SSC CGL Maths · Polynomials

Values, Remainders and Roots

The value of a polynomial at a number answers most questions: the remainder on division by x − a is the value at a, the sum of the roots can be read from two coefficients, and a long product of brackets can collapse to two terms.

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Why this matters

Six questions in the bank, four EASY and two MODERATE: three from the 2023 shifts, two from 2024 and one from 2025. Three of the six are the remainder theorem, which turns a long division into one substitution. The others are a careful substitution with negative numbers, a sum of roots read off the coefficients, and a product that folds up by the difference of squares.

Concept 1 of 4: Value of a polynomial at given numbers

To find the value, put the numbers in, one term at a time. The only danger is the signs: a negative number to an odd power stays negative, to an even power it turns positive. When every term has the same degree, there is a shortcut that turns a sign question into no work at all.

Definition

The degree of a term is the sum of its powers: $x^2y^3$ has degree 5. The degree of a polynomial is the highest degree among its terms.
Put the values in term by term and write each term's sign before adding: $(-a)^{\text{odd}} = -a^{\text{odd}}$ , $(-a)^{\text{even}} = a^{\text{even}}$ .
Same-degree shortcut: if every term has degree d, then changing the sign of every variable multiplies the value by $(-1)^d$ . So the value at $(-x, -y)$ is minus the value at $(x, y)$ when d is odd, and the same when d is even.
The shortcut is safe only when every term has the same degree. One constant term, or one term of a different degree, breaks it.

Same-degree sign rule

P(-x, -y) = (-1)^d \, P(x, y)

dthe degree shared by every term of P

Worked example

Find the value of

2x^3 - 3x^2y + 4y^3

when

x = 2

and

y = -1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 17 Jul 2023, Shift 4 · Q75Moderate

Example 1 · Polynomials · Values, Remainders and Roots

What is the value of

(27x^3 - 58x^2y + 31xy^2 - 8y^3)

, when

x = -5

and

y = -7

Losing the sign of an odd power

(-7)^3 = -343

and

(-5)^3 = -125

, but

(-7)^2 = 49

. Write each term's sign before you add; the wrong options are usually the right number with the wrong sign.

Using the sign rule when the degrees differ

P(-x, -y) = -P(x, y)

needs every term to have the same odd degree. For

x^3 + 1

the value at

x = 2

is 9, but at

x = -2

it is

-7

, not

-9

: the constant term does not change sign.

Concept 2 of 4: Remainder theorem and factor theorem

Dividing p(x) by x − a leaves a quotient and a number left over: p(x) = (x − a) × q(x) + r. Put x = a and the first part becomes zero, so what is left is r. The remainder is simply the value of the polynomial at a.

Definition

Remainder theorem: the remainder when p(x) is divided by $x - a$ is $p(a)$ . For $x + a$ use $p(-a)$ ; for $ax - b$ use $p\left(\frac{b}{a}\right)$ .
Factor theorem: $x - a$ is a factor of p(x) exactly when $p(a) = 0$ .
To make p(x) divisible by $x - a$ , subtract the remainder $p(a)$ (or add $-p(a)$ ).
Powers of x: $x^n - 1$ always has the factor $x - 1$ . $x^n + 1$ has the factor $x + 1$ when n is odd, and $x^n - 1$ has it when n is even.
Fast route: put the root of the divisor into p(x). No long division is ever needed for the remainder alone.

Remainder and factor theorems

p(x) \div (x - a) \Rightarrow \text{remainder} = p(a) \qquad (x - a) \text{ is a factor} \Leftrightarrow p(a) = 0

Worked example

Find the remainder when

x^3 + 2x^2 - 5x + 1

is divided by

x - 3

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 19 Sep 2024, Shift 1 · Q63Easy

Example 2 · Polynomials · Values, Remainders and Roots

What number should be subtracted from

x^3 - 4x^2 - 8x + 11

to make the number divisible by

(x + 2)

Putting x = a for the divisor x + a

The remainder on division by

x + a

p(-a)

, the value where the divisor is zero. For

x + 2

, put

x = -2

, not 2.

Subtracting the wrong number to make a polynomial divisible

Subtract the remainder,

p(a)

, not the number in the divisor. If dividing by

x + 4

leaves remainder 5, subtracting 5 makes it divisible; subtracting 4 does not.

Assuming x + 1 divides every xⁿ + 1

x = -1

x^n + 1

0

only when n is odd. For even n it is 2, so

x^{10} + 1

leaves remainder 2 on division by

x + 1

Drill 2 more on remainder theorem and factor theorem

Concept 3 of 4: Sum and product of the roots from the coefficients

If a polynomial has roots r and s, it is a multiple of (x − r)(x − s). Multiply that out and the roots appear in the coefficients: their sum sits in the x term and their product in the constant. So you can read the sum and product without solving.

Definition

Quadratic $ax^2 + bx + c = 0$ : sum of roots $= -\frac{b}{a}$ , product $= \frac{c}{a}$ .
Cubic $ax^3 + bx^2 + cx + d = 0$ with roots $p, q, r$ : $p + q + r = -\frac{b}{a}$ , $pq + qr + rp = \frac{c}{a}$ , $pqr = -\frac{d}{a}$ .
The signs alternate: minus, plus, minus. Always divide by the leading coefficient a.
A repeated root is counted each time it repeats.

Roots and coefficients

ax^2 + bx + c: \ \alpha + \beta = -\frac{b}{a},\ \alpha\beta = \frac{c}{a} \qquad ax^3 + bx^2 + cx + d: \ p + q + r = -\frac{b}{a},\ pq + qr + rp = \frac{c}{a},\ pqr = -\frac{d}{a}

Worked example

The roots of

2x^3 - 6x^2 + 4x - 18 = 0

are p, q and r. Find

p + q + r

and

pqr

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 1 · Q57Easy

Example 3 · Polynomials · Values, Remainders and Roots

If the cubic equation

a^3 - 4a^2 + 5a - 2 = 0

has the root

x

y

z

then find the value of

x + y + z

Reading the sum of roots with the coefficient's own sign

The sum of the roots is

-\frac{b}{a}

. For

x^2 - 13x + 40 = 0

it is 13, not

-13

Forgetting to divide by the leading coefficient

For

2x^2 - 8x + 3 = 0

the sum of the roots is

\frac{8}{2} = 4

, not 8. The rules hold for

ax^2 + bx + c

only after dividing by a.

Concept 4 of 4: A product of brackets that collapses by the difference of squares

Multiply (1 − p) by (1 + p) and you get 1 − p². Multiply that by (1 + p²) and you get 1 − p⁴. Each new bracket meets exactly the difference it needs, so a long chain folds into just two terms.

Definition

Difference of squares: $(a - b)(a + b) = a^2 - b^2$ .
Applied again and again: $(a - b)(a + b)(a^2 + b^2)(a^4 + b^4) \cdots (a^{2^{k-1}} + b^{2^{k-1}}) = a^{2^k} - b^{2^k}$ . The power doubles at each step: 2, 4, 8, 16, 32.
The chain needs a minus bracket to start it. If it is missing, multiply and divide by it.
Check an option by putting in a number that makes the options differ. $p = 0$ often gives the same value for several options; $p = 1$ or $p = 2$ usually separates them.

Difference-of-squares chain

(a - b)(a + b) = a^2 - b^2 \qquad (1 - p)(1 + p)(1 + p^2)(1 + p^4) = 1 - p^8

Worked example

Simplify

(1 + x)(1 + x^2)(1 + x^4)(1 - x)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 21 Jul 2023, Shift 3 · Q51Moderate

Example 4 · Polynomials · Values, Remainders and Roots

Simplify the following expression.

[(1 + p)(1 + p^2)(1 + p^4)(1 + p^8)(1 + p^{16})(1 - p) - 1]

Trying to collapse a chain with no minus bracket

(1 + p)(1 + p^2)(1 + p^4)

does not fold on its own. It equals

\frac{1 - p^8}{1 - p}

: put in the missing

(1 - p)

and divide by it.

Adding the exponents instead of doubling them

Each step squares the last power:

1 - p^2

1 - p^4

1 - p^8

. A chain ending in

(1 + p^{16})

gives

1 - p^{32}

, not

1 - p^{31}

, which comes from adding the powers 1, 2, 4, 8 and 16.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Value of a polynomial at given numbers
Same-degree sign rule
$P(-x, -y) = (-1)^d \, P(x, y)$
Remainder theorem and factor theorem
Remainder and factor theorems
$p(x) \div (x - a) \Rightarrow \text{remainder} = p(a) \qquad (x - a) \text{ is a factor} \Leftrightarrow p(a) = 0$
Sum and product of the roots from the coefficients
Roots and coefficients
$ax^2 + bx + c: \ \alpha + \beta = -\frac{b}{a},\ \alpha\beta = \frac{c}{a} \qquad ax^3 + bx^2 + cx + d: \ p + q + r = -\frac{b}{a},\ pq + qr + rp = \frac{c}{a},\ pqr = -\frac{d}{a}$
A product of brackets that collapses by the difference of squares
Difference-of-squares chain
$(a - b)(a + b) = a^2 - b^2 \qquad (1 - p)(1 + p)(1 + p^2)(1 + p^4) = 1 - p^8$

Watch out for (9)

Losing the sign of an odd power
→ Value of a polynomial at given numbers
Using the sign rule when the degrees differ
→ Value of a polynomial at given numbers
Putting x = a for the divisor x + a
→ Remainder theorem and factor theorem
Subtracting the wrong number to make a polynomial divisible
→ Remainder theorem and factor theorem
Assuming x + 1 divides every xⁿ + 1
→ Remainder theorem and factor theorem
Reading the sum of roots with the coefficient's own sign
→ Sum and product of the roots from the coefficients
Forgetting to divide by the leading coefficient
→ Sum and product of the roots from the coefficients
Trying to collapse a chain with no minus bracket
→ A product of brackets that collapses by the difference of squares
Adding the exponents instead of doubling them
→ A product of brackets that collapses by the difference of squares

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