Class 10Maths · AlgebraFull chapter

Polynomials

The whole chapter in one place — read the notes, plot a polynomial to see its zeroes on the graph, then test yourself. Clear notes, the key results in one sheet, an interactive grapher, practice and worked NCERT solutions.

In this chapter
Polynomials and Their ZeroesGeometrical Meaning of the Zeroes of a PolynomialRelationship Between Zeroes and Coefficients of a Quadratic PolynomialRelationship Between Zeroes and Coefficients of a Cubic PolynomialDivision Algorithm for Polynomials

Polynomials and Their Zeroes

Quick answer A polynomial's degree is its highest power of x, and a zero of a polynomial is any value of x that makes the polynomial equal to 0.

An algebraic expression of the form p(x) = anxn + ... + a1x + a0, where the powers of x are whole numbers, is a polynomial in one variable x. The highest power of x is called its degree.

  • A polynomial of degree 1, like 2x + 3, is linear.
  • A polynomial of degree 2, like x2 - 3x - 4, is quadratic.
  • A polynomial of degree 3, like x3 - 6x2 + 11x - 6, is cubic.

A real number k is a zero of the polynomial p(x) if p(k) = 0. To find zeroes, we set p(x) = 0 and solve.

Worked example. Take p(x) = x2 - 3x - 4. Setting p(x) = 0 and factorising: x2 - 3x - 4 = (x - 4)(x + 1) = 0. So x = 4 or x = -1. Check: p(4) = 16 - 12 - 4 = 0 and p(-1) = 1 + 3 - 4 = 0. Hence the zeroes are 4 and -1.

A key fact: a polynomial of degree n has at most n zeroes. So a linear polynomial has exactly one zero, a quadratic has at most two, and a cubic has at most three.

Zero of a polynomial p(k) = 0 k is a zero if substituting it makes the polynomial 0
Zero of a linear polynomial For p(x) = ax + b, zero = -b/a a ≠ 0; exactly one zero
Maximum number of zeroes number of zeroes ≤ degree n quadratic ≤ 2, cubic ≤ 3
Remember
  • Degree = the highest power of x in the polynomial.
  • Linear (degree 1), quadratic (degree 2), cubic (degree 3).
  • k is a zero of p(x) when p(k) = 0.
  • A degree-n polynomial has at most n zeroes.
  • Find zeroes by solving p(x) = 0, often by factorisation.

Geometrical Meaning of the Zeroes of a Polynomial

Quick answer The zeroes of a polynomial are exactly the x-coordinates of the points where its graph meets (cuts or touches) the x-axis.

Every polynomial can be drawn as a graph of y = p(x). Wherever the graph meets the x-axis, the value of y is 0, so at those points p(x) = 0. Therefore the x-coordinates of these meeting points are precisely the zeroes of the polynomial.

  • The graph of a linear polynomial y = ax + b is a straight line that cuts the x-axis at exactly one point, so it has one zero.
  • The graph of a quadratic polynomial is a parabola (opening upward when a > 0, downward when a < 0). It can cut the x-axis at two points (two distinct zeroes), touch it at one point (two equal zeroes), or not meet it at all (no real zero).
  • The graph of a cubic polynomial meets the x-axis at most three points, giving at most three zeroes.

Worked example. Consider p(x) = x2 - 3x - 4, whose zeroes we found to be 4 and -1. Its graph is an upward parabola that crosses the x-axis at the two points (-1, 0) and (4, 0). Those two x-coordinates, -1 and 4, are the zeroes. If instead a parabola only just touches the x-axis at one point, both zeroes are equal (as in y = x2, which touches at the origin, giving the zero 0 twice).

This is why the number of zeroes can never exceed the degree: a straight line, a parabola, or a cubic curve can only cross the x-axis a limited number of times.

Graph meets x-axis y = p(x) = 0 at the meeting points the x-coordinates there are the zeroes
Quadratic graph parabola: opens up if a > 0, down if a < 0 0, 1, or 2 x-intercepts
Number of x-intercepts number of real zeroes ≤ degree line 1, parabola ≤ 2, cubic ≤ 3
Remember
  • Zeroes are the x-coordinates where the graph meets the x-axis.
  • A linear graph (line) meets the x-axis at exactly one point.
  • A quadratic graph (parabola) meets the x-axis at 0, 1, or 2 points.
  • Touching the x-axis at one point means two equal zeroes.
  • A cubic graph meets the x-axis in at most three points.
See it on the graph. Open the function grapher — plot a polynomial and watch its zeroes appear exactly where the curve cuts the x-axis. Open the grapher

Relationship Between Zeroes and Coefficients of a Quadratic Polynomial

Quick answer For a quadratic ax² + bx + c with zeroes α and β, the sum of the zeroes is -b/a and the product is c/a.

Let p(x) = ax2 + bx + c (with a ≠ 0) have zeroes α and β. Then p(x) can be written in factor form as a(x - α)(x - β). Expanding gives a[x2 - (α + β)x + αβ] = ax2 - a(α + β)x + aαβ.

Comparing this with ax2 + bx + c, we match coefficients: -a(α + β) = b and aαβ = c. Dividing by a gives the two standard results:

  • Sum of zeroes: α + β = -b/a = -(coefficient of x)/(coefficient of x2).
  • Product of zeroes: αβ = c/a = (constant term)/(coefficient of x2).

Worked example. Take p(x) = x2 - 3x - 4, so a = 1, b = -3, c = -4. The relations predict α + β = -(-3)/1 = 3 and αβ = -4/1 = -4. We already know the actual zeroes are 4 and -1: their sum is 4 + (-1) = 3 and their product is 4 × (-1) = -4. Both match, confirming the results.

These relations also let us build a quadratic from its zeroes. If the sum of zeroes is S and the product is P, then a quadratic with those zeroes is k[x2 - Sx + P] for any non-zero constant k. For example, a quadratic whose zeroes have sum 3 and product -4 is x2 - 3x - 4.

Sum of zeroes α + β = -b/a -(coefficient of x)/(coefficient of x²)
Product of zeroes αβ = c/a (constant term)/(coefficient of x²)
Quadratic from zeroes k[x² - (α + β)x + αβ] k is any non-zero constant
Remember
  • For ax² + bx + c with zeroes α, β.
  • Sum of zeroes: α + β = -b/a.
  • Product of zeroes: αβ = c/a.
  • Comes from equating ax² + bx + c with a(x - α)(x - β).
  • To build a quadratic: k[x² - (sum)x + (product)].

Relationship Between Zeroes and Coefficients of a Cubic Polynomial

Quick answer For a cubic ax³ + bx² + cx + d with zeroes α, β, γ: the sum is -b/a, the sum of products taken two at a time is c/a, and the product is -d/a.

Let p(x) = ax3 + bx2 + cx + d (with a ≠ 0) have three zeroes α, β and γ. Writing p(x) = a(x - α)(x - β)(x - γ) and expanding, then comparing coefficients, gives three relationships:

  • Sum of the zeroes: α + β + γ = -b/a.
  • Sum of products taken two at a time: αβ + βγ + γα = c/a.
  • Product of the zeroes: αβγ = -d/a.

Worked example. Take p(x) = x3 - 6x2 + 11x - 6, so a = 1, b = -6, c = 11, d = -6. Factorising, p(x) = (x - 1)(x - 2)(x - 3), so the zeroes are 1, 2 and 3. Now check all three relations:

  1. Sum: 1 + 2 + 3 = 6, and -b/a = -(-6)/1 = 6.
  2. Sum of products two at a time: (1)(2) + (2)(3) + (3)(1) = 2 + 6 + 3 = 11, and c/a = 11/1 = 11.
  3. Product: (1)(2)(3) = 6, and -d/a = -(-6)/1 = 6.

All three match, confirming the results. Note the alternating signs (minus, plus, minus) in -b/a, c/a, -d/a.

Sum of zeroes α + β + γ = -b/a -(coefficient of x²)/(coefficient of x³)
Sum of products two at a time αβ + βγ + γα = c/a (coefficient of x)/(coefficient of x³)
Product of zeroes αβγ = -d/a -(constant term)/(coefficient of x³)
Remember
  • For ax³ + bx² + cx + d with zeroes α, β, γ.
  • Sum of zeroes: α + β + γ = -b/a.
  • Sum of products two at a time: αβ + βγ + γα = c/a.
  • Product of zeroes: αβγ = -d/a.
  • Signs alternate: -b/a, +c/a, -d/a.

Division Algorithm for Polynomials

Quick answer Dividing a polynomial p(x) by a non-zero polynomial g(x) gives a quotient q(x) and remainder r(x) such that p(x) = g(x)·q(x) + r(x), where r(x) = 0 or deg r(x) < deg g(x).

Just like whole numbers (Dividend = Divisor × Quotient + Remainder), polynomials obey a Division Algorithm. If p(x) and g(x) are any polynomials with g(x) ≠ 0, then we can find unique polynomials q(x) (the quotient) and r(x) (the remainder) such that:

p(x) = g(x) × q(x) + r(x), where either r(x) = 0 or the degree of r(x) is less than the degree of g(x).

The method is long division: at each step divide the highest-degree term of the current dividend by the highest-degree term of the divisor, multiply back, and subtract, until the remainder's degree drops below the divisor's degree.

Worked example. Divide p(x) = 3x3 + x2 + 2x + 5 by g(x) = x2 + 2x + 1.

  1. 3x3 ÷ x2 = 3x. Multiply: 3x(x2 + 2x + 1) = 3x3 + 6x2 + 3x. Subtract: (3x3 + x2 + 2x + 5) - (3x3 + 6x2 + 3x) = -5x2 - x + 5.
  2. -5x2 ÷ x2 = -5. Multiply: -5(x2 + 2x + 1) = -5x2 - 10x - 5. Subtract: (-5x2 - x + 5) - (-5x2 - 10x - 5) = 9x + 10.

Now the remainder 9x + 10 has degree 1, which is less than the divisor's degree 2, so we stop. Hence quotient q(x) = 3x - 5 and remainder r(x) = 9x + 10.

Verification. g(x)·q(x) + r(x) = (x2 + 2x + 1)(3x - 5) + (9x + 10) = (3x3 + x2 - 7x - 5) + (9x + 10) = 3x3 + x2 + 2x + 5 = p(x).

A useful special case is the Factor idea: if the remainder is 0, then g(x) is a factor of p(x). This helps find all zeroes of a cubic once one factor is known.

Division Algorithm p(x) = g(x) × q(x) + r(x) g(x) ≠ 0; r(x) = 0 or deg r(x) < deg g(x)
Number analogy Dividend = Divisor × Quotient + Remainder same structure as integer division
Factor condition if r(x) = 0 then g(x) is a factor of p(x) then p(x) = g(x) × q(x)
Remember
  • p(x) = g(x)·q(x) + r(x), with g(x) ≠ 0.
  • Remainder r(x) = 0, or deg r(x) < deg g(x).
  • Use polynomial long division, highest term first each step.
  • Stop when the remainder's degree is below the divisor's.
  • If remainder = 0, then g(x) is a factor of p(x).

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

p(k) = 0
Zero of a polynomial
For p(x) = ax + b, zero = -b/a
Zero of a linear polynomial
number of zeroes ≤ degree n
Maximum number of zeroes
y = p(x) = 0 at the meeting points
Graph meets x-axis
parabola: opens up if a > 0, down if a < 0
Quadratic graph
number of real zeroes ≤ degree
Number of x-intercepts
α + β = -b/a
Sum of zeroes
αβ = c/a
Product of zeroes
k[x² - (α + β)x + αβ]
Quadratic from zeroes
α + β + γ = -b/a
Sum of zeroes
αβ + βγ + γα = c/a
Sum of products two at a time
αβγ = -d/a
Product of zeroes
p(x) = g(x) × q(x) + r(x)
Division Algorithm
Dividend = Divisor × Quotient + Remainder
Number analogy
if r(x) = 0 then g(x) is a factor of p(x)
Factor condition

Test yourself

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0 correct · 0/12 answered
Q1 Zeroes of a polynomial easy

At most, how many zeroes can a quadratic polynomial have?

Q2 Geometrical meaning of zeroes easy

The zeroes of a polynomial p(x) are the x-coordinates of the points where the graph of y = p(x) meets the:

Q3 Zeroes of a polynomial easy

The zeroes of the polynomial x² − 2x − 8 are:

Q4 Relationship between zeroes and coefficients (quadratic) medium

If α and β are the zeroes of 2x² − 5x + 3, then α + β equals:

Q5 Relationship between zeroes and coefficients (quadratic) medium

A quadratic polynomial whose sum of zeroes is −3 and product of zeroes is 2 is:

Q6 Zeroes of a polynomial medium

If 2 is a zero of the polynomial x² + 3x + k, then the value of k is:

Q7 Relationship between zeroes and coefficients (cubic) medium

For the cubic polynomial x³ − 6x² + 11x − 6, the sum of its zeroes is:

Q8 Relationship between zeroes and coefficients (cubic) hard

If α, β, γ are the zeroes of x³ − 6x² + 11x − 6, then αβ + βγ + γα equals:

Q9 Division algorithm for polynomials medium

In the division algorithm p(x) = g(x)·q(x) + r(x), the degree of the remainder r(x) is:

Q10 Relationship between zeroes and coefficients (quadratic) hard

A quadratic polynomial whose zeroes are 2 + √3 and 2 − √3 is:

Q11 Relationship between zeroes and coefficients (quadratic) hard

If α and β are the zeroes of x² − 6x + 8, then 1/α + 1/β equals:

Q12 Geometrical meaning of zeroes medium

If the graph of y = p(x) does not intersect the x-axis at any point, then the number of real zeroes of p(x) is:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Find the zeroes of the quadratic polynomial x² − 2x − 8 and verify the relationship between the zeroes and the coefficients.Relationship between zeroes and coefficients (quadratic)

Given: p(x) = x2 − 2x − 8, with a = 1, b = −2, c = −8.

Finding the zeroes (splitting the middle term):

x2 − 4x + 2x − 8 = x(x − 4) + 2(x − 4) = (x − 4)(x + 2).

Setting each factor to 0: x = 4 or x = −2.

So the zeroes are 4 and −2.

Verification:

  • Sum of zeroes = 4 + (−2) = 2, and −b/a = −(−2)/1 = 2. Both equal.
  • Product of zeroes = 4 × (−2) = −8, and c/a = −8/1 = −8. Both equal.

Hence the relationship is verified.

2 Find the zeroes of 4x² − 4x + 1 and verify the relationship between the zeroes and the coefficients.Zeroes and coefficients

Given: p(x) = 4x2 − 4x + 1, with a = 4, b = −4, c = 1.

Factorising: 4x2 − 2x − 2x + 1 = 2x(2x − 1) − 1(2x − 1) = (2x − 1)(2x − 1).

So (2x − 1)2 = 0 gives x = 1/2 (a repeated zero).

The zeroes are 1/2 and 1/2.

Verification:

  • Sum = 1/2 + 1/2 = 1, and −b/a = −(−4)/4 = 1. Both equal.
  • Product = (1/2)(1/2) = 1/4, and c/a = 1/4. Both equal.

Hence the relationship is verified.

3 Find a quadratic polynomial the sum and product of whose zeroes are 1/4 and −1 respectively.Forming a polynomial from zeroes

Given: sum of zeroes = 1/4, product of zeroes = −1.

A quadratic polynomial is: k[x2 − (sum)x + (product)].

= k[x2 − (1/4)x + (−1)] = k[x2 − (1/4)x − 1].

Taking k = 4 to clear the fraction:

4x2 − x − 4.

Check: for 4x2 − x − 4, sum = −(−1)/4 = 1/4 and product = −4/4 = −1, as required.

4 Divide p(x) = x³ − 3x² + 5x − 3 by g(x) = x² − 2, and find the quotient and remainder.Division algorithm for polynomials

Given: p(x) = x3 − 3x2 + 5x − 3, g(x) = x2 − 2.

Step 1: Divide x3 by x2 to get x. Multiply: x(x2 − 2) = x3 − 2x.

Subtract: (x3 − 3x2 + 5x − 3) − (x3 − 2x) = −3x2 + 7x − 3.

Step 2: Divide −3x2 by x2 to get −3. Multiply: −3(x2 − 2) = −3x2 + 6.

Subtract: (−3x2 + 7x − 3) − (−3x2 + 6) = 7x − 9.

The degree of 7x − 9 is less than that of g(x), so we stop.

Quotient = x − 3 and Remainder = 7x − 9.

Check: (x2 − 2)(x − 3) + (7x − 9) = x3 − 3x2 − 2x + 6 + 7x − 9 = x3 − 3x2 + 5x − 3 = p(x).

5 Verify that 3, −1 and −1/3 are the zeroes of the cubic polynomial 3x³ − 5x² − 11x − 3, and then verify the relationship between the zeroes and the coefficients.Relationship between zeroes and coefficients (cubic)

Given: p(x) = 3x3 − 5x2 − 11x − 3, with a = 3, b = −5, c = −11, d = −3.

Checking each value is a zero:

  • 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(−1/3) = 3(−1/27) − 5(1/9) − 11(−1/3) − 3 = −1/9 − 5/9 + 33/9 − 27/9 = 0.

So 3, −1 and −1/3 are indeed the zeroes. Let α = 3, β = −1, γ = −1/3.

Verifying relationships:

  • α + β + γ = 3 − 1 − 1/3 = 5/3, and −b/a = −(−5)/3 = 5/3. Equal.
  • αβ + βγ + γα = (3)(−1) + (−1)(−1/3) + (−1/3)(3) = −3 + 1/3 − 1 = −11/3, and c/a = −11/3. Equal.
  • αβγ = (3)(−1)(−1/3) = 1, and −d/a = −(−3)/3 = 1. Equal.

Hence all three relationships are verified.

6 Find the zeroes of the quadratic polynomial 6x² − 7x − 3 and verify the relationship between the zeroes and the coefficients.Relationship between zeroes and coefficients (quadratic)

Given: p(x) = 6x2 − 7x − 3, with a = 6, b = −7, c = −3.

Splitting the middle term (product a×c = 6×(−3) = −18; factors −9 and 2 add to −7):

6x2 − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (2x − 3)(3x + 1).

Setting each factor to 0: x = 3/2 or x = −1/3.

The zeroes are 3/2 and −1/3.

Verification:

  • Sum = 3/2 + (−1/3) = 9/6 − 2/6 = 7/6, and −b/a = −(−7)/6 = 7/6. Equal.
  • Product = (3/2)(−1/3) = −1/2, and c/a = −3/6 = −1/2. Equal.

Hence the relationship is verified.

Previous-year board questions 4

Q1 Find the zeroes of the quadratic polynomial x² + 7x + 10 and verify the relationship between the zeroes and the coefficients. CBSE 2023 2 marks

Given: p(x) = x2 + 7x + 10, with a = 1, b = 7, c = 10.

Factorising: x2 + 5x + 2x + 10 = x(x + 5) + 2(x + 5) = (x + 5)(x + 2).

Setting each factor to 0: x = −5 or x = −2.

The zeroes are −5 and −2.

Verification:

  • Sum = −5 + (−2) = −7, and −b/a = −7/1 = −7. Equal.
  • Product = (−5)(−2) = 10, and c/a = 10/1 = 10. Equal.

Hence verified.

Q2 If α and β are the zeroes of the polynomial x² − 6x + 8, find the value of α² + β². CBSE 2020 2 marks

Given: p(x) = x2 − 6x + 8, with a = 1, b = −6, c = 8.

Using relationships:

  • α + β = −b/a = −(−6)/1 = 6.
  • αβ = c/a = 8/1 = 8.

Now: α2 + β2 = (α + β)2 − 2αβ.

= (6)2 − 2(8) = 36 − 16 = 20.

Q3 Find a quadratic polynomial whose zeroes are 5 + √2 and 5 − √2. CBSE 2019 3 marks

Given zeroes: α = 5 + √2, β = 5 − √2.

Sum of zeroes: α + β = (5 + √2) + (5 − √2) = 10.

Product of zeroes: αβ = (5 + √2)(5 − √2) = 52 − (√2)2 = 25 − 2 = 23.

A quadratic polynomial = x2 − (sum)x + (product).

= x2 − 10x + 23.

Q4 On dividing the polynomial x³ − 3x² + x + 2 by a polynomial g(x), the quotient and remainder were (x − 2) and (−2x + 4) respectively. Find g(x). CBSE 2018 4 marks

Given: p(x) = x3 − 3x2 + x + 2, quotient q(x) = x − 2, remainder r(x) = −2x + 4.

By the division algorithm: p(x) = g(x)·q(x) + r(x).

So g(x)·q(x) = p(x) − r(x).

= (x3 − 3x2 + x + 2) − (−2x + 4) = x3 − 3x2 + 3x − 2.

Therefore g(x) = (x3 − 3x2 + 3x − 2) ÷ (x − 2).

Dividing:

  • x3 ÷ x = x2; x2(x − 2) = x3 − 2x2; subtract to get −x2 + 3x − 2.
  • −x2 ÷ x = −x; −x(x − 2) = −x2 + 2x; subtract to get x − 2.
  • x ÷ x = 1; 1(x − 2) = x − 2; subtract to get 0.

So g(x) = x2 − x + 1.

Check: (x2 − x + 1)(x − 2) + (−2x + 4) = (x3 − 3x2 + 3x − 2) + (−2x + 4) = x3 − 3x2 + x + 2 = p(x).

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