Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
The degree of a polynomial p(x) is the highest power of x in it.
| Degree | Name | General form | Example |
|---|---|---|---|
| 1 | Linear | ax + b, a ≠ 0 | 2x − 3 |
| 2 | Quadratic | ax² + bx + c, a ≠ 0 | 2x² + 3x − 2/5 |
| 3 | Cubic | ax³ + bx² + cx + d, a ≠ 0 | 3x³ − 2x² + x − 1 |
p(k) is the value of p(x) when x is replaced by k. For p(x) = x² − 3x − 4, p(2) = 4 − 6 − 4 = −6.
A real number k is a zero of p(x) if p(k) = 0. Here p(−1) = 0 and p(4) = 0, so −1 and 4 are zeroes of x² − 3x − 4.
The graph of y = ax + b is a straight line. It meets the x-axis at exactly one point, (−b/a, 0), so a linear polynomial has exactly one zero.
The graph of y = ax² + bx + c is a parabola. It opens upward if a > 0 and downward if a < 0.
A polynomial of degree n has at most n zeroes, because its graph meets the x-axis at most n times. The number of zeroes is the number of points where the graph meets the x-axis, whether it crosses or just touches.
Drawing graphs of quadratic and cubic polynomials is not expected of students; reading them is.
If α and β are the zeroes of ax² + bx + c (a ≠ 0), then ax² + bx + c = a(x − α)(x − β). Comparing coefficients:
x² + 7x + 10 = (x + 2)(x + 5), so the zeroes are −2 and −5.
If the sum of zeroes is S and the product is P, a quadratic with these zeroes is
Example: S = −3, P = 2 gives x² + 3x + 2. If S or P is a fraction, choose k to clear it: S = 1/4, P = −1 gives 4x² − x − 4.
If α, β, γ are the zeroes of ax³ + bx² + cx + d (a ≠ 0):
For 2x³ − 5x² − 14x + 8 the zeroes are 4, −2 and 1/2.
With S = sum, Q = sum of products two at a time and P = product of the zeroes, a cubic is
| Polynomial | Zeroes | Relations |
|---|---|---|
| ax + b | one | zero = −b/a |
| ax² + bx + c | at most 2 | α + β = −b/a; αβ = c/a |
| ax³ + bx² + cx + d | at most 3 | α + β + γ = −b/a; αβ + βγ + γα = c/a; αβγ = −d/a |
Try each problem first, then open the solution to check your working.
Write the degree and type of each: (a) 5x³ − 2x + 7 (b) 3 − 4y (c) 2u² + u (d) 7
Use: Degree = highest power of the variable
(a) highest power 3: cubic
(b) highest power 1: linear
(c) highest power 2: quadratic
(d) 7 = 7x⁰: degree 0, a constant polynomial
Answer: (a) 3, cubic (b) 1, linear (c) 2, quadratic (d) 0, constant
For p(x) = 2x² − 5x + 3, find p(0), p(1) and p(−2). Is 1 a zero of p(x)?
Given: p(x) = 2x² − 5x + 3
Use: Replace x by the number
p(0) = 0 − 0 + 3 = 3
p(1) = 2 − 5 + 3 = 0
p(−2) = 8 + 10 + 3 = 21
Answer: 3, 0, 21; yes, 1 is a zero because p(1) = 0
Find the zero of (a) 3x − 7 (b) 5 − 2x.
Use: Zero of ax + b is −b/a
(a) a = 3, b = −7: zero = 7/3
(b) −2x + 5: a = −2, b = 5: zero = −5/(−2) = 5/2
Answer: (a) 7/3 (b) 5/2
Check whether 3, −3 and −2 are zeroes of p(x) = x³ + 2x² − 9x − 18.
Given: p(x) = x³ + 2x² − 9x − 18
p(3) = 27 + 18 − 27 − 18 = 0
p(−3) = −27 + 18 + 27 − 18 = 0
p(−2) = −8 + 8 + 18 − 18 = 0
Answer: All three are zeroes, and a cubic has at most three, so these are all its zeroes
The graph of y = p(x) is shown. Find the number of zeroes and the zeroes of p(x). Also find p(x) if it has the form x² + bx + c.
Given: Graph meets the x-axis at x = −1 and x = 3
Use: Zeroes = x-coordinates of the meeting points
Two meeting points, so two zeroes: −1 and 3.
p(x) = (x + 1)(x − 3) = x² − 2x − 3
Answer: 2 zeroes, −1 and 3; p(x) = x² − 2x − 3
The graph of the cubic y = q(x) is shown. How many zeroes does q(x) have?
Given: Crosses at x = −2, touches at x = 1
Use: Every meeting point, crossing or touching, gives a zero
The graph meets the x-axis at two points: (−2, 0) and (1, 0).
Answer: 2 zeroes: −2 and 1 (1 is a repeated zero)
A parabola opens upward and its lowest point is (2, 3). How many zeroes does the polynomial have? What if it opens downward with highest point (2, 3)?
Opening upward from a lowest point above the x-axis, the curve never comes down to the axis: no zeroes.
Opening downward from a highest point above the axis, it comes down and crosses the axis twice.
Answer: Upward: no zero; downward: two zeroes
Find the zeroes of x² + x − 12 and verify the relationship between the zeroes and the coefficients.
Given: a = 1, b = 1, c = −12
Use: α + β = −b/a, αβ = c/a
x² + x − 12 = x² + 4x − 3x − 12 = (x + 4)(x − 3)
Zeroes: −4 and 3
Sum = −1 = −(1)/1 ✓ Product = −12 = −12/1 ✓
Answer: −4 and 3
Find the zeroes of 5x² − 8x − 4 and verify the relationships.
Given: a = 5, b = −8, c = −4
Use: Split −8x into two terms with product 5 × (−4) = −20
5x² − 10x + 2x − 4 = 5x(x − 2) + 2(x − 2) = (5x + 2)(x − 2)
Zeroes: 2 and −2/5
Sum = 2 − 2/5 = 8/5 = −(−8)/5 ✓ Product = −4/5 ✓
Answer: 2 and −2/5
Find the zeroes of 3x² − 27 and verify.
Given: a = 3, b = 0, c = −27
3x² − 27 = 3(x² − 9) = 3(x − 3)(x + 3)
Zeroes: 3 and −3
Sum = 0 = −0/3 ✓ Product = −9 = −27/3 ✓
Answer: 3 and −3
If α and β are the zeroes of x² − 5x + 6, find (a) α² + β² (b) 1/α + 1/β, without finding α and β.
Given: α + β = 5, αβ = 6
Use: α² + β² = (α + β)² − 2αβ
(a) 25 − 12 = 13
(b) (α + β)/αβ = 5/6
Answer: (a) 13 (b) 5/6
One zero of x² + kx − 6 is 2. Find k and the other zero.
Given: p(2) = 0
Use: Product of zeroes = −6
4 + 2k − 6 = 0, so k = 1
2 × β = −6, so β = −3 (check: sum = −1 = −k ✓)
Answer: k = 1; other zero −3
If the sum of the zeroes of kx² − 6x + 2 is 3, find k and the product of the zeroes.
Given: sum = 3
Use: Sum = −b/a = 6/k
6/k = 3, so k = 2
Product = c/a = 2/2 = 1
Answer: k = 2; product 1
One zero of 3x² − 8x + 2k + 1 is seven times the other. Find the zeroes and k.
Given: zeroes α and 7α
Use: Sum = 8/3, product = (2k + 1)/3
α + 7α = 8/3, so α = 1/3; zeroes 1/3 and 7/3
Product = 7/9 = (2k + 1)/3, so 2k + 1 = 7/3 and k = 2/3
Answer: Zeroes 1/3 and 7/3; k = 2/3
Find a quadratic polynomial whose zeroes have sum −3/2 and product −1. Find its zeroes.
Given: S = −3/2, P = −1
Use: k(x² − Sx + P)
x² + (3/2)x − 1; take k = 2: 2x² + 3x − 2
2x² + 4x − x − 2 = (2x − 1)(x + 2), zeroes 1/2 and −2
Answer: 2x² + 3x − 2; zeroes 1/2 and −2
Find a quadratic polynomial with zeroes 2 + √3 and 2 − √3.
Use: S and P of the zeroes
S = (2 + √3) + (2 − √3) = 4
P = (2 + √3)(2 − √3) = 4 − 3 = 1
Answer: x² − 4x + 1
Find a quadratic polynomial with zeroes −3 and 5/2.
S = −3 + 5/2 = −1/2; P = −15/2
x² + (1/2)x − 15/2; multiply by 2
Answer: 2x² + x − 15
Show that 1, −1 and 3 are zeroes of x³ − 3x² − x + 3 and verify the three relations.
Given: a = 1, b = −3, c = −1, d = 3
p(1) = 1 − 3 − 1 + 3 = 0; p(−1) = −1 − 3 + 1 + 3 = 0; p(3) = 27 − 27 − 3 + 3 = 0
Sum = 3 = −b/a ✓
Two at a time: (1)(−1) + (−1)(3) + (3)(1) = −1 = c/a ✓
Product = −3 = −d/a ✓
Answer: Verified
Find a cubic polynomial whose zeroes have sum 2, sum of products two at a time −7, and product −14. Find its zeroes.
Given: S = 2, Q = −7, P = −14
Use: x³ − Sx² + Qx − P
x³ − 2x² − 7x + 14
= x²(x − 2) − 7(x − 2) = (x − 2)(x² − 7)
Answer: x³ − 2x² − 7x + 14; zeroes 2, √7, −√7
Two zeroes of x³ − 6x² + 11x − 6 are 1 and 2. Find the third zero.
Given: α = 1, β = 2
Use: α + β + γ = −b/a = 6
γ = 6 − 1 − 2 = 3
Check with the product: 1 × 2 × 3 = 6 = −(−6)/1 ✓
Answer: 3
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The number of zeroes is the number of points where the graph meets the x-axis.
(i) The graph is a line parallel to the x-axis and never meets it: 0 zeroes.
(ii) It meets the x-axis at one point: 1 zero.
(iii) It meets the x-axis at three points: 3 zeroes.
(iv) It meets the x-axis at two points: 2 zeroes.
(v) It meets the x-axis at four points: 4 zeroes.
(vi) It meets the x-axis at three points (it touches the axis at the middle one): 3 zeroes.
Answer: (i) 0 (ii) 1 (iii) 3 (iv) 2 (v) 4 (vi) 3
(i) x² − 2x − 8 = x² − 4x + 2x − 8 = (x − 4)(x + 2). Zeroes: 4 and −2.
Sum of zeroes = 4 + (−2) = 2; −b/a = −(−2)1 = 2 ✓
Product of zeroes = 4 × (−2) = −8; c/a = −81 = −8 ✓
(ii) 4s² − 4s + 1 = (2s − 1)². Zeroes: 1/2 and 1/2.
Sum = 12 + 12 = 1; −b/a = −(−4)4 = 1 ✓
Product = 12 × 12 = 14; c/a = 14 ✓
(iii) In standard form 6x² − 7x − 3 = 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Zeroes: 3/2 and −1/3.
Sum = 32 − 13 = 76; −b/a = −(−7)6 = 76 ✓
Product = 32 × (−13) = −12; c/a = −36 = −12 ✓
(iv) 4u² + 8u = 4u(u + 2). Zeroes: 0 and −2.
Sum = −2; −b/a = −84 = −2 ✓ Product = 0; c/a = 04 = 0 ✓
(v) t² − 15 = (t − √15)(t + √15). Zeroes: √15 and −√15.
Sum = 0; −b/a = −01 = 0 ✓ Product = −15; c/a = −151 = −15 ✓
(vi) 3x² − x − 4 = 3x² − 4x + 3x − 4 = x(3x − 4) + 1(3x − 4) = (x + 1)(3x − 4). Zeroes: 4/3 and −1.
Sum = 43 − 1 = 13; −b/a = −(−1)3 = 13 ✓
Product = −43; c/a = −43 ✓
Answer: (i) 4, −2 (ii) 1/2, 1/2 (iii) 3/2, −1/3 (iv) 0, −2 (v) √15, −√15 (vi) 4/3, −1
A quadratic whose zeroes have sum S and product P is k(x² − Sx + P), k ≠ 0. We choose k to clear fractions.
(i) x² − 14x − 1; with k = 4: 4x² − x − 4
(ii) x² − √2x + 13; with k = 3: 3x² − 3√2x + 1
(iii) x² − 0·x + √5: x² + √5
(iv) x² − x + 1
(v) x² + 14x + 14; with k = 4: 4x² + x + 1
(vi) x² − 4x + 1
Answer: (i) 4x² − x − 4 (ii) 3x² − 3√2x + 1 (iii) x² + √5 (iv) x² − x + 1 (v) 4x² + x + 1 (vi) x² − 4x + 1