Polynomials

Every algebraic expression you have ever simplified — $4x + 2$ , $2y^2 - 3y + 4$ , $5x^3 - 4x^2 + x - \tfrac{1}{2}$ — belongs to a family called polynomials, and every polynomial hides the same quiet question: for which values of the variable does the whole expression collapse to zero? Those special input values are the polynomial's zeroes, and this chapter turns them from a curiosity spotted by trial and error into one of the most dependable tools in board-level algebra. The chapter opens by re-sorting what you already know about degree — linear, quadratic, cubic — and then asks a picture-based question first: what does a zero actually look like on a graph? The answer is beautifully simple. Plot $y = p(x)$ and its zeroes are exactly the points where the curve touches or crosses the x-axis. A straight line can only cross once, so a linear polynomial has exactly one zero; a parabola can cross twice, touch once, or miss the axis entirely, so a quadratic polynomial has two, one, or no real zero; a cubic's curve can weave across the axis up to three times. This single geometric fact — that a degree- $n$ polynomial has at most $n$ zeroes — governs everything that follows. The chapter's real payoff, though, is algebraic: once you factorise a quadratic or a cubic and read off its zeroes, those zeroes turn out to be locked to the polynomial's coefficients by fixed, provable formulas — a sum here, a product there, no factoring required if you already know the answer you are checking. You will prove these relationships from the factor theorem, use them to build a polynomial backwards from just its sum and product of zeroes, and verify them on both quadratic and cubic examples. By the end, "find the zeroes" and "use the coefficients" become two directions of the very same road.

Key takeaways

A polynomial's degree — the highest power of the variable with a nonzero coefficient — names it: degree $1$ is a linear polynomial $ax+b$ , degree $2$ a quadratic polynomial $ax^2+bx+c$ , degree $3$ a cubic polynomial $ax^3+bx^2+cx+d$ , always with $a \neq 0$ .
The value of a polynomial $p(x)$ at $x=k$ is $p(k)$ , found by substituting $k$ for $x$ ; a real number $k$ is called a zero of the polynomial if $p(k) = 0$ .
The zero of a linear polynomial $ax+b$ is $k = -b/a$ , that is, $-\text{(constant term)}/\text{(coefficient of }x\text{)}$ .
Geometrically, the zeroes of $p(x)$ are precisely the x-coordinates of the points where the graph of $y=p(x)$ meets the x-axis — an x-intercept.
A quadratic's graph is a parabola: it can cross the x-axis at two distinct points, touch it at one repeated point, or miss it completely, so a quadratic polynomial has two, one, or no real zero — never more than two.
A degree- $n$ polynomial has at most $n$ zeroes; a cubic polynomial's graph can cross the x-axis at up to three points.
For a quadratic polynomial with zeroes $\alpha, \beta$ : sum of zeroes $\alpha+\beta = -b/a$ and product of zeroes $\alpha\beta = c/a$ , both provable from the factor theorem.
For a cubic polynomial with zeroes $\alpha, \beta, \gamma$ : $\alpha+\beta+\gamma = -b/a$ , $\alpha\beta+\beta\gamma+\gamma\alpha = c/a$ , and $\alpha\beta\gamma = -d/a$ .
Given only a required sum and product, splitting the middle term in reverse — the template $k(x^2 - (\text{sum})x + \text{product})$ — builds an entire family of quadratic polynomials sharing those zeroes.
Key idea: a real number k is a zero of p(x) exactly when p(k) = 0, which is exactly where the graph of y = p(x) meets the x-axis, and for quadratic and cubic polynomials the zeroes are tied to the coefficients by fixed sum-and-product relations derived from the factor theorem.

The core insight

A real number $k$ is a zero of a polynomial $p(x)$ exactly when $p(k)=0$ , which is exactly where the graph of $y=p(x)$ crosses the x-axis, and for a quadratic or cubic polynomial the zeroes are locked to the coefficients by fixed sum-and-product relations proved from the factor theorem.

What this chapter covers

The chapter moves in three connected stages. It begins by recalling polynomials in one variable and their degree, naming linear, quadratic and cubic polynomials by their general forms, and defining precisely what the value of a polynomial and a zero of a polynomial mean. It then turns to the geometrical meaning of zeroes: plotting $y=p(x)$ for linear, quadratic and cubic examples shows that zeroes are nothing but the x-coordinates where the graph meets the x-axis, which immediately explains why a linear polynomial has exactly one zero, a quadratic at most two, and a cubic at most three — and, more generally, why a degree- $n$ polynomial can never have more than $n$ zeroes. The chapter closes with its algebraic centrepiece, the relationship between zeroes and coefficients: for a quadratic polynomial $ax^2+bx+c$ the sum and product of the zeroes come out as simple ratios of $a$ , $b$ , $c$ , and the same pattern extends, with one extra relation, to a cubic polynomial $ax^3+bx^2+cx+d$ . Along the way you revisit splitting the middle term to factorise quadratics, and learn to run the sum-and-product relations backwards to construct a polynomial from its zeroes' sum and product alone.

The core ideas

Polynomials and their degree

A polynomial in one variable is built by adding and subtracting whole-number powers of that variable, each multiplied by a fixed number called a coefficient; the term with no variable attached is the constant term. The highest power present, with a nonzero coefficient, is the polynomial's degree, and degree gives every one-variable polynomial its name.

ax+b \ (\text{degree }1),\qquad ax^2+bx+c \ (\text{degree }2),\qquad ax^3+bx^2+cx+d \ (\text{degree }3), \qquad a \neq 0.

where: the first is a linear polynomial, the second a quadratic polynomial — "quadratic" comes from quadrate, meaning square, since $x^2$ is involved — and the third a cubic polynomial; in every case $a$ , the coefficient of the highest power, cannot be zero, or the degree would drop. Expressions with a variable in a denominator or under a root, such as $\dfrac{1}{x-1}$ or $\sqrt{x+2}$ , are not polynomials at all, no matter how they are rearranged.

The value of a polynomial and the zero of a polynomial

If $p(x)$ is a polynomial and $k$ is any real number, replacing $x$ by $k$ throughout $p(x)$ gives a single number called the value of the polynomial at $x=k$ , written $p(k)$ . For $p(x) = x^2-3x-4$ , the value at $x=2$ is $p(2) = 2^2 - 3(2) - 4 = -6$ , and the value at $x=0$ is $p(0) = -4$ .

p(k) = 0 \iff k \text{ is a zero of } p(x).

where: a zero of the polynomial is precisely a value of $x$ that makes the polynomial evaluate to $0$ ; for $p(x)=x^2-3x-4$ , checking $x=-1$ gives $p(-1) = 1+3-4 = 0$ and checking $x=4$ gives $p(4) = 16-12-4=0$ , so $-1$ and $4$ are both zeroes of this quadratic polynomial. For the simplest case, the linear polynomial $ax+b$ , setting $p(k)=ak+b=0$ and solving gives the zero directly: $k = -b/a$ , or in words, minus the constant term divided by the coefficient of $x$ .

Geometrical meaning of zeroes

Every polynomial $p(x)$ can be graphed as $y=p(x)$ , and the zeroes of the polynomial turn out to have an exact geometric identity: they are the x-coordinates of the points where that graph meets the x-axis. For a linear polynomial $ax+b$ ( $a \neq 0$ ), the graph of $y=ax+b$ is always a straight line, and a straight line that is not itself horizontal or vertical crosses the x-axis at exactly one point, $(-b/a,\,0)$ — matching the algebraic zero found above.

y = 2x+3 \quad \text{crosses the x-axis at} \quad \left(-\tfrac{3}{2},\,0\right).

where: this is the specific case NCERT works through — the line passes through $(-2,-1)$ and $(2,7)$ , and its single x-intercept at $x=-3/2$ is exactly the zero of $2x+3$ found algebraically by solving $2x+3=0$ .

Figure

A straight line y equals two x plus three crossing the x axis at the single zero of the linear polynomial

How many zeroes can a polynomial have

For a quadratic polynomial $ax^2+bx+c$ ( $a\neq 0$ ), the graph of $y=ax^2+bx+c$ is always a parabola, opening upward if $a>0$ and downward if $a<0$ . Three, and only three, things can happen where this curve meets the x-axis, and each case fixes how many real zeroes the quadratic has.

\text{two distinct zeroes} \quad\text{or}\quad \text{two equal (repeated) zeroes} \quad\text{or}\quad \text{no real zero}.

where: the first case is a parabola cutting the x-axis at two separate points; the second is a parabola that just touches the x-axis at its vertex, so the two points of case one coincide into one; the third is a parabola sitting entirely above or entirely below the x-axis, never touching it at all. This is exactly why a quadratic polynomial has at most two zeroes — it can never have three or more, since a parabola cannot cross a straight line more than twice.

Figure

Three parabolas showing a quadratic polynomial with two distinct zeroes two equal zeroes and no real zero

A cubic polynomial's graph extends this pattern one step further: it can weave across the x-axis at up to three points, as with $x^3-4x$ , whose zeroes are $-2$ , $0$ and $2$ . In general, the graph of $y=p(x)$ for a polynomial of degree $n$ meets the x-axis at at most $n$ points, so a polynomial of degree $n$ has at most $n$ zeroes — this single geometric remark covers linear, quadratic, cubic and every higher-degree polynomial at once.

Relationship between zeroes and coefficients — quadratic polynomials

Factorising $2x^2-8x+6$ by splitting the middle term gives $2x^2-6x-2x+6 = 2x(x-3)-2(x-3) = 2(x-1)(x-3)$ , so its zeroes are $1$ and $3$ . Checking the sum and product against the coefficients reveals a pattern that holds for every quadratic polynomial, not just this one.

\alpha+\beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}.

where: $\alpha$ and $\beta$ are the two zeroes of $p(x)=ax^2+bx+c$ ; the sum of zeroes equals minus the coefficient of $x$ divided by the coefficient of $x^2$ , and the product of zeroes equals the constant term divided by the coefficient of $x^2$ . For $2x^2-8x+6$ : sum $=1+3=4=-(-8)/2$ and product $=1\times3=3=6/2$ , both matching the formula exactly; the general proof of why this always works is given in the first derivation below.

Relationship between zeroes and coefficients — cubic polynomials

The same idea extends to a cubic polynomial, with one extra relation to account for the extra zero. For $p(x)=2x^3-5x^2-14x+8$ , the zeroes turn out to be $4$ , $-2$ and $\tfrac{1}{2}$ .

\alpha+\beta+\gamma = -\frac{b}{a}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a}, \qquad \alpha\beta\gamma = -\frac{d}{a}.

where: $\alpha$ , $\beta$ , $\gamma$ are the three zeroes of $p(x)=ax^3+bx^2+cx+d$ ; besides the sum and the product of all three zeroes, there is now a middle relation — the sum of the products of the zeroes taken two at a time. For $2x^3-5x^2-14x+8$ : sum $=4+(-2)+\tfrac12=\tfrac52=-(-5)/2$ , sum of pairwise products $=4(-2)+(-2)(\tfrac12)+(\tfrac12)(4)=-8-1+2=-7=(-14)/2$ , and product $=4\times(-2)\times\tfrac12=-4=-8/2$ — every one of the three formulas checks out.

Derivation ARelationship between the zeroes and coefficients of a quadratic polynomial

Section 2.3 of the chapter observes, from worked examples like $2x^2-8x+6$ and $3x^2+5x-2$ , that the sum and product of a quadratic polynomial's zeroes always come out as tidy ratios of its coefficients. This derivation proves that observation once and for all for the general quadratic $ax^2+bx+c$ with $a \neq 0$ , using nothing beyond the factor theorem and comparing coefficients, so the same two formulas can be trusted for every quadratic you will ever meet, not just the ones checked by hand.

Step 1. Since $\alpha$ and $\beta$ are zeroes of $p(x)=ax^2+bx+c$ , the factor theorem says $x-\alpha$ and $x-\beta$ are factors of $p(x)$ , so $ax^2+bx+c = k(x-\alpha)(x-\beta)$ for some nonzero constant $k$ .

Step 2. Expand the right-hand side: $k(x-\alpha)(x-\beta) = k\left[x^2-(\alpha+\beta)x+\alpha\beta\right] = kx^2 - k(\alpha+\beta)x + k\alpha\beta$ .

Step 3. Compare the coefficients of $x^2$ , $x$ , and the constant term on both sides of $ax^2+bx+c = kx^2-k(\alpha+\beta)x+k\alpha\beta$ : matching gives $a=k$ , $b=-k(\alpha+\beta)$ , and $c=k\alpha\beta$ .

Step 4. Solve for the sum and product using $a=k$ . From $b=-k(\alpha+\beta)=-a(\alpha+\beta)$ , dividing by $-a$ gives $\alpha+\beta=-\dfrac{b}{a}$ . From $c=k\alpha\beta=a\alpha\beta$ , dividing by $a$ gives $\alpha\beta=\dfrac{c}{a}$ . These are exactly the two relations used throughout the chapter, and they hold for every quadratic polynomial with real coefficients, whether its zeroes are rational, irrational, or not real at all — a quick check against Example 2's zeroes $-2,-5$ of $x^2+7x+10$ gives sum $-7=-7/1$ and product $10=10/1$ , both matching.

Go deeper · Derivation B · 1 minRelationship between the zeroes and coefficients of a cubic polynomial

Derivation BRelationship between the zeroes and coefficients of a cubic polynomial

The chapter states, and checks only for the specific case $2x^3-5x^2-14x+8$ , that a cubic polynomial $ax^3+bx^2+cx+d$ with zeroes $\alpha,\beta,\gamma$ obeys three linked relations. This derivation supplies the missing general argument, extending the same factor-theorem-and-compare-coefficients method used for the quadratic to one extra factor, so that any cubic's zero-coefficient relationship rests on a proved rule rather than a pattern spotted in a single example.

Step 1. Since $\alpha,\beta,\gamma$ are zeroes of $p(x)=ax^3+bx^2+cx+d$ , the factor theorem gives $ax^3+bx^2+cx+d = k(x-\alpha)(x-\beta)(x-\gamma)$ for some nonzero constant $k$ .

Step 2. Multiply the first two factors: $(x-\alpha)(x-\beta) = x^2-(\alpha+\beta)x+\alpha\beta$ . Multiplying this by the third factor $(x-\gamma)$ gives $x^3-(\alpha+\beta)x^2+\alpha\beta x - \gamma x^2 + \gamma(\alpha+\beta)x - \alpha\beta\gamma$ , which regroups as $x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma$ .

Step 3. Multiply this whole expression by $k$ : $kx^3 - k(\alpha+\beta+\gamma)x^2 + k(\alpha\beta+\beta\gamma+\gamma\alpha)x - k\alpha\beta\gamma$ . Comparing this with $ax^3+bx^2+cx+d$ term by term gives $a=k$ , $b=-k(\alpha+\beta+\gamma)$ , $c=k(\alpha\beta+\beta\gamma+\gamma\alpha)$ , and $d=-k\alpha\beta\gamma$ .

Step 4. Divide each equation by $a=k$ to isolate the symmetric combinations of zeroes: $\alpha+\beta+\gamma=-\dfrac{b}{a}$ , $\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{c}{a}$ , and $\alpha\beta\gamma=-\dfrac{d}{a}$ . Checking against $2x^3-5x^2-14x+8$ with zeroes $4,-2,\tfrac12$ : sum $=\tfrac52=-(-5)/2$ , pairwise sum $=-7=(-14)/2$ , and product $=-4=-8/2$ — all three formulas confirmed.

Go deeper · Derivation C · 1 minBuilding a quadratic polynomial from a given sum and product of zeroes

Derivation CBuilding a quadratic polynomial from a given sum and product of zeroes

Example 4 of the chapter shows one instance of running the sum-and-product relations backwards, building a quadratic polynomial whose zeroes have sum $-3$ and product $2$ . This derivation generalises that single worked case into a template that works for any given sum and product directly, without reinventing the factoring argument each time, and it explains why the answer is never unique but always a family of polynomials differing only by an overall constant multiple.

Step 1. From the first derivation above, any quadratic polynomial with zeroes $\alpha$ and $\beta$ can be written as $p(x) = k\left[x^2-(\alpha+\beta)x+\alpha\beta\right]$ for a nonzero constant $k$ . If we are only told the required sum $s=\alpha+\beta$ and product $q=\alpha\beta$ , the individual zeroes $\alpha$ and $\beta$ are not needed at all — only these two numbers.

Step 2. Substitute the given sum and product directly into the template: $p(x) = k\left(x^2-sx+q\right)$ . This formula already answers the question in full generality; taking the simplest choice $k=1$ gives one valid polynomial, $p(x)=x^2-sx+q$ .

Step 3. Apply the template to Example 4's data, $s=-3$ and $q=2$ : $p(x)=x^2-(-3)x+2=x^2+3x+2$ , matching the textbook's own answer exactly.

Step 4. Explain why the answer is a whole family, not one polynomial: since $k$ can be any nonzero real number, $k(x^2-sx+q)$ describes infinitely many polynomials sharing the same pair of zeroes, differing only by scale — for instance $2x^2+6x+4$ or $-x^2-3x-2$ both have zeroes summing to $-3$ and multiplying to $2$ . This is exactly why NCERT states that any quadratic polynomial fitting the given sum and product "will be of the form $k(x^2+3x+2)$ ."

Worked examples

Board level - zeroes of a quadratic by splitting the middle term (Example 2 pattern)

Find the zeroes of the quadratic polynomial $x^2+7x+10$ , and verify the relationship between the zeroes and the coefficients.

Step 1. Split the middle term $7x$ as a sum of two terms whose product is $1\times x^2 \times 10 = 10x^2$ : the pair $2x$ and $5x$ works, since $2\times5=10$ . So $x^2+7x+10 = x^2+2x+5x+10$ .

Step 2. Factorise by grouping: $x^2+2x+5x+10 = x(x+2)+5(x+2) = (x+2)(x+5)$ .

Step 3. The polynomial is zero when $x+2=0$ or $x+5=0$ , giving $x=-2$ or $x=-5$ .

Step 4. Verify against the coefficients $a=1$ , $b=7$ , $c=10$ : sum of zeroes $=-2+(-5)=-7=-\dfrac{7}{1}=-\dfrac{b}{a}$ , and product of zeroes $=(-2)(-5)=10=\dfrac{10}{1}=\dfrac{c}{a}$ . Answer: the zeroes are -2 and -5, and both relations check out.

Board level - zeroes of a quadratic with irrational roots (Example 3 pattern)

Find the zeroes of the polynomial $x^2-3$ , and verify the relationship between the zeroes and the coefficients.

Step 1. Recognise the difference-of-squares identity $a^2-b^2=(a-b)(a+b)$ with $a=x$ and $b=\sqrt{3}$ , since $3=(\sqrt3)^2$ : $x^2-3 = x^2-(\sqrt3)^2 = (x-\sqrt3)(x+\sqrt3)$ .

Step 2. The polynomial is zero when $x-\sqrt3=0$ or $x+\sqrt3=0$ , giving $x=\sqrt3$ or $x=-\sqrt3$ .

Step 3. Verify against the coefficients $a=1$ , $b=0$ (there is no $x$ term), $c=-3$ : sum of zeroes $=\sqrt3+(-\sqrt3)=0=-\dfrac{0}{1}=-\dfrac{b}{a}$ .

Step 4. Product of zeroes $=\sqrt3\times(-\sqrt3)=-3=\dfrac{-3}{1}=\dfrac{c}{a}$ . Answer: the zeroes are √3 and -√3, and the relations hold even though the zeroes are irrational.

Board HOTS / foundation JEE-NEET - cubic zeroes and a symmetric-function shortcut (Example 5 pattern)

Verify that $3$ , $-1$ and $-\tfrac13$ are the zeroes of the cubic polynomial $p(x)=3x^3-5x^2-11x-3$ , verify the relationship between the zeroes and the coefficients, and then find $\alpha^2+\beta^2+\gamma^2$ without squaring each zero separately.

Step 1. Check each candidate directly: $p(3)=3(27)-5(9)-11(3)-3=81-45-33-3=0$ ; $p(-1)=3(-1)-5(1)-11(-1)-3=-3-5+11-3=0$ ; $p(-\tfrac13)=3(-\tfrac1{27})-5(\tfrac19)-11(-\tfrac13)-3=-\tfrac19-\tfrac59+\tfrac{11}3-3=-\tfrac23+\tfrac{11}3-3=3-3=0$ . All three are confirmed zeroes.

Step 2. With $a=3$ , $b=-5$ , $c=-11$ , $d=-3$ , compute the three coefficient ratios: $\alpha+\beta+\gamma=-\dfrac{b}{a}=\dfrac53$ ; $\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{c}{a}=-\dfrac{11}3$ ; $\alpha\beta\gamma=-\dfrac{d}{a}=1$ .

Step 3. Match these against direct computation from $3,-1,-\tfrac13$ : sum $=3-1-\tfrac13=\tfrac53$ ✓; pairwise sum $=3(-1)+(-1)(-\tfrac13)+(-\tfrac13)(3)=-3+\tfrac13-1=-\tfrac{11}3$ ✓; product $=3(-1)(-\tfrac13)=1$ ✓ — every relation confirmed without needing a fourth computation.

Step 4. Now use the identity $\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)$ , built purely from the two relations already found in Step 2: $\left(\dfrac53\right)^2-2\left(-\dfrac{11}3\right)=\dfrac{25}9+\dfrac{22}3=\dfrac{25}9+\dfrac{66}9=\dfrac{91}9$ . A direct check confirms it: $3^2+(-1)^2+\left(-\tfrac13\right)^2=9+1+\tfrac19=\dfrac{91}9$ . Answer: the zeroes are 3, -1, -1/3, and α² + β² + γ² = 91/9. Timing note: budget about three minutes. The shortcut is the whole point — once you have the sum and the pairwise-product sum from the coefficient relations, $\alpha^2+\beta^2+\gamma^2$ needs no individual squaring at all, which is exactly how this identity is used in competitive-exam polynomial questions where the zeroes themselves are never asked for. The trap is sign errors: $\alpha\beta+\beta\gamma+\gamma\alpha$ already carries whatever sign $c/a$ has, so doubling it and subtracting means double-checking the sign twice, not once.

Common traps

The most common slip is confusing the value of a polynomial with its zero — $p(0)$ is simply the constant term, not automatically the answer to "find the zero," and a zero can be any real number $k$ with $p(k)=0$ , not necessarily $x=0$ itself. Many students also assume a quadratic polynomial must always have exactly two real zeroes, forgetting Case (iii): a parabola sitting entirely above or below the x-axis has no real zero at all, and Case (ii) gives only one distinct value when the two zeroes coincide. Sign errors are rampant in the coefficient relations: writing $\alpha+\beta = b/a$ instead of $-b/a$ , or forgetting the extra minus sign in the cubic's product relation $\alpha\beta\gamma = -d/a$ , are the single most frequent marks lost in this chapter. A related error is misreading which coefficient goes where once a polynomial is not written with $x$ terms in descending order — always rewrite in standard form $ax^n+\cdots$ before reading off $a$ , $b$ , $c$ (and $d$ ). Finally, students often forget that "degree $n$ has at most $n$ zeroes" is a ceiling, not a guarantee — a cubic can have just one real zero, with the other two not real, and that is still entirely consistent with the theorem.

You should be able to…

State the general form and degree of a linear, quadratic, and cubic polynomial.
Compute the value of a polynomial $p(x)$ at a given $x=k$ , and decide whether $k$ is a zero.
Find the zero of a linear polynomial $ax+b$ directly from its coefficients.
Explain the geometrical meaning of a zero as an x-intercept of $y=p(x)$ .
Describe the three possible cases for how a parabola meets the x-axis, and connect each to the number of real zeroes of a quadratic.
State and prove the relationship $\alpha+\beta=-b/a$ , $\alpha\beta=c/a$ for a quadratic polynomial.
State and verify the three relationships between the zeroes and coefficients of a cubic polynomial.
Factorise a quadratic polynomial by splitting the middle term and read off its zeroes.
Construct a quadratic polynomial from a given sum and product of zeroes.
Recognise that a degree- $n$ polynomial has at most $n$ zeroes, and that this is a ceiling, not a guarantee.

Ask the tutor: Want to work through building a quadratic polynomial from a given sum and product of zeroes, step by step, then check it against the factor theorem?

Why is a zero called a zero if the value of x itself usually isn't zero?

The name refers to the output, not the input. A zero of $p(x)$ is a value of $x$ that makes the polynomial's value equal to zero, that is, $p(k)=0$ — the input $k$ can be any real number, positive, negative, fractional or irrational, like $\sqrt3$ in this chapter's worked example. It is called a zero of the polynomial because substituting it makes the whole expression vanish, not because the number itself is zero. Only in special cases, such as $p(x)=x^2-x$ , does $x=0$ happen to also be a zero.

Can a polynomial ever have more zeroes than its degree?

No — this is one of the chapter's guaranteed ceilings. The graph of $y=p(x)$ for a degree- $n$ polynomial can meet the x-axis at, at most, $n$ points, so a linear polynomial has at most one zero, a quadratic at most two, and a cubic at most three. A quadratic can have fewer real zeroes than its degree allows — one repeated zero, or none at all — but it can never have three. If you ever compute three different real solutions for a quadratic equation, that is a sure sign of an arithmetic error somewhere upstream.

Why does the sum-of-zeroes formula have a minus sign but the product formula doesn't?

Both signs fall straight out of comparing coefficients after expanding $k(x-\alpha)(x-\beta) = kx^2 - k(\alpha+\beta)x + k\alpha\beta$ . The $x$ -term naturally carries a minus sign because it comes from $-(\alpha+\beta)$ inside the factored form, while the constant term comes from $+\alpha\beta$ with no extra sign attached. Matching coefficients then forces $b=-k(\alpha+\beta)$ , so the sum picks up a minus sign when solved for, while $c=k\alpha\beta$ leaves the product sign-free. It is not an arbitrary rule to memorise — it is a direct, checkable consequence of multiplying out two linear factors, proved in full in the first derivation.

My quadratic has no real zeroes at all - does that mean I made a mistake?

Not necessarily. A quadratic polynomial whose parabola lies entirely above or entirely below the x-axis genuinely has no real zero — this is Case (iii) from the geometrical picture, just as valid as the other two cases. Before assuming an error, sketch or reason about the shape: if the parabola opens upward ( $a>0$ ) and its lowest point (vertex) still sits above the x-axis, or it opens downward and its highest point sits below the x-axis, no real zero exists, and that is the correct final answer, not a sign that something went wrong in your factorisation.