NCERT Solutions 10th Maths Chapter 2 Polynomials Exercise 2.4

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Exercise 2.4
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 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, a=1/2, fl=1 and ?=-2
Now,
-b/a = a+fl+?
? 1/2 = 1/2 + 1 - 2
? 1/2 = 1/2
c/a = afl+fl?+?a
? -5/2 = (1/2 ◊ 1) + (1 ◊ -2) + (-2 ◊ 1/2)
? -5/2 = 1/2 - 2 - 1
? -5/2 = -5/2
-d/a = afl?
? -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 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, a=2, fl=1 and ?=1
Now,
-b/a = a+fl+?
? 4/1 = 2 + 1 + 1
? 4 = 4
c/a = afl+fl?+?a
? 5/1 = (2 ◊ 1) + (1 ◊ 1) + (1 ◊ 2)
? 5 = 2 + 1 + 2
? 5 = 5
-d/a = afl?
? 2/1 = (2 ◊ 1 ◊ 1)
? 2 = 2
Thus, the relationship between zeroes and the coefficients are verified.
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, ñ7, ñ14 respectively.
Answer
Let the polynomial be ax3 + bx2 + cx + d and the zeroes be a, fl and ?
Then, a + fl + ? = -(-2)/1 = 2 = -b/a
afl + fl? + ?a = -7 = -7/1 = c/a
afl? = -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
3. If the zeroes of the polynomial x3 ñ 3x2 + x + 1 are añb, a, a+b, find a and b.
Answer
Since, (a - b), a, (a + b) are the zeroes of the polynomial x3 ñ 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 = ±v2
Hence, a = 1 and b = ±v2
4. If two zeroes of the polynomial x4 ñ 6x3 ñ 26x2 + 138x ñ 35 are 2±v3, find other zeroes.
Answer
2+v3 and 2-v3 are two zeroes of the polynomial p(x) = x4 ñ 6x3 ñ 26x2 + 138x ñ 35.
Let x = 2±v3
So, x-2 = ±v3
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)
? (x + 5) and (x - 7) are other factors of p(x).
? - 5 and 7 are other zeroes of the given polynomial.
5. If the polynomial x4 ñ 6x3 + 16x2 ñ 25x + 10 is divided by another polynomial x2 ñ 2x + k, the remainder comes out to be x + a, find k and a.
Answer
On dividing x4 ñ 6x3 + 16x2 ñ 25x + 10 by x2 ñ 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