Go To: Exercise 2.1 Exercise 2.2 Exercise 2.3
NCERT Solutions Maths Class 10 Exercise 2.4 (Optional)
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:
(ii) x3 − 4x2 + 5x − 2;
2, 1, 1
Solution: (i) 2x3 + x2 − 5x + 2; 1/2, 1, −2
Comparing the given polynomial with the standard form of a cubic polynomial ax3 + bx2 + cx + d, we get
a = 2, b = 1, c = −5 and d = 2
= 
= 
= 0
P(1) = 2(1)3 + (1)2 – 5(1) + 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 2x3 + x2 – 5x + 2.
Now, α + β + γ
= 
And αβ + βγ + γα
= (1/2)(1) + (1)( −2) + (−2)(1/2)
= ½ – 2 – 1 = −5/2 = c/a
And αβγ = ½ × 1 × (−2) = −1 = −2/2 = −d/a
(ii) x3 − 4x2 + 5x − 2; 2, 1, 1
Comparing the given polynomial with the standard form of a cubic polynomial ax3 + bx2 + cx + d, we get
a = 1, b = −4, c = 5 and d = −2
= 8 – 16 + 10 − 2 = 0
p(1) = (1)3 −4(1)2
+ 5(1) −2
= 1 – 4 + 5 − 2 = 0
Thus, 2, 1 and 1 are the
zeroes of x3 − 4x2 + 5x – 2.
Now, α + β + γ = 2 + 1 + 1 = 4 = –(–4)/1 = –b/a
And αβ + βγ + γα = (2)(1)+(1)(1)+(1)(2)
= 2 + 1 + 2 = 5/1 = c/a
And αβγ = 2 × 1 × 1 = 2 = –(–2)/1 = –d/a
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 are 2, −7, −14 respectively.
Solution: Let the cubic polynomial be ax3 + bx2 + cx + d and its zeroes be α, β and γ.
Then α + β + γ = 2 = –(–2)/1
= –b/a and αβ
+ βγ + γα = –7 = –7/1 = c/a
And αβγ = –14
= –14/1 = –d/a
Here, a = 1, b = −2,
c = −7 and d = 14
Hence, the cubic polynomial will be x3 − 2x2 − 7x + 14.
3. If the zeroes of the
polynomial x3 − 3x2 + x + 1 are a − b, a, a + b, find a and b.
Solution: Since (a − b), a, (a + b) are the zeroes of
the polynomial x3 − 3x2
+ x + 1,
therefore, α + β + γ = a – b + a + a + b = –(–3)/1 = 3
⇒ 3a
= 3
⇒ a = 1
And αβ + βγ + γα = (a – b)a + a(a + b) + (a + b)(a
– b) = 1/1 = 1
⇒ a2 – ab + a2 + ab + a2 − b2 = 1
⇒ 3a2 − b2 = 1
⇒ 3(1)2
– b2 = 1 (Since, a = 1)
⇒ 3 – b2 = 1
⇒ b2 = 2
⇒ b = ±2
Hence, a = 1 and b
= ±2.
4. If two zeroes of the
polynomial x4 – 6x3 – 26x2
+ 138x – 35 are 2 ± √3, find other zeroes.
Solution: Let
p(x) = x4
– 6x3 –26x2 +138x – 35
Since (2 ± √3) are two zeroes of the polynomial p(x), x = 2 ± √3 will satisfy the given polynomial.
Thus, x – 2
= ±√3
Squaring both sides, we get
x2 – 4x + 4 = 3
x2 –
4x + 1 = 0
Now, we divide the polynomial p(x) by x2 – 4x
+ 1 to
obtain the other zeroes.
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 the other
factors of p(x).
Thus, (x + 5)(x – 7) = 0
⇒ x = –5, 7
Hence, –5 and 7 are the other
zeroes of the given polynomial.
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.
Solution: Let us divide the given
polynomial x4 –
6x3 + 16x2 – 25x + 10 by x2 – 2x + k, we get
Remainder = (2k − 9)x
– (8 – k)k + 10
On comparing this remainder with the given remainder, i.e. x + a, we get
2k – 9 = 1
2k = 10
k = 5
And –(8 − k)k + 10 = a
a = −(8 − 5)5 + 10 = −5
Hence, k = 5 and a = −5.