Access Answers for SAT Maths Chapter 2 – Polynomials
Exercise 2.1
1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number
of zeroes of p(x), in each case.
Solutions:
Graphical method to find zeroes:-
Total number of zeroes in any polynomial equation = total number of times the curve intersects x-
axis.
(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does
not cut it at any point.
(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at
only one point.
(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at
any three points.
(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at
two points.
(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at
four points.
(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at
three points.
, Exercise 2.2 Page: 33
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the
zeroes and the coefficients.
Solutions:
(i) x2–2x –8
⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)
Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2)
Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x 2)
Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)
(ii) 4s2–4s+1
⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)
Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s 2)
Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )
(iii) 6x2–3–7x
⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)
Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x 2)
Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2 )
(iv) 4u2+8u
⇒ 4u(u+2)
Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).
Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u 2)
Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )
(v) t2–15
⇒ t2 = 15 or t = ±√15
Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)
Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t 2)
Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )
(vi) 3x2–x–4
⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)
Exercise 2.1
1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number
of zeroes of p(x), in each case.
Solutions:
Graphical method to find zeroes:-
Total number of zeroes in any polynomial equation = total number of times the curve intersects x-
axis.
(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does
not cut it at any point.
(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at
only one point.
(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at
any three points.
(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at
two points.
(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at
four points.
(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at
three points.
, Exercise 2.2 Page: 33
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the
zeroes and the coefficients.
Solutions:
(i) x2–2x –8
⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)
Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2)
Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x 2)
Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)
(ii) 4s2–4s+1
⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)
Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s 2)
Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )
(iii) 6x2–3–7x
⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)
Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x 2)
Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2 )
(iv) 4u2+8u
⇒ 4u(u+2)
Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).
Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u 2)
Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )
(v) t2–15
⇒ t2 = 15 or t = ±√15
Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)
Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t 2)
Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )
(vi) 3x2–x–4
⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)
