NCERT Solutions for Class 10 Maths Polynomials Exercise 2.4
NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.4
In this page we have NCERT Solutions for Class 10 Maths Polynomials for
EXERCISE 2.4 on pages 36 and 37 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.
If $k_1$, $k_2$, $k_3$ are the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$, then
Question 1.
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) 2x3+x2-5x+ 2; 1/2, 1, -2
(ii)x3- 4x2+5x -2; 2, 1, 1 Answer
(i)p(x) = 2x3+x2-5x+ 2
Now for verification of zeroes, putting the given value in x.
P (1/2) = 2(1/2)3+(1/2)2- 5(1/2)+ 2
=(2×1/8)+ 1/4 - 5/2+ 2
= 1/4+ 1/4 - 5/2+ 2
= 1/2 - 5/2+ 2 = 0
P (1) = 2(1)3+(1)2- 5(1)+ 2
=(2×1)+ 1 - 5 + 2
= 2+ 1 - 5 + 2 = 0
P (-2) = 2(-2)3+(-2)2- 5(-2)+ 2
=(2 × -8)+ 4 + 10+ 2
= -16 + 16= 0
Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.
Comparing the given polynomial with ax3+bx2+cx+ d, we get a=2, b=1, c=-5, d=2
Also,k1=1/2, k2=1 and k3=-2
Now,
-b/a = k1 + k2 +k3
⇒ 1/2 = 1/2+ 1 - 2
⇒ 1/2 = 1/2
c/a =k1k2+k2k3+k1k3
⇒ -5/2 = (1/2 × 1)+ (1 × -2)+ (-2 × 1/2)
⇒ -5/2 = 1/2 - 2 - 1
⇒ -5/2 = -5/2
-d/a = k1k2k3
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1
Thus, the relationship between zeroes and the coefficients are verified.
(ii) p(x) =x3- 4x2+5x -2
Now for verification of zeroes, putting the given value in x.
p(2) = 23- 4(2)2+ 5(2)- 2
= 8 - 16 +10 - 2
= 0
p(1) = 13- 4(1)2+5(1)- 2
= 1 - 4+5 - 2
= 0
p(1) = 13- 4(1)2+5(1)- 2
= 1 - 4+5 - 2
= 0
Thus, 2, 1 and 1 are the zeroes of the given polynomial.
Comparing the given polynomial with ax3+bx2+cx+ d, we get a=1, b=-4, c=5, d=-2
Also,k1=2, k2=1 and k3=1
Now,
-b/a = k1 + k2 +k3
⇒ 4/1 = 2+ 1 + 1
⇒ 4 = 4
c/a =k1k2+k2k3+k1k3
⇒ 5/1 = (2 × 1)+ (1 × 1)+ (1 × 2)
⇒ 5 = 2 + 1+ 2
⇒ 5 = 5
-d/a = k1k2k3
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2
Thus, the relationship between zeroes and the coefficients are verified.
Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively. Answer
Let the polynomial beax3+bx2+cx+ dand the zeroes be k1, k2 and k3
Then, k1 + k2 +k3 = -(-2)/1 = 2 = -b/a
k1k2+k2k3+k1k3= -7 = -7/1 = c/a
k1k2k3 = -14 = -14/1 = -d/a
We can take any value of a and then find the values of b,c and d
Taking a = 1
b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will bex3- 2x2 -7x+ 14
Question 3.
If the zeroes of the polynomialx3– 3x2+x+ 1 are a–b, a, a+b, find a and b. Answer
Since, (a - b), a, (a + b) are the zeroes of the polynomialx3– 3x2+x+ 1.
Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3
⇒ 3a = 3 ⇒ a =1
∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1
a2- ab + a2+ ab + a2- b2= 1
⇒ 3a2- b2=1
Putting the value of a,
⇒ 3(1)2- b2= 1
⇒ 3 - b2= 1
⇒ b2= 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2
Question 4
If two zeroes of the polynomial x4– 6x3– 26x2+ 138x – 35 are 2±√3,find other zeroes. Answer
2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4– 6x3– 26x2+ 138x – 35.
So, $(x- 2 + \sqrt 3)(x -2 - \sqrt 3)$ is a factor of the polynomial
Multiplying them
$x^2- 4x + 1$ is a factor of polynomial p(x)
Another Method to get this would be
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2- 4x + 4 = 3,
⇒ x2- 4x + 1= 0
Now, dividing p(x) by x2- 4x + 1
p(x) = x4- 6x3- 26x2+ 138x - 35
= (x2- 4x + 1) (x2- 2x - 35)
= (x2- 4x + 1) (x2- 7x + 5x - 35)
= (x2- 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x2- 4x + 1) (x + 5) (x - 7)
So (x + 5) and (x - 7) are other factors of p(x).
Therefore
- 5 and 7 are other zeroes of the given polynomial.
Question 5.
If the polynomial x4– 6x3+ 16x2– 25x + 10 is divided by another polynomial x2– 2x + k, the remainder comes out to be x + a, find k and a.
Answer
On dividing x4– 6x3+ 16x2– 25x + 10 by x2– 2x + k
Remainder = (2k - 9)x - (8 - k)k + 10
But the remainder is given as x+ a.
On comparing their coefficients,
2k - 9 = 1
⇒ k = 10
⇒ k = 5 and,
-(8-k)k +10 = a
⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5
Hence, k = 5 and a = -5
