RS Aggarwal Solutions Chapter 2 Polynomials Exercise  2B Class 10 Maths
Chapter Name  RS Aggarwal Chapter 2 Polynomials 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 2B Solutions
1. Verify that 3, 2, 1 are the zeros of the cubic polynomial p(x) = (x^{3}– 2x^{2}– 5x + 6) and verify the relation between it zeros and coefficients.
Solution
The given polynomial is p(x) = (x^{3 }– 2x^{2 }– 5x + 6)
∴ p(3) = (3^{3 }– 2×3^{2}  5×3 + 6) = (27–18–15 + 6) = 0
p(2) = [(–2^{3}) – 2×(–2)^{2 }– 5×(–2) + 6] = (–8 –8 +10+6) = 0
p(1) = (1^{3 }– 2×1^{2 }– 5×1 +6) = (12–5+6) = 0
∴ 3, –2 and 1are the zeroes of p(x),
Let Î± = 3, Î² = –2 and Î³ = 1. Then we have:
(Î± + Î² + Î³) = (3 – 2 + 1) = 2 = (coefficient of x^{2})/(coefficient of x^{3})
(Î±Î² + Î²Î³ + Î³Î±) = (6 – 2 + 3) = 5/1 = (coefficient of x)/(coefficient of x^{3})
Î±Î²Î³ = {(3 × (2) × 1} = 6/1 = (coefficient term)/(coefficient of x^{3})
2. Verify that 5, 2 and ^{1}_{3}are the zeroes of the cubic polynomial p(x) = (3x^{3 }– 10x^{2 }– 27x + 10) and verify the relation between its zeroes and coefficients.
Solution
p(x) = (3x^{3 }– 10x^{2 }– 27x + 10)
p(5) = (3×5^{3 }– 10×5^{2 }– 27×5 + 10) = (375–250–135+10) = 0
p(–2) = [3×(–2^{3}) – 10×(–2^{2}) – 27×(–2) + 10] = (–24 – 40 + 54 + 10) = 0
p(1/3) = {3×(1/3)^{3 }– 10×(1/3)^{2 }– 27× 1/3 + 10 } = (3× 1/27 – 10× 1/9 – 9 + 10)
= (1/9 – 10/9 + 1) = (1 – 10 – 9)/9 = (0/9) = 0
∴ 5, 2 and 1/3 are zeros of p(x).
Let Î± = 5, Î² = –2 and Î³ = 1/3. Then we have:
(Î± + Î² + Î³) = (5 – 2 + 1/3) = 10/3 = (coefficient of x^{2})/(coefficient of x^{3})
(Î±Î² + Î²Î³ + Î³Î±) = (10 – 2/3 + 5/3) = 27/3 = (coefficient of x)/(coefficient of x^{3})
Î±Î²Î³ = {5 × (2) × 1/3} = 10/3 = (coefficient term)/(coefficient of x^{3})
3. Find a cubic polynomial whose zeroes are 2, 3 and 4.
Solution
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x^{3 }– (a + b + c)x^{2 }+ (ab + bc + ca)x – abc …(1)
Let a = 2, b = –3 and c = 4
Substituting the values in 1, we get
x^{3 }– (2 – 3 + 4)x^{2 }+ (– 6 – 12 + 8)x – (–24)
⇒ x^{3 }– 3x^{2 }– 10x + 24
4. Find a cubic polynomial whose zeroes are 1/2, 1 and –3.
Solution
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x^{3 }– (a + b + c)x^{2 }+ (ab + bc + ca)x – abc …(1)
Let a = 1/2, b = 1 and c = –3
Substituting the values in (1), we get
x^{3 }– (1/2 + 1 − 3)x^{2 }+ (1/2 – 3 – 3/2)x – (3/2)
⇒ x^{3 }– (3/2)x^{2} – 4x + 3/2
⇒ 2x^{3 }+ 3x^{2} – 8x + 3
5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as 5, 2 and 24 respectively.
Solution
We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as
x^{3 }– (sum of the zeroes)x^{2 }+ (sum of the product of the zeroes taking two at a time)x – product of zeroes
Therefore, the required polynomial is
x^{3}– 5x^{2 }– 2x + 24
6. If f(x) = x^{3} – 3x + 5x – 3 divided by g(x) = x^{2} – 2
Solution
Quotient q(x) = x – 3Remainder r(x) = 7x – 9
7. If f(x) = x^{4 }– 3x^{2 }+ 4x + 5 is divided by g(x)= x^{2 }– x + 1
Solution
Remainder r(x) = 8
8. If f(x) = x^{4 }– 5x + 6 is divided by g(x) = 2 – x^{2}.
Solution
We can write
f(x) as x^{4 }+ 0x^{3 }+ 0x^{2 }– 5x + 6 and g(x) as – x^{2 }+ 2
Quotient q(x) = – x^{2}– 2
Remainder r(x) = –5x + 10
9. By actual division, show that x^{2 }– 3 is a factor of 2x^{4 }+ 3x^{3 }– 2x^{2 }– 9x – 12.
Solution
Let f(x) = 2x^{4 }+ 3x^{3}– 2x^{2}– 9x – 12 and g(x) as x^{2 }– 3
Quotient q(x) = 2x^{2 }+ 3x + 4
Remainder r(x) = 0
Since, the remainder is 0.
Hence, x^{2}– 3 is a factor of 2x^{4 }+ 3x^{3}– 2x^{2}– 9x – 12
10. On dividing 3x^{3 }+ x^{2 }+ 2x + 5 is divided by a polynomial g(x), the quotient and remainder are (3x – 5) and (9x + 10) respectively. Find g(x).
Solution
By using division rule, we have
Dividend = Quotient × Divisor + Remainder
∴ 3x^{3 }+ x^{2 }+ 2x + 5 = (3x – 5)g(x) + 9x + 10
⇒ 3x^{3 }+ x^{2 }+ 2x + 5 – 9x – 10 = (3x – 5)g(x)
⇒ 3x^{3 }+ x^{2 }– 7x – 5 = (3x – 5)g(x)
⇒ g(x) = (3x^{3 }+ x^{2} – 7x – 5)/(3x – 5)
∴ g(x) = x^{2 }+ 2x + 1
11. Verify division algorithm for the polynomial f(x) = (8 + 20x + x^{2 }– 6x^{3}) by g(x) = (2 + 5x – 3x^{2}).
Solution
We can write f(x) as – 6x^{3 }+ x^{2 }+ 20x + 8 and g(x) as –3x^{2 }+ 5x + 2
Quotient = 2x + 3
Remainder = x + 2
By using division rule, we have
Dividend = Quotient × Divisor + Remainder
∴ – 6x^{3 }+ x^{2 }+ 20x + 8 = (– 3x^{2 }+ 5x + 2) (2x + 3) + x + 2
⇒ – 6x^{3 }+ x^{2 }+ 20x + 8 = –6x^{3 }+ 10x^{2 }+ 4x – 9x^{2 }+ 15x + 6 + x + 2
⇒ – 6x^{3 }+ x^{2 }+ 20x + 8 = – 6x^{3 }+ x^{2 }+ 20x + 8
12. It is given that –1 is one of the zeroes of the polynomial x^{3 }+ 2x^{2}– 11x – 12. Find all the zeroes of the given polynomial.
Solution
Let f(x) = x^{3 }+ 2x^{2 }– 11x – 12
Since – 1 is a zero of f(x), (x + 1) is a factor of f(x).
On dividing f(x) by (x + 1), we get
f(x) = x^{3 }+ 2x^{2}– 11x – 12
= (x + 1)(x^{2 }+ x – 12)
= (x + 1){x^{2 }+ 4x – 3x – 12}
= (x + 1){x(x + 4) – 3(x + 4)}
= (x + 1)(x – 3)(x + 4)
∴ f(x) = 0 ⇒ (x + 1)(x – 3)(x + 4) = 0
⇒ (x + 1) = 0 or (x – 3) = 0 or (x + 4) = 0
⇒ x = – 1 or x = 3 or x = – 4
Thus, all the zeroes are – 1, 3 and – 4.
13. If 1 and –2 are two zeroes of the polynomial (x^{3}– 4x^{2}– 7x + 10), find its third zero.
Solution
Let f(x) = x^{3}– 4x^{2}– 7x + 10
Since 1 and –2 are the zeroes of f(x), it follows that each one of (x – 1) and (x + 2) is a factor of f(x).
Consequently, (x – 1)(x + 2) = (x^{2 }+ x – 2) is a factor of f(x).
On dividing f(x) by (x^{2 }+ x – 2), we get:
⇒ (x – 1)(x + 2)(x – 5) = 0
⇒ x = 1 or x = – 2 or x = 5
Hence, the third zero is 5.
14. If 3 and –3 are two zeroes of the polynomial (x^{4 }+ x^{3}– 11x^{2}– 9x + 18), find all the zeroes of the given polynomial.
Solution
Let x^{4 }+ x^{3 }– 11x^{2 }– 9x + 18
Since 3 and – 3 are the zeroes of f(x), it follows that each one of (x + 3) and (x – 3) is a factor of f(x).
Consequently, (x – 3) (x + 3) = (x^{2}– 9) is a factor of f(x).
On dividing f(x) by (x^{2}– 9), we get:
f(x) = 0 ⇒ (x^{2 }+ x – 2) (x^{2 }– 9) = 0
⇒ (x^{2 }+ 2x – x – 2) (x – 3) (x + 3)
⇒ (x – 1) (x + 2) (x – 3) (x + 3) = 0
⇒ x = 1 or x = 2 or x = 3 or x = 3
Hence, all the zeroes are 1, 2, 3 and 3.
15. If 2 and 2 are two zeroes of the polynomial (x^{4 }+ x^{3}– 34x^{2}– 4x + 120), find all the zeroes of the given polynomial.
Solution
Let f(x) = x^{4 }+ x^{3}– 34x^{2}– 4x + 120
Since 2 and 2 are the zeroes of f(x), it follows that each one of (x – 2) and (x + 2) is a factor of f(x).
Consequently, (x – 2) (x + 2) = (x^{2}– 4) is a factor of f(x).
On dividing f(x) by (x^{2}– 4), we get:
f(x) = 0
⇒ (x^{2 }+ x – 30) (x^{2 }– 4) = 0
⇒ (x^{2 }+ 6x – 5x – 30) (x – 2) (x + 2)
⇒ [x(x + 6) – 5(x + 6)] (x – 2) (x + 2)
⇒ (x – 5) (x + 6) (x – 2) (x + 2) = 0
⇒ x = 5 or x = 6 or x = 2 or x = 2
Hence, all the zeroes are 2, 2, 5 and 6.
16. Find all the zeroes of (x^{4 }+ x^{3 }– 23x^{2 }– 3x + 60), if it is given that two of its zeroes are √3 and –√3.
Solution
Let f(x) = x^{4 }+ x^{3 }– 23x^{2 }– 3x + 60
Since √3 and –√3 are the zeroes of f(x), it follows that each one of (x – √3) and (x + √3) is a factor of f(x).
Consequently, (x – √3) (x + √3) = (x^{2 }– 3) is a factor of f(x).
On dividing f(x) by (x^{2 }– 3), we get:
⇒ (x^{2 }+ x – 20) (x^{2 }– 3) = 0
⇒ (x^{2 }+ 5x – 4x – 20)(x^{2 }– 3)
⇒ [x(x + 5) – 4(x + 5)](x^{2 }– 3)
⇒ (x – 4) (x + 5) (x – √3)(x + √3) = 0
⇒ x = 4 or x = 5 or x = √3 or x = √3
Hence, all the zeroes are √3, √3, 4 and 5.
17. Find all the zeroes of (2x^{4 }– 3x^{3 }– 5x^{2 }+ 9x – 3), it is being given that two of its zeroes are √3 and –√3.
Solution
The given polynomial is f(x) = 2x^{4 }– 3x^{3 }– 5x^{2 }+ 9x – 3
Since √3 and –√3 are the zeroes of f(x), it follows that each one of (x – √3) and (x + √3) is a factor of f(x).
Consequently, (x – √3) (x + √3) = (x^{2 }– 3) is a factor of f(x).
On dividing f(x) by (x^{2 }– 3), we get:
f(x) = 0
⇒ 2x^{4 }– 3x^{3 }– 5x^{2 }+ 9x – 3 = 0
⇒ (x^{2 }– 3) (2x^{2 }– 3x + 1) = 0
⇒ (x^{2 }– 3) (2x^{2 }– 2x – x + 1) = 0
⇒ (x – √3) (x + √3) (2x – 1) (x – 1) = 0
⇒ x = √3 or x = √3 or x = 1/2 or x = 1
Hence, all the zeroes are √3, √3, 1/2 and 1.
18. Obtain all other zeroes of (x^{4 }+ 4x^{3 }– 2x^{2}– 20x – 15) if two of its zeroes are √5 and –√5.
Solution
The given polynomial is f(x) = x^{4 }+ 4x^{3}– 2x^{2}– 20x – 15.
Since (x – √5) and (x + √5) are the zeroes of f(x) it follows that each one of (x – √5) and
(x + √5) is a factor of f(x).
Consequently, (x – √5) (x + √5) = (x^{2 }– 5) is a factor of f(x).
On dividing f(x) by (x^{2 }– 5), we get:
⇒ x^{4 }+ 4x^{3 }– 7x^{2 }– 20x – 15 = 0
⇒ (x^{2 }– 5) (x^{2 }+ 4x + 3) = 0
⇒ (x – √5) (x + √5) (x + 1) (x + 3) = 0
⇒ x = √5 or x = √5 or x = 1 or x = 3
Hence, all the zeroes are √5, √5, 1 and 3.
19. Find all the zeroes of polynomial (2x^{4}– 11x^{3 }+ 7x^{2 }+ 13x – 7), it being given that two of its zeroes are (3 + √2) and (3 – √2).
Solution
The given polynomial is f(x) = 2x^{4 }– 11x^{3 }+ 7x^{2 }+ 13x – 7.
Since (3 + √2) and (3 – √2) are the zeroes of f(x) it follows that each one of (x + 3 + √2) and (x + 3 – √2) is a factor of f(x).
Consequently, [(x – (3 + √2)] [(x – (3 – √2)] = [(x – 3)  √2] [(x – 3) + √2] = [(x – 3)^{2 }– 2] = x^{2 }– 6x + 7, which is a factor of f(x).
On dividing f(x) by (x^{2 }– 6x + 7), we get:
f(x) = 0
⇒ 2x^{4 }– 11x^{3 }+ 7x^{2 }+ 13x – 7 = 0
⇒ (x^{2 }– 6x + 7) (2x^{2 }+ x – 7) = 0
⇒ (x + 3 + √2) (x + 3 – √2) (2x – 1) (x + 1) = 0
⇒ x = –3 – √2 or x = –3 + √2 or x = ^{1}_{2}or x = 1
Hence, all the zeroes are (–3 – √2), (–3 + √2), 1/2 and 1.