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RD Chapter 2 Polynomials Ex MCQS Solutions

Question - 21 : -

If two of the zeros of the cubic polynomial ax3 + bx2 + cx + d are each equal to zero, then thethird zero is

Answer - 21 : -

(c) 

Two of the zeros of the cubic polynomial ax3 + bx2 + cx + d are each equal to zero
Let α, β and γ are its zeros, then


Third zero will be -b/a

Question - 22 : -

If two zeros of x3 + x2 – 5x – 5 are √5 and – √5 then its third zerois
(a) 1
(b) -1
(c) 2
(d) -2

Answer - 22 : - (b)


Question - 23 : -

The product of the zeros of x3 + 4x2 + x – 6 is
(a) – 4
(b) 4
(c) 6
(d) – 6

Answer - 23 : - (c)


Question - 24 : -

What should be added to the polynomial x2 – 5x + 4, so that 3 is the zero of the resultingpolynomial ?
(a) 1
(b) 2
(c) 4
(d) 5

Answer - 24 : - (b)

3 is the zero of the polynomial f(x) = x2 – 5x + 4
x – 3 is a factor of f(x)
Now f(3) = (3)2 – 5 x 3 + 4 = 9 – 15 + 4 = 13 – 15 = -2
-2 is to be subtracting or 2 is added

Question - 25 : -

What should be subtracted to the polynomial x2 – 16x + 30, so that 15 is the zero of the resultingpolynomial ?
(a) 30
(b) 14
(b) 15
(d) 16

Answer - 25 : - (c)

15 is the zero of polynomial f(x) = x2 – 16x + 30
Then f(15) = 0
f(15) = (15)2 – 16 x 15 + 30 = 225 – 240 + 30 = 255 – 240 = 15
15 is to be subtracted

Question - 26 : -

A quadratic polynomial, the sum of whose zeroes is 0 and onezero is 3, is
(a) x2 – 9
(b) x2 + 9
(c) x2 + 3
(d) x2 – 3

Answer - 26 : -

(a) In a quadratic polynomial
Let α and β be its zeros
and α + β = 0
and one zero = 3
3 + β = 0 ⇒ β = -3 .
Second zero = -3
Quadratic polynomial will be
(x – 3) (x + 3) ⇒ x2 – 9

Question - 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

Answer - 27 : - (a)


=> γ = -1
Third zero = -1

Question - 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

Answer - 28 : - (b)


Question - 29 : -

If x + 2 is a factor 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

Answer - 29 : -

(b) x + 2 is a factor of x2 + ax + 2b and a +b = 4
x + 2 is one of the factor
x = – 2 is its one zero
f(-2) = 0
=> (-2)2 + a (-2) + 2b = 0
=> 4 – 2a + 2b = 0
=> 2a – 2b = 4
=> a – b = 2
But a + b = 4
Adding we get, 2a = 6 => a = 3
and a + b = 4 => 3 + b = 4 => b = 4 – 3 = 1
a = 3, b = 1

Question - 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

Answer - 30 : -

(c) Divisor = – x2 + x – 1, Quotient= x – 2 and
Remainder = 3, Therefore
Polynomial = Divisor x Quotient+Remainder
= (-x2 + x – 1) (x – 2) +3
= – x3 + x2 – x + 2x2 – 2x + 2 + 3
= – x3 + 3x2 – 3x + 5

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