Polynomials

A polynomial in $x$ is an expression of the form $p(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0,$ where the $a_i$ are real numbers, $a_n\ne 0$ and $n$ is a non-negative integer. The degree is the highest power of $x$ that appears.

What CBSE examines here (2026-27). Zeroes of a polynomial, their geometric meaning, and the relationship between zeroes and coefficients of quadratic polynomials. The division algorithm for polynomials was removed from this chapter — it is in the appendix at the end, marked as background only.

1. Zeroes of a Polynomial

A real number $k$ is a zero of $p(x)$ if $p(k)=0$.

Geometrically, the zeroes are the $x$-coordinates of the points where the graph $y=p(x)$ meets the $x$-axis. That is the whole idea, and it is worth saying precisely why: $p(x)=0$ means the height of the curve is zero, and height zero is the $x$-axis.

• A linear polynomial has exactly one zero — a straight line that is not horizontal crosses the axis once.
• A quadratic has at most two. A parabola can cut the axis twice, touch it once, or miss it entirely.
• A polynomial of degree $n$ has at most $n$ zeroes.

Two traps live here. First, students count crossings of the y-axis; those are never zeroes. Second, a curve that touches the axis and turns back still counts as meeting it — that is a repeated zero, not a missing one.

Reading a graph. The parabola opens upward when the coefficient of $x^2$ is positive and downward when it is negative. So a downward parabola sitting entirely below the axis has no real zeroes, and you can say so without any algebra.

2. Relationship Between Zeroes and Coefficients

For a quadratic $p(x)=a{x}^2+bx+c$ with zeroes $\alpha$ and $\beta$:

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

Note the minus sign on the sum — it is the single most common slip in this chapter. Sanity-check it on something you can factorise by eye: $(x-2)(x-3)=x^2-5x+6$ has zeroes 2 and 3, sum 5, and $-b/a=5$. The 5 appears in the polynomial as $-5x$.

Going backwards — building a polynomial

If you know the sum $S$ and product $P$ of the zeroes, then

$p(x)=k\left[x^2-Sx+P\right]$

for any non-zero constant $k$. The $k$ is not a detail to ignore: scaling a polynomial does not move its zeroes, so there are infinitely many correct answers. Pick $k$ to clear fractions and you get the tidy one.

Worked example — zeroes 3 and $-2$. Sum $=1$, product $=-6$, so $p(x)=x^2-x-6$.

Worked example — sum $\frac{1}{4}$, product $-1$. $x^2-\frac{1}{4}x-1$, and multiplying by 4 gives $4x^2-x-4$.

Using the relationships without finding the zeroes

Many board questions never ask you to solve anything — they hand you the sum and product and test whether you can manipulate them.

Example. If $\alpha ,\beta$ are zeroes of $x^2-5x+k$ and $\alpha -\beta =1$, find $k$.

From the coefficients, $\alpha +\beta =5$. With $\alpha -\beta =1$, adding gives $2\alpha =6$, so $\alpha =3$ and $\beta =2$. Then $k=\alpha \beta =6$.

Example. If $\alpha ,\beta$ are zeroes of $x^2-px-c$, find $(\alpha +1)(\beta +1)$.

Expand first: $(\alpha +1)(\beta +1)=\alpha \beta +(\alpha +\beta )+1$. Now substitute $\alpha \beta =-c$ and $\alpha +\beta =p$ to get $1+p-c$. No zeroes were ever computed — expanding before substituting is the trick.

Key results to memorise

• Zeroes of $p(x)$ = $x$-coordinates where $y=p(x)$ meets the $x$-axis.
• Degree $n$ ⇒ at most $n$ zeroes.
• Quadratic: $\alpha +\beta =-b/a$ and $\alpha \beta =c/a$.
• Building from zeroes: $p(x)=k[x^2-(\text{sum})x+(\text{product})]$.
• Discriminant $b^2-4ac$: positive ⇒ two distinct zeroes, zero ⇒ one repeated zero, negative ⇒ no real zeroes.

Practice (try before checking)

1. Find the zeroes of $4x^2-4x-3$ and verify both relationships.
2. Find a quadratic polynomial whose zeroes have sum $-3$ and product $2$.
3. If $\alpha ,\beta$ are the zeroes of $x^2-6x+8$, find $\frac{1}{\alpha }+\frac{1}{\beta }$ without solving for the zeroes.
4. For what value of $k$ does $x^2-kx+9$ have exactly one (repeated) zero?
5. A parabola $y=p(x)$ opens downward and lies entirely below the $x$-axis. How many real zeroes does $p(x)$ have?

Answers

1. Split the middle term with $-6$ and $+2$: $4x^2-6x+2x-3=(2x-3)(2x+1)$, so the zeroes are $\frac{3}{2}$ and $-\frac{1}{2}$. Sum $=1=-(-4)/4$ ✓; product $=-\frac{3}{4}=-3/4$ ✓.
2. $x^2-(-3)x+2=x^2+3x+2$. Check by factorising: $(x+1)(x+2)$ has zeroes $-1,-2$, sum $-3$ ✓, product 2 ✓.
3. $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{6}{8}=\frac{3}{4}$.
4. One repeated zero means $b^2-4ac=0$, so $k^2=36$ and $k=\pm 6$.
5. None. It never meets the $x$-axis, so there is no $x$ with $p(x)=0$.

Common mistakes

1. Dropping the minus sign. The sum is $-b/a$, not $b/a$. Rebuild it from $(x-2)(x-3)$ whenever you doubt it.
2. Not writing the polynomial in standard form first. For $6x^2-3-7x$ you must reorder to $6x^2-7x-3$ before reading off $a$, $b$, $c$ — otherwise you will use $b=-3$.
3. Counting y-axis crossings as zeroes. Only the $x$-axis matters.
4. Forgetting that a repeated zero counts twice. For $(2s-1{)}^2$ the zeroes are $\frac{1}{2}$ and $\frac{1}{2}$, so the sum is 1, not $\frac{1}{2}$.
5. Writing $x^2+Sx+P$ when building a polynomial. The middle sign is negative.

CBSE exam focus

The chapter is small and carries fewer marks than its neighbours, but the coefficient relationships reappear inside Quadratic Equations — so time spent here pays twice.

Quick self-test

1. How many zeroes does a polynomial of degree 5 have at most? (Five.)
2. Zeroes of $x^2-15$? (±√15 — a difference of two squares.)
3. Build a quadratic with sum 0 and product $\sqrt{5}$. (Sum 0 kills the middle term: $x^2+\sqrt{5}$.)
4. Zeroes of $4u^2+8u$? (Factor out $4u$: 0 and −2.)
5. If $\alpha \beta =-\frac{4}{3}$ and $\alpha +\beta =\frac{1}{3}$, name a possible polynomial. ($3x^2-x-4$.)

Appendix — beyond the current syllabus

Not examinable in CBSE 2026-27. The division algorithm for polynomials was removed from this chapter during rationalisation, and the coefficient relationships for cubic polynomials go beyond the quadratic case the syllabus asks for. Both are kept here because they are genuinely useful later — polynomial division underpins the factor theorem and turns up again in higher classes — but do not budget board-exam time for them.

The division algorithm for polynomials

For polynomials $p(x)$ and $g(x)$ with $g(x)\ne 0$, there are unique polynomials $q(x)$ and $r(x)$ with

$p(x)=g(x)q(x)+r(x),r(x)=0\text{ or }\mathrm{deg}r(x)<\mathrm{deg}g(x).$

It is the polynomial version of integer division: quotient and remainder, with the remainder forced to be "smaller" than the divisor — smaller in degree rather than in size.

Example. Dividing $3x^3+x^2+2x+5$ by $x^2+2x+1$ gives

$3x^3+x^2+2x+5=(x^2+2x+1)(3x-5)+(9x+10),$

and $\mathrm{deg}(9x+10)=1<2=\mathrm{deg}g(x)$, as required.

Zeroes and coefficients of a cubic

For $p(x)=a{x}^3+b{x}^2+cx+d$ with zeroes $\alpha ,\beta ,\gamma$:

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

Note the alternating signs. For $x^3-2x^2-5x+6$, whose zeroes are 1, −2 and 3: the sum is 2 and $-b/a=2$ ✓, while the product is −6 and $-d/a=-6$ ✓.

Real Numbers supplies the factorisation habits this chapter leans on. Next, Pair of Linear Equations in Two Variables moves from one unknown to two.