# Polynomials-NCERT Solutions

Class X Math
NCERT Solutions for Polynomials
Exercise– 2.1
Q1.  The graphs of y = p(x) are given in the figure given below. for some polynomials p (x). Find the number of zeroes of p (x), in each case Sol.  (i)  The given graph is parallel to x-axis. It does not intersect the x-axis.
•  It has no zeroes.
(ii)  The given graph intersects the x-axis at one point only.
•  It has one zero.
(iii)  The given graph intersects the x-axis at three points.
•  It has three zeroes.
(iv)  The given graph intersects the x-axis at two points.
•  It has two zeroes.
(v)  The given graph intersects the x-axis at four points.
•  It has four zeroes.
(vi)  The given graph meets the x-axis at three points.
•  It has three zeroes.
NCERT TEXTBOOK QUESTIONS SOLVED
Exercise– 2.2
Q1.  Find 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
Sol.  (i)  x2 – 2x – 8
We have p(x) = x2 – 2x – 8
= x2 + 2x – 4x – 8 = x (x + 2) – 4 (x + 2)
= (x – 4) (x + 2)
For p(x) = 0, we have
(x – 4) (x + 2) = 0
Either x – 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = – 2
∴ The zeroes of x2 2x – 8 are 4 and –2.
Now, sum of the zeroes   Thus, relationship between zeroes and the coefficients in x2 – 2x – 8 is verified.
(ii)  4s2 – 4s + 1
We have p(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1 = 2S (2S – 1) –1 (2s – 1)
= (2s – 1) (2s – 1)
For p(s) = 0, we have, Now, Thus, the relationship between the zeroes and coefficients in the polynomial 4s2 – 4s + 1 is verified.
(iii)  6x2 – 3 – 7x
We have
p (x) = 6x2 – 3 – 7x
= 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x (2x – 3) + 1 (2x – 3)
= (3x + 1) (2x – 3)
For p (x) = 0, we have, Thus, the relationship between the zeroes and the coefficients in the polynomial 6x2 – 3 – 7x is verified.
(iv)  4u2 + 8u
We have, f(u) = 4u2 + 8u = 4u (u + 2)
For f(u) = 0,
Either 4u = 0 ⇒ u = 0
or u + 2 = 0 ⇒ u = –2
∴ The zeroes of 4u2 + 8u and 0 and –2
Now, 4u2 + 8u can be written as 4u2 + 8u + 0. Thus, the relationship between the zeroes and the coefficients in the polynomial 4u2 + 8u is verified.
(v)  t2 – 15
We have, For f(t) = 0, we have  Now, we can write t2 – 15 as t2 + 0t – 15. Thus, the relationship between the zeroes and the coefficients in the polynomial t2 – 15 is verified.
(vi)  3x2 – x – 4
We have,
f(x) = 3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x (x + 1) –4 (x + 1)
= (x + 1) (3x – 4)
For f(x) = 0 ⇒ (x + 1) (3x – 4) = 0
Either (x + 1) = 0 ⇒ x = – 1  Thus, the relationship between the zeroes and the coefficients in 3x2 – x – 4 is verified.
Q2.  Find a quadratic polynomial each with the given numbers as sum and product of its zeroes respectively: Sol.
Note:
A quadratic polynomial whose zeroes are α and β is given by
p(x) = {x2 – (α + β) x + αβ}
p(x) = {x2 – (sum of the zeroes) x – (product of the zeroes)}
(i)  Since, sum of the zeroes, Product of the zeroes, α β = –1
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ Since, have same zeroes, is the required quadratic polynomial.
(ii)  Since, sum of the zeroes, Product of zeroes,  Since, have same zeroes, is required quadratic polynomial.
(iii)  Since, sum of zeroes, (α + β) 0
Product of zeroes, αβ = 5
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ
= x2 – (0) x + 5
= x2 + 5
(iv)  Since, sum of the zeroes, (α + β) = 1
Product of the zeroes = 1
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ
= x2 – (1) x + 1
= x2 – x + 1
(v)  Since, sum of the zeroes, Product of the zeroes ∴ The required quadratic polynomial is
x2 – (α + β) x + αβ  Since, and (4x2 + x + 1) have same zeroes, the required quadratic polynomial is (4x2 + x + 1).
(vi)  Since, sum of the zeroes, (α + β) = 4
Product of the zeroes, αβ = 1
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ
= x2 – (4)x + 1
= x2 – 4x + 1
Exercise– 2.3
Q1.  Q1. Divide the polynomial p (x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i)  p (x) = x3 – 3x2 + 5x – 3, g (x) = x2 – 2
(ii)  p (x) = x4 – 3x2 + 4x + 5, g (x) = x2 + 1 – x
(iii)  p (x) = x3 – 5x + 6, g (x) = 2 – x2
Sol.  Here, dividend p (x) = x3 – 3x2 + 5x – 3
divisor g(x) = x2 – 2
∴ We have Thus, the quotient = (x – 3) and remainder = (7x – 9)
(iii)  Here, dividend p(x) = x4 – 3x2 + 4x + 5
and divisor g(x) = x2 + 1 – x
= x2 – x + 1
∴ We have Thus, the quotient is (x2 + x – 3) and remainder = 8
(iii)  Here, divided, p(x) = x4 – 5x + 6
and divisor, g(x) = 2 – x2
∴ We have Thus, the quotient = –x2 – 2 and remainder = –5x + 10.
Q2.  Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i)  t2 – 3;                 2t4 + 3t3 – 2t2 – 9t – 12
(ii)  x2 + 3x + 1;                                  3x4 + 5x3 – 7x2 + 2x + 2
(iii)  x3 + 3x + 1;                                  x5 + 4x3 + x + 3x + 1
Sol.  (i)  Dividing 2t4 + 3t3 – 2t2 – 9t – 12 by t2 – 3,
We have: ∴ Remainder = 0
∴ (t2 – 3) is a factor of 2t4 + 3t3 – 2t2 – 9t – 12.
(ii)  Dividing 3x4 + 5x3 – 7x2 + 2x + 2 by
x2 + 3x + 1, we have: ∴ Remainder = 0.
∴ x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2.
(iii)  Dividiing x5 + 4x3 + x + 3x + 1, we get: ∴ The remainder = 2, i.e., remainder ≠ 0
∴ x3 – 3x + 1 is not a factor of
x5 – 4x3 + x2 + 3x + 1.
Q3.  Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are Sol.  We have p(x) = 3x4 + 6x3 – 2x2 – 10x – 5 Now, let us divide 3x4 + 6x3 – 2x2 – 10x – 5 by  ∴ 3x4 + 6x3 – 2x2 – 10x – 5  For p(x) = 0, we have or 3x + 3 = 0 ⇒ x = –1
or x + 1 = 0 ⇒ x = –1
Thus, all the other zeroes of the given polynomial are –1 and –1.
Q4.  On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4 respectively. Find g(x).
Sol.  Here,
Dividened p(x) = x3 – 3x2 + x + 2
Divisor = g(x)
Quotient = (x – 2)
Remainder = (–2x + 4)
Since,
(Quotient – Divisor) + Remainder = Dividend
∴ [(x – 2)] – g(x)] + [(–2x + 4)] = x3 – 3x2 + x + 2
⇒ (x – 2) – g(x) = x3 – 3x2 + x + 2 – (–2x + 4)
= x3 – 3x2 + x + 2 + 2x – 4
= x3 – 3x2 + 3x – 2 Now, dividing x3 – 3x2 + 3x – 2 by x – 2,
We have Thus, the required divisor g(x) = x2 – x + 1.
Q5.  Give example of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i)  deg p(x) = deg q (x)
(ii)  deg q(x) = deg r (x)
(iii)  deg r (x) = 0
Sol. We can have
(i)  p(x) = 3x2 – 6x + 27
g(x) = 3
q(x) = x2 – 2x + 9
r(x) = 0
⇒ p(x) = q(x) × g(x) + r(x).
(ii)  p(x) = 2x3 – 2x2 + 2x + 3
g(x) = 2x2 – 1
q(x) = x – 1
r(x) = 3x + 2
⇒ p(x) = q(x) × g(x) + r(x)
(iii)  p(x) = 2x3 – 4x2 + x + 4
g(x) = 2x2 + 1
q(x) = x – 2
r(x) = 6
⇒ p(x) = q(x) × g(x) + r(x)
Exercise– 2.4
Q1.  Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) (i)  x3 – 4x2 + 5x – 2; 2, 1, 1
Sol.  (i)  ∵ p(x) = 2x3 + x2 – 5x + 2  Again,
p(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2
= (2 + 2 + 1) – 5
= 5 – 5 = 0
⇒ 1 is a zero of p(x).
Also,
p(–2) = 2(–2)3 + (–2)2 –5 (–2) + 2
= 2(–8) + (4) + 10 + 2
= –16 + 4 + 10 + 2
= –16 + 16 = 0
⇒ –2 is a zero of p(x).
Relationship
∵p(x) = 2x3 + x2 – 5x + 2
∴Comparing it with ax3 + bx2 + cx + d, we have:
a = 2, b = 1, c = – 5 and d = 2
Also 1 and – 2 are the zeroes of p(x)  Sum of product of zeroes taken in pair: Thus, the relationship between the co-efficients and the zeroes of p(x) is verified.
(ii)  Here, p(x) = x3 – 4x2 + 5x – 2
p(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2
= 18 – 18 = 0
⇒ 2 is a zero of p(x).
Again p(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
= 6 – 6 = 0
⇒ 1 is a zero of p(x).
∴2, 1, 1 are zeroes of p(x).
Now, comparing
p(x) = x3 – 4x2 + 5x – 2 with ax3 + bx2 +cx + d = 0, we have
a = 1, b = –4, c = 5 and d = –2
∵2, 1, 1 are the zeroes of p(x)
Let
α = 2
β = 1
γ = 1
Relationship, α + β + γ = 2 + 1+ 1 = 4 Sum of product of zeroes taken in pair:
αβ + βγ + γα = 2(1) + 1(1) + 1(2)
= 2 + 1 + 2 = 5 ⇒ αβ + βγ + γα =
Product of zeroes = αβγ = (2) (1) (1) = 2 Thus, the relationship between the zeroes and the co-efficients of p(x) is verified.
Q2.  Find the cubic polynomial with the sum, of the products of its zeroes taken two at a time and the product of its zeroes as 2, –7, –14 respectively.
Sol.  Let the required cubic polynomial be ax3 + bx2 + cx + d and its zeroes be α, β and γ Since, ∴The required cubic polynomial
= 1x3 + (–2)x2 + (–7)x + 14
= x3 – 2x2 – 7x + 14
Q3.  If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a and a + b then find ‘a’ and ‘b’.
Sol.  We have
p(x) = x3 – 3x2 + x + 1
comparing it with Ax3 + Bx2 + Cx + D.
We have
A = 1, B = –3, C = 1 and D = 1
∵It is given that (a – b), a and (a + b) are the zeroes of the polynomial.
∴Let,
α = (a – b)
β = a
and γ = (a + b) ⇒ (a – b) + a + (a + b) = 3
⇒ 3a = 3 ⇒ (a – b) × a × (a + b) = –1
⇒ (1– b) ×1× (1 + b) = –1
⇒ 1– b2 = –1 Q4.  If two zeroes of the polynomial x4 6x3 – 26x2 + 138x – 35 are find other zeroes.
Sol.  Here, p(x) = x4 – 6x3 – 26x2 + 138x – 35.
∴Two of the zeroes of p (x) are:  Now, dividing p (x) by x2 – 4x + 1, we have: ∴(x2 – 4x + 1)(x2 – 2x – 35) = p(x)
⇒ (x2 – 4x + 1) (x – 7) (x + 5) = p(x)
i.e., (x – 7) and (x + 5) are other factors of p(x).
∴7 and –5 are other zeroes of the given polynomial.
Q5.  If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be (x + α ), find k and α.
Sol.  Applying the division algorithm to the polynomials x4 – 6x3 + 16x2 – 25x + 10 and x2 – 2x + k, we have: ∴Remainder = (2k – 9) x – k(8 – k) + 10
But the remainder = x + α
Therefore, coparing them, we have: and
α = –k(8 – k) + 10
= –5(8 – 5) + 10
= –5(3) + 10
= –15 + 10
= –5
Thus, k = 5 and α = –5