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NCERT Solutions for Class 10 Maths Polynomials Exercise 4<



In this page we have NCERT Solutions for Class 10 Maths Polynomials for EXERCISE 4 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
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 - 5+ 2; 1/2, 1, -2
(ii) x3 - 4x2 + 5x - 2; 2, 1, 1
Answer
(i) p(x) = 2x3 + x2 - 5+ 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 + c+ d, we get a=2, b=1, c=-5, d=2
Also, k1=1/2, k2=1 and k3=-2
NCERT Solutions  for Class 10 Maths Polynomials
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 + c+ d, we get a=1, b=-4, c=5, d=-2
Also, k1=2, k2=1 and k3=1
NCERT Solutions  for Class 10 Maths Polynomials
 
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 be ax3 + bx+ cx + d and the zeroes be k1, k2 and k3
NCERT Solutions  for Class 10 Maths Polynomials
Then, k1 + k2 +k3 = -(-2)/1 = 2 = -b/a
k1k2+k2k3+k1k3= -7 = -7/1 = c/a
k1k2k3 = -14 = -14/1 = -d/a
∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be x3 - 2x2  - 7x + 14

Question 3.
If the zeroes of the polynomial x3 – 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 polynomial x3 – 3x2 + x + 1.
NCERT Solutions  for Class 10 Maths PolynomialsTherefore, 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 - b= 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.
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
 
NCERT Solutions  for Class 10 Maths Polynomials
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
NCERT Solutions  for Class 10 Maths Polynomials
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 
 
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