ’P(x)-P(x) is formed by negating every coefficient; the sum is the zero polynomial, which is why "no degree" has to be allowed.

The degree of a product

Multiplication is the well-behaved case. The leading term of a product is the product of the two leading terms, and multiplying two non-zero numbers can never give zero, so the top term never disappears. Hence the degree of a product is the sum of the degrees:

deg⁑(P(x) Q(x))=deg⁑P(x)+deg⁑Q(x).\deg\big(P(x)\,Q(x)\big) = \deg P(x) + \deg Q(x).

This lets you read off the degree, the leading coefficient and the constant term of a big product without expanding it. For (2x2+1)(x3βˆ’4)(5x5+3)(2x^2 + 1)(x^3 - 4)(5x^5 + 3) the degree is 2+3+5=102 + 3 + 5 = 10, the leading coefficient is 2Γ—1Γ—5=102 \times 1 \times 5 = 10, and the constant term is 1Γ—(βˆ’4)Γ—3=βˆ’121 \times (-4) \times 3 = -12.

Identically equal polynomials and equating coefficients

Two polynomials P(x)P(x) and Q(x)Q(x) are identically equal, written P(x)≑Q(x)P(x) \equiv Q(x), if they are equal for every value of xx, not just for a few. The single most useful fact in this whole topic follows from that:

If two polynomials are identically equal, then their corresponding coefficients are equal.

This is what powers "equating coefficients": you write the same polynomial two different ways, set the coefficient of each power on one side equal to the coefficient of that power on the other, and solve the resulting equations for the unknowns. It is the standard method for splitting a fraction into partial fractions later in the course, and it turns one polynomial identity into a whole system of ordinary equations.

The distinction between ≑\equiv (true for all xx) and == (true for the particular xx you are solving for) matters. An equation like x2βˆ’5x+6=0x^2 - 5x + 6 = 0 is true only for x=2x = 2 and x=3x = 3; an identity like (xβˆ’2)(xβˆ’3)≑x2βˆ’5x+6(x - 2)(x - 3) \equiv x^2 - 5x + 6 is true for every xx, and only an identity lets you equate coefficients.

Even and odd polynomials

A function is even if P(βˆ’x)=P(x)P(-x) = P(x) for all xx (symmetric in the yy-axis) and odd if P(βˆ’x)=βˆ’P(x)P(-x) = -P(x) for all xx (symmetric in the origin). For a polynomial these reduce to a clean statement about which powers appear, because replacing xx with βˆ’x-x flips the sign of every odd power and leaves every even power alone:

  • a polynomial is even exactly when only even powers have non-zero coefficients (for example 3x4βˆ’5x2+23x^4 - 5x^2 + 2);
  • a polynomial is odd exactly when only odd powers have non-zero coefficients (for example 2x5βˆ’3x3+x2x^5 - 3x^3 + x).

So if P(x)=a4x4+a3x3+a2x2+a1x+a0P(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 is even, then comparing P(βˆ’x)P(-x) with P(x)P(x) forces a3=a1=0a_3 = a_1 = 0; if it is odd, the even-power coefficients a4,a2,a0a_4, a_2, a_0 are all zero. Most polynomials are neither even nor odd.

How exam questions ask about the language of polynomials

The wording is predictable once you know the moves to match it to.

  • "State the degree, leading coefficient, leading term and constant term." Read them straight off the highest-power term, after first expanding if the polynomial is given in factored or bracketed form. Watch a leading coefficient that is negative or fractional, and remember monic means leading coefficient exactly 11.
  • "What is the degree of P(x)+Q(x)P(x) + Q(x) / P(x)Q(x)P(x)Q(x)?" For a product, add the degrees, no expansion needed. For a sum, take the larger degree, unless the two degrees are equal, in which case warn that the leading terms might cancel and the degree could drop.
  • "Find the constants so that ... ≑\equiv ... for all xx." Equate coefficients of each power to get a system of equations, then solve. The phrase "for all xx" is the signal that this is an identity, not an equation to be solved for xx.
  • "Show that b=d=0b = d = 0 if P(x)P(x) is even" (or the odd version). Write out P(βˆ’x)P(-x), set it identically equal to P(x)P(x) (or to βˆ’P(x)-P(x)), and equate coefficients; the odd-power (or even-power) coefficients are forced to zero.
  • "Give an example of two polynomials of degree 22 whose sum has degree 00." This is testing whether you understand same-degree cancellation: choose two quadratics whose x2x^2 and xx terms cancel but whose constants do not, such as x2+1x^2 + 1 and βˆ’x2+1-x^2 + 1, summing to 22.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFor the polynomial P(x)=βˆ’x5+4x2βˆ’7P(x) = -x^5 + 4x^2 - 7, state (i) the degree, (ii) the leading term, (iii) the leading coefficient, (iv) the constant term, and (v) whether P(x)P(x) is monic.
Show worked solution β†’
Find the leading term
The term of highest power with a non-zero coefficient is βˆ’x5-x^5, so this is the leading term.
Read off degree and leading coefficient
The degree is the power of the leading term, 55. The leading coefficient is the coefficient of that term, which is βˆ’1-1 (the coefficient of βˆ’x5-x^5).
Read off the constant term
The constant term is the term with no xx, which is βˆ’7-7.
Decide if monic
A polynomial is monic only when its leading coefficient is 11. Here the leading coefficient is βˆ’1-1, so P(x)P(x) is not monic.
Answer
(i) degree 55; (ii) leading term βˆ’x5-x^5; (iii) leading coefficient βˆ’1-1; (iv) constant term βˆ’7-7; (v) not monic.
foundation2 marksWrite down the monic polynomial whose degree, leading coefficient and constant term are all equal to each other.
Show worked solution β†’

Use the monic condition first. Monic means the leading coefficient is 11. The three quantities must all be equal, so they must all equal 11.

Match degree and constant to that value. The degree must be 11 and the constant term must be 11. A degree-11 monic polynomial with constant term 11 is

P(x)=x+1.P(x) = x + 1.

Check. Degree 11, leading coefficient 11, constant term 11: all three equal 11, as required. (No other polynomial works, because monic forces the common value to be 11, which then fixes both the degree and the constant term.)

core4 marksLet P(x)=3x2βˆ’2x+5P(x) = 3x^2 - 2x + 5 and Q(x)=x3+4xβˆ’1Q(x) = x^3 + 4x - 1. Find P(x)+Q(x)P(x) + Q(x) and P(x)Γ—Q(x)P(x) \times Q(x), and state the degree of each.
Show worked solution β†’

Add by collecting like terms. Line up matching powers and add coefficients:

P(x)+Q(x)=x3+3x2+(βˆ’2+4)x+(5βˆ’1)=x3+3x2+2x+4.P(x) + Q(x) = x^3 + 3x^2 + (-2 + 4)x + (5 - 1) = x^3 + 3x^2 + 2x + 4.

The highest power present is x3x^3, so the degree of the sum is 33. (This matches the rule: the degrees are 22 and 33, and since they differ the sum takes the larger, 33.)

Multiply each term of PP by each term of QQ. Expanding and collecting:

P(x)Γ—Q(x)=3x5βˆ’2x4+17x3βˆ’11x2+22xβˆ’5.P(x) \times Q(x) = 3x^5 - 2x^4 + 17x^3 - 11x^2 + 22x - 5.

State the product degree. The degree of a product is the sum of the degrees, 2+3=52 + 3 = 5, which agrees with the leading term 3x2Γ—x3=3x53x^2 \times x^3 = 3x^5.

Answer. P+Q=x3+3x2+2x+4P + Q = x^3 + 3x^2 + 2x + 4 (degree 33); PΓ—Q=3x5βˆ’2x4+17x3βˆ’11x2+22xβˆ’5P \times Q = 3x^5 - 2x^4 + 17x^3 - 11x^2 + 22x - 5 (degree 55).

core3 marksFind aa and bb if (aβˆ’b)x2+(2a+b)x=7xβˆ’x2(a - b)x^2 + (2a + b)x = 7x - x^2 for all values of xx.
Show worked solution β†’

Match coefficients of like powers. Because the two sides are equal for all xx, the coefficient of each power must agree. Comparing x2x^2 coefficients and then xx coefficients:

aβˆ’b=βˆ’1,2a+b=7.a - b = -1, \qquad 2a + b = 7.

Solve the pair by adding. Adding the two equations eliminates bb:

(aβˆ’b)+(2a+b)=βˆ’1+7β€…β€ŠβŸΉβ€…β€Š3a=6β€…β€ŠβŸΉβ€…β€Ša=2.(a - b) + (2a + b) = -1 + 7 \implies 3a = 6 \implies a = 2.

Back-substitute. From aβˆ’b=βˆ’1a - b = -1 with a=2a = 2:

2βˆ’b=βˆ’1β€…β€ŠβŸΉβ€…β€Šb=3.2 - b = -1 \implies b = 3.

Check. aβˆ’b=2βˆ’3=βˆ’1a - b = 2 - 3 = -1 and 2a+b=4+3=72a + b = 4 + 3 = 7: both match, so a=2a = 2 and b=3b = 3.

exam4 marksThe monic cubic P(x)=(xβˆ’1)(x2+px+q)P(x) = (x - 1)(x^2 + px + q) is identically equal to x3βˆ’2x2βˆ’5x+6x^3 - 2x^2 - 5x + 6. Find pp and qq, and hence write P(x)P(x) as a product of its factor (xβˆ’1)(x - 1) and a quadratic.
Show worked solution β†’

Expand the given factored form. Multiplying out (xβˆ’1)(x2+px+q)(x - 1)(x^2 + px + q) term by term:

(xβˆ’1)(x2+px+q)=x3+(pβˆ’1)x2+(qβˆ’p)xβˆ’q.(x - 1)(x^2 + px + q) = x^3 + (p - 1)x^2 + (q - p)x - q.

Equate coefficients with the target. Matching against x3βˆ’2x2βˆ’5x+6x^3 - 2x^2 - 5x + 6 power by power gives

pβˆ’1=βˆ’2,qβˆ’p=βˆ’5,βˆ’q=6.p - 1 = -2, \qquad q - p = -5, \qquad -q = 6.

Solve, starting from the simplest equation
From βˆ’q=6-q = 6 we get q=βˆ’6q = -6. From pβˆ’1=βˆ’2p - 1 = -2 we get p=βˆ’1p = -1.
Verify with the spare equation
The middle relation must also hold: qβˆ’p=βˆ’6βˆ’(βˆ’1)=βˆ’5q - p = -6 - (-1) = -5, which matches the coefficient of xx. So the values are consistent.
Write the result
With p=βˆ’1p = -1 and q=βˆ’6q = -6,

P(x)=(xβˆ’1)(x2βˆ’xβˆ’6).P(x) = (x - 1)(x^2 - x - 6).

(Equating coefficients used all three conditions and the redundant one confirmed the answer, which is the usual exam safety check.)

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