Chapter 2 - Polynomials, RD Sharma Solutions - (Part-7) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 2.61

Ques.26. A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is
(a) x² − 9
(b) x² + 9
(c) x² + 3
(d) x² − 3
Ans. Since α and β are the zeros of the quadratic polynomials such that
0 = α + β
If one of zero is 3 then
α + β = 0
3 + β = 0
β = 0 - 3
β = -3
Substituting β = -3 in α + β = 0 we get
α - 3 = 0
α = 3
Let S and P denote the sum and product of the zeros of the polynomial respectively then
S = α + β
S = 0
P = αβ
P = 3 x - 3
P = -9
Hence, the required polynomials is
= (x² - Sx + P)
= (x² - 0x - 9)
= x² - 9
Hence, the correct choice is a.

Ques.27. If two zeroes of the polynomial x³ + x² − 9x − 9 are 3 and −3, then its third zero is
(a) −1
(b) 1
(c) −9
(d) 9
Ans. Let α = 3 and β = -3 be the given zeros and γ be the third zero of the polynomial x³ + x² - 9x - 9 then 
By using α + β + γ = ((-Coefficient of x²/Coefficient of x³)
α + β + γ = (-1)/1
α + β + γ = -1
Substituting α =3 and β = -3 in α + β + γ = -1, we get
3 - 3 + γ = -1
γ = -1
Hence, the correct choice is a.

Ques.28. If √5 and −√5 are two zeroes of the polynomial x³ + 3x² − 5x − 15, then its third zero is
(a) 3
(b) −3
(c) 5
(d) −5
Ans. Let α = √5 and β = -√5 be the given zeros and γ be the third zero of the polynomial x³ + 3x² - 5x - 15. Then,
By using α + β + γ = ((-Coefficient of x²/Coefficient of x³)
α + β + γ = (-3)/1
α + β + γ = -3
Substituting α = √5 and β = -√5 in α + β + γ = -3
We get

Hence, the correct choice is b.

Ques.29. If x + 2 is a factor of x² + ax + 2b and a + b = 4, then
(a) a = 1, b = 3
(b) a = 3, b = 1
(c) a = −1, b = 5
(d) a = 5, b = −1
Ans. Given that x + 2 is a factor of x² + ax + 2b and a + b = 4
 
By solving -4 = -2a + 2b and a + b = 4 by elimination method we get
Multiply a + b = 4 by 2 we get, 
2a + 2b = 8. So
 
By substituting b = 1 in a + b = 4 we get
a + 1 = 4
a = 4 - 1
a = 3
Then a = 3, b = 1
Hence, the correct choice is b.

Ques.30. The polynomial which when divided by −x² + x − 1 gives a quotient x − 2 and remainder 3, is
(a) x³ − 3x² + 3x − 5
(b) −x³ − 3x² − 3x − 5
(c) −x³ + 3x² − 3x + 5
(d) x³ − 3x² − 3x + 5
Ans. We know that
f(x) = g(x)q(x) + r(x)
=(-x2 + x - 1)(x - 2) + 3
f(x) = g(x)q(x) + r(x)
= (-x² + x - 1)(x - 2) + 3
= -x³ + x² - x + 2x + 2 + 3
= -x³ + x² + 2x² - x - 2x + 2 + 3
= -x³ + 3x² - 3x + 5
Therefore,
The polynomial which when divided by -x² + x - 1 gives a quotient x - 2 and remainder 3, is-x³ + 3x² - 3x + 5
Hence, the correct choice is c.

Ques.31. The number of polynomials having zeroes  −2 and 5  is
(a) 1
(b) 2
(c) 3
(d) more than 3
Ans. Polynomials having zeros −2 and 5 will be of the form
p(x) = a(x + 2)ⁿ (x − 5)^m
Here, n and m can take any value from 1, 2, 3,...
Thus, the number of polynomials will be more than 3.
Hence, the correct answer is option D.

Ques.32. The zeroes of the quadratic polynomial x² + 99x + 127 are
(a) both positive
(b) both negative
(c) both equal
(d) one positive and one negative
Ans. Let f(x) = x² + 99x + 127.
Product of the zeroes of f(x) = 127 × 1 = 127 [Product of zeroes = c/a when f(x) = ax² + bx + c]
Since the product of zeroes is positive, we can say that it is only possible when both zeroes are positive or both zeroes are negative.
Also, sum of the zeroes =  –99
 [Sum of zeroes = −b/a when f(x) = ax² + bx + c]
The sum being negative implies that both zeroes are positive is not correct.
So, we conclude that both zeroes are negative.
Hence, the correct answer is option B.

Ques.33. If the zeroes of the quadratic polynomial x² + (a + 1) x + b. are 2 and −3, then
(a)  a = −7, b = −1
(b) a = 5 , b = −1
(c)  a =2 , b = −6
(d)  a = 0 , b = −6
Ans. The given quadratic equation is x² + (a + 1)x + b = 0.
Since the zeroes of the given equation are 2 and –3.
So,
α = 2 and β = −3
Now,
Sum of zeroes = -(Coefficient of x /Coefficient of x²)
⇒ 2 + (−3) = -((a+1)/1)
⇒ −1 = −a − 1
⇒ a = 0
Product of zeroes = Constant term/Coefficient of x²
⇒ 2 × (−3) = b/1
⇒ b = −6
So, a = 0 and b = −6
Hence, the correct answer is option D.

Ques.34. Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d  is zero , the product of the other two zeroes is
(a) ((-c)/a)
(b) c/a
(c) 0
(d) ((-b)/a)
Ans. Let p(x) = ax³ + bx² + cx + d.
Now 0 is the zero of the polynomial.
So, p(0) = 0.
⇒ a(0)³ + b(0)² + c(0) + d = 0
⇒ d = 0
So,
p(x) = ax³ + bx² + cx = x(ax² + bx + c)
Putting p(x) = 0, we get
x = 0 or ax² + bx + c = 0 .....(1)
Let α, β be the other zeroes of  ax² + bx + c = 0.
So, αβ = c/a
Hence, the correct answer is option B.

Ques.35. The zeroes of the quadratic polynomial x² + ax + a, a ≠ 0,
(a) cannot be both positive
(b) cannot both be negative
(c) are always unequal
(d) are always equal
Ans. Let f(x) = x² + ax + a.
Product of the zeroes of f(x) = a [Product of zeroes = c/a when f(x) = ax² + bx + c]
Since the product of zeroes is positive, so the zeroes must be either both positive or both negative.
Also, sum of the zeroes = –a [Sum of zeroes = -(b/a) when f(x) = ax² + bx + c]
So, the sum of the zeroes is negative, so the zeroes cannot be both positive.
Hence, the correct answer is option B.

Ques.36. If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is −1, then the product of other two zeroes is
(a) b − a + 1
(b) b − a − 1
(c) a − b + 1
(d) a − b − 1
Ans. Let  p(x)=x³ + ax² + bx + c.
Now, −1 is a zero of the polynomial.
So, p(−1) = 0.
⇒ (−1)³ + a(−1)² + b(−1) + c = 0
⇒ −1 + a −b + c = 0
⇒ a − b + c = 1
⇒ c = 1 − a + b
Now, if α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then product of zeroes is given by αβγ = -(d/a)
So, for the given polynomial, p(x) = x³ + ax² + bx + c
αβ(−1) = ((-c)/1) = ((-(1 - a - b))/1)
⇒ αβ = 1 − a + b
Hence, the correct answer is option A.

Ques.37. Given that two of the zeroes of the cubic polynomial ax³ + bx² + cx + d are 0, the third zero is
(a) -(b/a)
(b) b/a
(c) c/a
(d) -(d/a)
Ans. Let the polynomial be f(x)= ax³ + bx² + cx + d.
Suppose the two zeroes of f (x) are α = 0 and β = 0.
We know that,
Sum of the zeros,
α + β + γ = ((-b)/a)
⇒ 0 + 0 + γ = ((-b)/a)
⇒γ = ((-b)/a)
Hence, the correct answer is option A.

Ques.38. If one zero of the quadratic polynomial x² + 3x + k  is 2, then the value of k is
(a) 10
(b) −10
(c) 5
(d) −5
Ans. Let the given polynomial be f(x) = x² + 3x + k.
Since 2 is one of the zero of the given plynomial, so (x − 2) will be a factor of the given polynomial.
Now, f(2) = 0
⇒2² + 3 × 2 + k = 0
⇒ 4 + 6 + k = 0
⇒ k = −10
Hence, the correct answer is option B.

Ques.39. If the zeroes of the quadratic polynomial  ax² + bx + c, c ≠ 0  are equal, then
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c  and a have the same sign
(d) c and b have the same sign
Ans. Let the given quadratic polynomial be f(x) = ax² + bx + c.
Suppose α and β be the zeroes of the given polynomial.
Since α and β are equal so they will have the same sign i.e. either both are positive or both are negative.
So, αβ > 0
But, αβ= c/a
∴ c/a > 0, which is possible only when both have same sign
Hence, the correct answer is option C.

Ques.40. If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it
(a) has no linear term and constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.
Ans. Let the quadratic polynomial be f(x) = x² + ax + b.
Now, the zeroes are α and −α.
So, the sum of the zeroes is zero.
∴ α + (−α) = ((-a)/1) = -a
⇒ a = 0
So, the polynomial becomes f(x) = x² + b, which is not linear
Also, the product of the zeros,
αβ = b/1 = b
⇒ α(−α) = b
⇒ −α² = b
Thus, the constant term is negative.
Hence, the correct answer is option A.