Key Concepts & Formulae at a Glance
- Zero of a polynomial:A real number k is a zero of p(x) if p(k) = 0.
- Geometrical meaning:The zeroes of p(x) are the x-coordinates of the points where the graph of y = p(x) intersects the x-axis.
- Maximum zeroes:A polynomial of degree n has at most n zeroes (linear: 1, quadratic: at most 2, cubic: at most 3).
- Quadratic: sum & product:For ax²+bx+c with zeroes α, β: α+β = −ba, αβ = ca.
- Cubic: sum & product:For ax³+bx²+cx+d with zeroes α, β, γ: α+β+γ=−ba, αβ+βγ+γα=ca, αβγ=−da.
- Constructing a quadratic:Given sum S and product P of zeroes, one such polynomial is x² − Sx + P.
Exercise 2.1
1 Question
1The graphs of y = p(x) are given below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
(i)
0 zeroes
(ii)
2 zeroes
(iii)
3 zeroes
(iv)
2 zeroes
(v)
4 zeroes
(vi)
3 zeroes
Note: these are schematic illustrations showing the correct number of x-axis crossings for each case, not exact reproductions of the textbook's figures.
The number of zeroes of p(x) equals the number of points where the graph of y = p(x) crosses (or touches) the x-axis (geometrical meaning of zeroes).
(i) 0 (ii) 2 (iii) 3 (iv) 2 (v) 4 (vi) 3
Exercise 2.2
2 Questions
1Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
(i) x² − 2x − 8 (ii) 4s² − 4s + 1 (iii) 6x² − 3 − 7x
(iv) 4u² + 8u (v) t² − 15 (vi) 3x² − x − 4
(i) x² − 2x − 8
x² − 2x − 8 = x² − 4x + 2x − 8 = x(x − 4) + 2(x − 4)
= (x − 4)(x + 2) (splitting the middle term)
Zeroes: x = 4 or x = −2
Sum of zeroes = 4 + (−2) = 2 = −−21 = −coefficient of xcoefficient of x²
Product of zeroes = 4 × (−2) = −8 = −81 = constant termcoefficient of x²
= (x − 4)(x + 2) (splitting the middle term)
Zeroes: x = 4 or x = −2
Sum of zeroes = 4 + (−2) = 2 = −−21 = −coefficient of xcoefficient of x²
Product of zeroes = 4 × (−2) = −8 = −81 = constant termcoefficient of x²
Zeroes: 4, −2 — sum = 2, product = −8, both verified.
(ii) 4s² − 4s + 1
4s² − 4s + 1 = 4s² − 2s − 2s + 1 = 2s(2s − 1) − 1(2s − 1)
= (2s − 1)(2s − 1) (splitting the middle term)
Zeroes: s = 12, 12 (equalrepeated zero)
Sum of zeroes = 12 + 12 = 1 = −−44 = −coefficient of scoefficient of s²
Product of zeroes = 12 × 12 = 14 = constant termcoefficient of s²
= (2s − 1)(2s − 1) (splitting the middle term)
Zeroes: s = 12, 12 (equalrepeated zero)
Sum of zeroes = 12 + 12 = 1 = −−44 = −coefficient of scoefficient of s²
Product of zeroes = 12 × 12 = 14 = constant termcoefficient of s²
Zeroes: 12, 12 — sum = 1, product = 14, both verified.
(iii) 6x² − 3 − 7x
Rewriting in standard form: 6x² − 7x − 3
6x² − 7x − 3 = 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3)
= (2x − 3)(3x + 1) (splitting the middle term)
Zeroes: x = 32 or x = −13
Sum of zeroes = 32 + (−13) = 76 = −−76 = −coefficient of xcoefficient of x²
Product of zeroes = 32 × (−13) = −12 = −36 = constant termcoefficient of x²
6x² − 7x − 3 = 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3)
= (2x − 3)(3x + 1) (splitting the middle term)
Zeroes: x = 32 or x = −13
Sum of zeroes = 32 + (−13) = 76 = −−76 = −coefficient of xcoefficient of x²
Product of zeroes = 32 × (−13) = −12 = −36 = constant termcoefficient of x²
Zeroes: 32, −13 — sum = 76, product = −12, both verified.
(iv) 4u² + 8u
4u² + 8u = 4u(u + 2) (taking 4u common)
Zeroes: u = 0 or u = −2
Sum of zeroes = 0 + (−2) = −2 = −84 = −coefficient of ucoefficient of u²
Product of zeroes = 0 × (−2) = 0 = 04 = constant termcoefficient of u²
Zeroes: u = 0 or u = −2
Sum of zeroes = 0 + (−2) = −2 = −84 = −coefficient of ucoefficient of u²
Product of zeroes = 0 × (−2) = 0 = 04 = constant termcoefficient of u²
Zeroes: 0, −2 — sum = −2, product = 0, both verified.
(v) t² − 15
t² − 15 = (t − √15)(t + √15) (difference of squares, a²−b²=(a−b)(a+b))
Zeroes: t = √15 or t = −√15
Sum of zeroes = √15 + (−√15) = 0 = −01 = −coefficient of tcoefficient of t²
Product of zeroes = √15 × (−√15) = −15 = −151 = constant termcoefficient of t²
Zeroes: t = √15 or t = −√15
Sum of zeroes = √15 + (−√15) = 0 = −01 = −coefficient of tcoefficient of t²
Product of zeroes = √15 × (−√15) = −15 = −151 = constant termcoefficient of t²
Zeroes: √15, −√15 — sum = 0, product = −15, both verified.
(vi) 3x² − x − 4
3x² − x − 4 = 3x² − 4x + 3x − 4 = x(3x − 4) + 1(3x − 4)
= (3x − 4)(x + 1) (splitting the middle term)
Zeroes: x = 43 or x = −1
Sum of zeroes = 43 + (−1) = 13 = −−13 = −coefficient of xcoefficient of x²
Product of zeroes = 43 × (−1) = −43 = −43 = constant termcoefficient of x²
= (3x − 4)(x + 1) (splitting the middle term)
Zeroes: x = 43 or x = −1
Sum of zeroes = 43 + (−1) = 13 = −−13 = −coefficient of xcoefficient of x²
Product of zeroes = 43 × (−1) = −43 = −43 = constant termcoefficient of x²
Zeroes: 43, −1 — sum = 13, product = −43, both verified.
2Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
(i) 14, −1 (ii) √2, 13 (iii) 0, √5
(iv) 1, 1 (v) −14, 14 (vi) 4, 1
If S = sum of zeroes and P = product of zeroes, one such quadratic polynomial is x² − Sx + P (since α+β=−ba and αβ=ca, taking a=1). Any real multiple k(x² − Sx + P) also works.
(i) S = 14, P = −1
x² − (14)x + (−1) = x² − (14)x − 1
Multiplying through by 4 to clear the fraction: 4x² − x − 4
Multiplying through by 4 to clear the fraction: 4x² − x − 4
One such polynomial: 4x² − x − 4
(ii) S = √2, P = 13
x² − √2 x + 13
Multiplying through by 3: 3x² − 3√2 x + 1
Multiplying through by 3: 3x² − 3√2 x + 1
One such polynomial: 3x² − 3√2 x + 1
(iii) S = 0, P = √5
x² − (0)x + √5 = x² + √5
One such polynomial: x² + √5
(iv) S = 1, P = 1
x² − x + 1
One such polynomial: x² − x + 1
(v) S = −14, P = 14
x² − (−14)x + 14 = x² + (14)x + 14
Multiplying through by 4: 4x² + x + 1
Multiplying through by 4: 4x² + x + 1
One such polynomial: 4x² + x + 1
(vi) S = 4, P = 1
x² − 4x + 1
One such polynomial: x² − 4x + 1
Frequently Asked Questions
The zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y = p(x) intersects the x-axis.
At most n zeroes — a linear polynomial (degree 1) has exactly one zero, a quadratic (degree 2) has at most two, a cubic (degree 3) has at most three, and so on.
For a quadratic polynomial ax² + bx + c with zeroes α and β: the sum of zeroes α+β = −ba, and the product of zeroes αβ = ca.
Since α+β = −ba and αβ = ca, taking a = 1 gives the polynomial x² − (sum of zeroes)x + (product of zeroes). Any real multiple of this polynomial also satisfies the same conditions.
Yes — a printable Formula Card covering all 7 units, including Polynomials, is available for ₹99.
Continue to the Next Chapter
Chapter 3 (Pair of Linear Equations in Two Variables) builds on the graphical thinking introduced here.