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**Question 1)** If the polynomial f(x) = x

^{4} -6x

^{3} + 16x

^{2} – 25x + 10 is divided by another polynomial x

^{2} -2x + k, the remainder comes out to be x + a, find k and a.

Answer
Dividend = Divisor X Quotient + Remainder

Dividend - Remainder = Divisor X Quotient

Dividend - Remainder is always divisible by the divisor.

Now, it is given that f(x) when divided by x

^{2} -2x + k leaves (x + a) as remainder.

So f(x) -(x+a) is divided by x

^{2} -2x + k

x

^{4} -6x

^{3} + 16x

^{2} – 25x + 10 -(x+a) =x

^{4} -6x

^{3} + 16x

^{2} – 26x + 10 -a

Now doing the division

Now remainder should be zero

(-10 + 2k)x + (10 - a - 8k + k

^{2}) = 0

-10 + 2k = 0 or k=5

Now (10 -a - 8k + k

^{2}) = 0

or 10 - a - 8 (5) + 5

^{2} = 0

or - a - 5 = 0

a =-5

**Question 2)** Find all the zeroes of the polynomial x

^{4} – 3x

^{3 }+ 6x – 4, if two of its zeroes are √2 and -√2

**Question 3)** If p and q are he zeroes of the quadratic polynomial f(x) = x

^{2} – 2x + 3, find a polynomial whose roots are:

- p + 2, q + 2
- (p-1)/(p+1) , (q-1)/(q+1)

**Question 4)** For what value of k, -7 is the zero of the polynomial 2x

^{2} + 11x + (6k – 3)? Also find the other zero of the polynomial

**Question 5)** What must be added to f(x) = 4x

^{4 }+ 2x

^{3} – 2x

^{2} + x – 1 so that the resulting polynomial is divisible by g(x) = x

^{2} + 2x -3?

**Question 6)** Find k so that x

^{2} + 2x + k is a factor of 2x

^{4} + x

^{3} – 14 x

^{2} + 5x + 6. Also find all the zeroes of the two polynomials.

**Question 7)** If the zeroes of the quadratic polynomial ax

^{2} + 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

Answer
Given that, the zeroes of the quadratic polynomial ax^{2} + bx + c , c ≠ 0 are equal

Value of discriminant (D) has to be zero

b^{2}– 4ac = 0

b^{2} = 4ac

Since L.H.S. (b^{2}) can never be negative

R.H.S. also can never be negative.

a and c must have same sign

**Question 8)** Find the zeroes of 2x

^{3} – 11x

^{2} + 17x – 6.

**Question 9)** If (x - 2) and [x – ½ ] are the factors of the polynomials qx

^{2} + 5x + r prove that q = r

Answer
4q+10+r=0 -(1)

q/4 +5/2 +r=0 or q+10+4r=0 -(2)

Subtracting 1 from 2

3q-3r=0

q=r

**Question 10)** Find k so that the polynomial x

^{2} + 2x + k is a factor of polynomial 2x

^{4} + x

^{3} – 14x

^{2} + 5x + 6. Also, find all the zeroes of the two polynomials.

Answer
For x

^{2} + 2x + k is a factor of polynomial 2x

^{4} + x

^{3} – 14x

^{2} + 5x + 6, it should be able to divide the polynomial without any remainder

Comparing the coefficient of x we get.

21+7k=0 or k=-3

So x

^{2} + 2x + k becomes x

^{2} + 2x -3 = (x-1)(x+3)

Now

2x

^{4} + x

^{3} – 14x

^{2} + 5x + 6= (x

^{2} + 2x -3)(2x

^{2}-3x-8+2k)

=(x

^{2} + 2x -3)(2x

^{2}-3x-2)

=(x-1)(x+3)(x-2)(2x+1)

or x= 1,-3,2,=-1/2

**Question 11)** On dividing p(x) = x

^{3} – 3x

^{2} + x + 2 by a polynomial q(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).

**Question 12)** a, b, c are zeroes of cubic polynomial x

^{3} – 2x

^{2 }+ qx – r. If a + b = 0 then show that 2q = r.

**Question 13)** Find the remainder when x

^{51} +51 is divided by (x+1).

**Question 14)** a,b and c are zeroes of polynomial x

^{3 }+ px

^{2} + qx + 2 such that a b + 1 = 0. Find the value of 2p + q + 5.

Answer
By cubic polynomial equation, we have

a+b+c=-p

ab+bc+ac=q

abc=-2

Now given ab+1=0 or ab=-1

So abc=-2

(-1)c=-2 or c=2

2p + q + 5

=-2(a+b+c) +(ab+bc+ac) +5

=-2(a+b+2) +[-1+2(a+b)] +5

=-4 -1+5=0

**Question 15)** Find the quadratic polynomial, the sum and product of whose zeroes are 4 and 1, respectively

Answer
2) √2 , -√2, 2,1

4) -3, 3/2, -7

5) 61x – 65

11) x^{2} – x +1

Class 10 Maths
Class 10 Science