## Chapter 2 Polynomials R.D. Sharma Solutions for Class 10th Math MCQ's

**Multiple Choice Questions**

1. If Î±, Î² are the zeros of the polynomial f(x) = x

^{2}Â + x + 1 , then 1/Î± + 1/ Î² =
(a) 1

(b) -1

(c) 0

(d) None of these

Since Î± and Î² are the zeros of the quadratic polynomial f(x) = x

**Solution**Since Î± and Î² are the zeros of the quadratic polynomial f(x) = x

^{2}Â + x + 1
Î± + Î² = - coefficient of x/coefficient of x

^{2}
= -1/1 = 1

Î± Ã— Î² = constant term / coefficient of x

^{2}
= 1/1 = 1

We have

1/Î± + 1/Î² =Â Â

= Î²+Î±/ Î±Î²

= -1/1

= -1

The value of 1/Î± + 1/Î² is -1

Hence , the correct choice is b .

2.Â If Î±, Î² are the zerosÂ Â of the polynomial p(x) = 4x

2.Â If Î±, Î² are the zerosÂ Â of the polynomial p(x) = 4x

^{2}Â + 3x + 7, then 1/Î± + 1/Î² is equal toÂ
(a) 7/3

(b) â€“ 7/3

(c) 3/7

(d) -3/7Â

**Solution**

Since Î± and Î² are the zeros of the quadratic polynomial p(x) = 4x

Î± + Î² = -coefficient of x / coefficient of x

= -3/4

= Î±Î² = Constant term / Coefficient of x

= 7/4

We have

We are given f(x) = (k

^{2}Â + 3x + 7Î± + Î² = -coefficient of x / coefficient of x

^{2}= -3/4

= Î±Î² = Constant term / Coefficient of x

^{2}= 7/4

We have

3. If one zero of the polynomial f(x) = (k

^{2}Â + 4)x^{2}Â + 13x + 4k is reciprocal of the other, then k =Â
(a) 2

(b) -2

(c) 1

(d) -1

**Solution**

^{2}Â + 4)x^{2}Â 13x + 4k then
Î± + Î² = -Coefficient of x/ Coefficient of x

^{2}Â Â
= -13/k

^{2}Â + 4
Î± Ã—Â Â Î² = Constant term / Coefficient of x

^{2}Â Â
= 4k/k

^{2}Â + 4
One root of the polynomial is reciprocal of the other. Then , we have

Î± Ã—Â Â Î² = 1

= 4k/k

^{2}+4
= 1

= k

^{2}Â â€“ 4k + 4 = 0
= (k-2)

^{2}Â = 0
= k = 2

Hence the correct choice is a .

4. If the sum of the zeros of the polynomial f(x) = 2x

4. If the sum of the zeros of the polynomial f(x) = 2x

^{3}Â âˆ’ 3kx^{2}Â + 4x âˆ’ 5 is 6, then the value of k is
(a) 2

(b) 4

(c) âˆ’2

(d) âˆ’4

(b) 4

(c) âˆ’2

(d) âˆ’4

**Solution**

Let Î±, Î² be the zeros of the polynomial f(x) = 2x

^{3}Â â€“ 3kx^{2}Â + 4x -5 and we are that Î± + Î² + Î³ = 6 .
Then,

Î± + Î² + Î³ = -Coefficient of x/Coefficient of x2

= - (-3k)/2 = 3k/2

It is given that

Î± + Î² + Î³ = 6

Substituting Î± + Î² + Î³ = 3k/2 , we get

+3k/2 = 6

+ 3k = 6 Ã— 2

+3k = 12

k = 12/+3

k = +4

The value of k is 4 .

Hence, the correct alternative is b .

5. If Î± andÂ Â Î² are the zeros of the polynomial f(x) = x

^{2}Â + px + q, thenÂ Â a polynomial havingÂ Â 1/Î± and 1/ Î² is its zero is
(a) x

_{2}Â + qx + p
(b) x

_{2}Â â€“ px + q
(c) qx

_{2}Â + px + 1
(d) px

_{2}Â + qx + 1Â**Solution**

Let Î± andÂ Â Î² are the zeros of the polynomial f(x) = x

^{2}Â + px + q . Then ,

Î± + Î² = -Coefficient of x/ Coefficient of x

^{2}
= -p/1

= -p

And

Î±Î² = Constant term/Coefficient of x

^{2}
= q/1

= q

Let S and R denote respectively the sum and the product of the zeros of a polynomial

Whose zeros are 1/Î± and 1/Î² . then

S = 1/Î± + 1/Î²

= Î±+ Î²/ Î±Î²

= -p/q

R =Â Â 1/Î± Ã— 1/Î²

= 1/ Î±Î²

= 1/q

Hence, the required polynomial g(x) whose sum and product of zeros are S and R is given by

x

^{2}Â Â - Sx + R = 0
x

^{2}Â + P/q x + 1/q = 0
qx

^{2}Â + Px + 1/q = 0
qx

^{2}Â + Px + 1
So g(x) = qx

^{2}Â + Px + 1
Hence , the correct choice is c .

6. If Î±,Î² are the zeros of polynomial f(x) = x

^{2}Â â€“ p(x+1) â€“ c , then (Î± + 1) (Î² + 1) =
(a) c â€“ 1

(b) 1 â€“ c

(c) c

(d) 1 + c

**Solution**

Since, Î± and Î² are the zeros of quadratic polynomial

f(x) = x

^{2}Â â€“ p(x+1)-c
= x

^{2}
Î± + Î² = -Coefficient of x/Coefficient of x

^{2}
= -(-p/1)

= p

Î±Ã—Î² = Constant term/Coefficient of x

^{2}
= -p-c/1

= -p-c

We have

(Î± +1)( Î²+1)

=Â Â Î±Î² + Î² + Î± + 1

= Î±Î² + (Î±+Î²) + 1

= - p - c + (p) + 1

= - p - c + (p) + 1

= -c + 1

= 1 â€“c

The value of (Î± +1)( Î²+1) is 1 â€“ c .

Hence, the correct choice is b .

7. If Î±, Î² are the zeros of the polynomial f(x) = x

7. If Î±, Î² are the zeros of the polynomial f(x) = x

^{2}Â âˆ’ p(x + 1) âˆ’ c such that (Î± +1) (Î² + 1) = 0, then c =
(a) 1

(b) 0

(c) âˆ’1

(d) 2

(b) 0

(c) âˆ’1

(d) 2

**Solution**

Since, Î± and Î² are the zeros of quadratic polynomial

f(x) = x

^{2}Â â€“ p(x+1) â€“ c
f(x) = x

^{2}Â â€“ px â€“ p â€“ c
Î±+Î² = -Coefficient of x/Coefficient of x

^{2}
= -(-p/1)

= p

Î±Î² = Constant term/Coefficient of x

^{2}
= -p - c/1

= -p-c

We have

0 = (Î±+1)(Î²+1)

0 =Â Â Î±Î² + (Î±+Î²) + 1

0 = -p-c+p+1

0 = -c + 1

c = 1

The value of c is 1 .

Hence, the correct alternative is a.

8. If f(x) = ax

(a) c = 0

(b) c > 0

(c) c < 0

(d) None of these

^{2}Â + bx + c has no real zeros and a + b + c < 0, then(a) c = 0

(b) c > 0

(c) c < 0

(d) None of these

**Solution**

If f(x) = ax

^{2}Â + bx + c has no real zeros and a + b + c < 0 then c < 0

Hence, the correct choice is c .

Â 9. If the diagram in Fig. 2.22 shows the graph of the polynomial f(x) = ax

(a) a > 0, b < 0 and c > 0

(b) a < 0, b < 0 and c < 0

(c) a < 0, b > 0 and c > 0

(d) a < 0, b > 0 and c < 0

Â 9. If the diagram in Fig. 2.22 shows the graph of the polynomial f(x) = ax

^{2}Â + bx + c, then(b) a < 0, b < 0 and c < 0

(c) a < 0, b > 0 and c > 0

(d) a < 0, b > 0 and c < 0

**Solution**
Clearly, f(x) = ax

Therefore, f(x) = ax

Hence , the correct choice is a .

^{2}Â + bxÂ Â + c represent a parabola opening upwar .Therefore, f(x) = ax

^{2}Â + bx + c cuts Y axis at P which lies on OY. Putting x = 0 in y = ax^{2}Â + bx + c, we get y = c. So the coordinates of P is (0,c). Clearly , P lies on OY. Therefore c > 0Hence , the correct choice is a .

10. Figure 2.23 show the graph of the polynomial f(x) = ax

(a) a < 0, b > 0 and c > 0

(b) a < 0, b < 0 and c > 0

(c) a < 0, b < 0 and c < 0

(d) a > 0, b > 0 and c < 0

Clearly, f(x) = ax

y = ax

The vertex (-b/2a, -D/4a) of the parabola is in the second quadrant. Therefore -b/2a<0, b<0

Therefore a<0,b<0, and c>0

Hence , the correct choice is b

11. If the product of zeros of the polynomial f(x) ax

(a) 3/2

(b) -3/2

(c) 2/3

(d) -2/3

^{2}Â + bx + c for which(b) a < 0, b < 0 and c > 0

(c) a < 0, b < 0 and c < 0

(d) a > 0, b > 0 and c < 0

**Solution**Clearly, f(x) = ax

^{2}Â + bx + c represent a parabola opening downwards. Therefore, a<0y = ax

^{2}Â + bx + c cuts y - axis at P which lies on OY. Putting x = 0 in y = ax^{2}Â + bx + c, we get y = c.So the coordinates P are (0,c). Clearly , P lies on OY. Therefore c > 0The vertex (-b/2a, -D/4a) of the parabola is in the second quadrant. Therefore -b/2a<0, b<0

Therefore a<0,b<0, and c>0

Hence , the correct choice is b

11. If the product of zeros of the polynomial f(x) ax

^{3}Â âˆ’ 6x^{2}Â + 11x âˆ’ 6 is 4, then a =(a) 3/2

(b) -3/2

(c) 2/3

(d) -2/3

**Solution**
Since Î± and Î² are the zeros of quadratic polynomial f(x) = ax

Î±Î² = -Constant term/Coefficient of x

So we have

4 = -(-6/a)

4 = 6/a

4a = 6

a = 6/4

a = 3Ã—2/2Ã—2

a = 3/2

The value of a is 3/2

Hence, the correct alternative is a .

^{3}Â - 6x^{2}Â + 11x - 6Î±Î² = -Constant term/Coefficient of x

^{2}So we have

4 = -(-6/a)

4 = 6/a

4a = 6

a = 6/4

a = 3Ã—2/2Ã—2

a = 3/2

The value of a is 3/2

Hence, the correct alternative is a .

12. If zeros of the polynomial f(x) = x

^{3}Â âˆ’ 3px^{2}Â + qx âˆ’ r are in A.P., then
(a) 2p

^{3}Â = pq âˆ’ r
(b) 2p

^{3}Â = pq + r
(c) p

^{3}Â = pq âˆ’ r
(d) None of these

**Solution**

Let a-d,a,a+d be the zeros of the polynomial f(x) = x

^{3}Â - 3px^{2}Â + qx - r then
Sum of zeros = -Coefficient ofx

^{2}/Coefficient of x^{3}
(a-d)+a+(a+d) = -(-3p)/1

a-d + a + a + d = 3p

3a = 3p

a = 3/3p

a = p

Since a is a zero of the polynomial f(x)

Therefore,

f(a) = 0

a

^{3}Â - 3pa^{3}Â + qa - r = 0
Substituting a = p . we get

p

^{3}Â - 3p(p)^{2}Â + qÃ—p-r=0
p

-2p^{3}Â - 3p^{3}Â + qp - r = 0^{3 + qp - r = 0}qp - r = 2p

^{3}

Hence, the correct choice is a.

^{}13. If the product of two zeros of the polynomial f(x) = 2x

^{3}Â + 6x

^{2}Â âˆ’ 4x + 9 is 3, then its third zero isÂ

^{}(a) 3/2

(b) -3/2

(c) 9/2

(d) -9/2

^{}

**Solution**

^{Â }

^{}

^{Let Î±,Î²,Î³ be the zeros of polynomial f(x) = 2x3Â + 6x2Â - 4x + 9 such that Î±Î² = 3We have,Î±Î²Î³ = Constant term/Coefficient of x3= -9/2PuttingÂ Î±Î² = 3 in Î±Î²Î³ = -9/2, we getÎ±Î²Î³ = -9/23Î³ = -9/2Î³ =Â -9/2 Ã— 1/3Î³ = -3/2Therefore, the value of third zero is -3/2Hence, the correct alternative is b.Â }

^{}14. If the polynomial f(x) = ax

^{3}Â + bx âˆ’ c is divisible by the polynomial g(x) = x

^{2}Â + bx + c, then ab =

(a) 1

(b) 1/c

(c) -1

(d) -1/c

**Solution**

We have to find the value of ab

Given f(x) = ax

^{3}Â + bx - c is divisible by the polynomial g(x) = x^{2}Â + bx + c
We must have

bx - acx + ab

^{2}x + abc - c = 0, for all x
So put x = 0 in this equation

x(b-ac+ab

^{2})+c(ab-1) = 0
c(ab-1) = 0

Since câ‰ 0 , so

ab - 1 = 0

â‡’ ab = 1

Hence, the correct alternative is a.

15. If the polynomial f(x) = ax

^{3}Â + bx âˆ’ c is divisible by the polynomial g(x) = x^{2}Â + bx + c, then c =
(a) b

(b) 2b

(c) 2b

^{2}
(d) âˆ’2b

**Solution**

We have to find the value of c

Given f(x) = ax

^{3}Â + bx -c is divisible by the polynomial g(x) = x^{2}Â + bx + c
We must have,

bx - acx + ab

^{2}x + abc - c = 0 for all x
x(b-ac+b

^{2}) + c(ab - 1) = 0 ...(1)
c(ab-1) = 0

Since câ‰ 0, so

ab - 1 = 0

ab = 1

Now in the equation (1) the condition is true for all x. So put x =1 and also we have ab = 1

Therefore we have

b-ac+ab

^{2}=0
b+ab

^{2}Â -ac=0
b(1+ab)-ac=0

Substituting a = 1/b and ab = 1 we get,

b(1+1) - 1/bÃ—c=0

2b - 1/bÃ—c = 0

-1/bÃ—c = -2b

c=2bÃ—b/1

c=2b

^{2}
Hence, the correct alternative is c .

16. If one root of the polynomial f(x) = 5x

^{2}Â + 13x + k is reciprocal of the other, then the value of k is
(a) 0

(b) 5

(c) 1/6

(d) 6

**Solution**

If one zero of the polynomial f(x) = 5x

^{2}Â + 13x + k is reciprocal of the other . So Î² = 1/Î± â‡’ Î±Î² = 1
Now we have

Î±Ã—Î² = Constant term/Coefficient of x

^{2}
= k/5

Since Î±Î² = 1

Therefore we have

Î±Î² = k/5

1= k/5

â‡’ k = 5

Hence the correct choice is b .

17. If Î±,Î²,Î³ are the zeros of the polynomial f(x) ax

^{3}Â + bx^{2}Â + cx + d, then 1/Î± + 1/Î² + 1/Î³ =
(a) -b/d

(b) c/d

(c) -c/d

(d) -c/a

**Solution**

We have to find the value of 1/Î± + 1/Î² + 1/Î³ =

Given Î±,Î²,Î³ be the zeros of the polynomial f(x) = ax

^{3}Â + bx^{2}Â + cx + d
we know that

Î±Î² +Î²Î³ + Î³Î± = coefficient of x/coefficient of x

^{3}Â
= c/a

Î±Î²Î³ = -Constant term/Coefficient of x

^{3}
= -d/a

So

1/Î± + 1/Î² + 1/Î³ = Î²Î³ + Î±Î³ + Î±Î²/Î±Î²Î³

1/Î± + 1/Î² + 1/Î³ = c/a/-d/a

1/Î± + 1/Î² + 1/Î³ = c/a Ã— (-a/d)

1/Î± + 1/Î² + 1/Î³ = -c/d

Hence, the correct choice is c.Â

^{}

18. If Î±, Î², Î³ are the zeros of the polynomial f(x) = ax

^{3}Â + bx^{2}Â + cx + d, then Î±^{2}Â + Î²^{2}Â + Î³^{2}Â =^{Â }
(a) b

^{2}-ac/a^{2}
(b) b

^{2}-2ac/a
(c) b

^{2}Â + 2ac/b^{2}
(d) b

^{2}Â - 2ac/a^{2}^{}

**Solution**

^{Â }

^{}

We have to find the value of Î±

^{2}Â + Î²^{2}Â + Î³^{2}
Given Î±,Î²,Î³ be the zeros of the polynomial f(x) - ax

^{3}Â + bx^{2}Â + cx + d
Î±+Î²+Î³ = -Coefficient of x

^{2}/Coefficient of x^{3}
= -b/a

Î±Î² + Î²Î³ + Î³Î± = Coefficient of x/Coefficient of x

^{3}
= c/a

Now

Î±

^{2}Â + Î²^{2}Â + Î³^{2}Â = (Î±+Î²+Î³)-2(Î±Î² + Î²Î³ + Î³Î±)
Î±

^{2}Â + Î²^{2}Â + Î³^{2}Â = (-b/a)^{2}Â -2(c/a)
Î±

^{2}Â + Î²^{2}Â + Î³^{2}Â = b^{2}/Î±^{2}Â - 2c/a
Î±

^{2}Â + Î²^{2}Â + Î³^{2}Â = b^{2}/Î±^{2}Â - 2ca/Î±^{2}
Î±

^{2}Â + Î²^{2}
+ Î³

^{2}Â = b^{2}-2ac/a^{2}
The value of Î±

^{2}Â + Î²^{2}Â + Î³^{2}Â = b^{2}-2ac/a^{2}
Hence, the correct choice is d.

^{Â }^{}19. If Î±, Î², Î³ are the zeros of the polynomial f(x) = x

^{3}Â - px

^{2}Â + qx - r , then 1/Î±Î² + 1/Î²Î³ + 1/Î³Î± =

(a) r/p

(b) p/r

(c) -p/r

(d) -r/p

**Solution**

If Î±, Î², Î³ are the zeros of the polynomial f(x) = x

^{3}Â - px^{2}Â + qx - r , then 1/Î±Î² + 1/Î²Î³ + 1/Î³Î± =
We have to find the value of 1/Î±Î² + 1/Î²Î³ + 1/Î³Î±

Given Î±,Î²,Î³ be the zeros of the polynomial f(x) = x

^{3}Â - px^{2}Â + qx - r
Î±+Î²+Î³ = -Coefficient of x

^{2}/Coefficient of x^{3}
= -(p)/1

= p

Î±Î²Î³ = -Constant term/Coefficient of x

^{3}
= -(r)/1

= r

Now we calculate the expression

1/Î±Î² + 1/Î²Î³ + 1/Î³Î± = Î³/Î±Î²Î³ + Î±/Î±Î²Î³ + Î²/Î±Î²Î³

1/Î±Î² + 1/Î²Î³ + 1/Î³Î± = Î±+Î³+Î²/Î±Î²Î³

1/Î±Î² + 1/Î²Î³ + 1/Î³Î± = p/r

Hence, the correct choice is b.

^{}20. If Î±,Î² are the zeros of the polynomial f(x) = ax

^{2}Â + bx + c, then 1/Î± + 1/Î² =

(a) b

^{2}-2ac/a^{2}
(b) b

^{2}-2ac/c^{2}
(c)b

^{2}+2ac/a^{2}
(d) b

^{2}+2ac/c^{2}**Solution**

We have to find the value of 1/Î±

^{2}+ 1/Î²^{2}
Given Î± and Î² are the zeros of the quadratic polynomial f(x) = ax

^{2}Â + bx + c
Î±+Î² = -Coefficient of x/Coefficient of x

^{3}
= -b/a

Î±Î² = Constant term/Coefficient of x

^{3}
= c/a

We have,

21. If two of the zeros of the cubic polynomial ax

^{3}Â + bx^{2}Â + cx + d are each equal to zero, then the third zero is
(a) -d/a

(b) c/a

(c) -b/a

(d) b/a

**Solution**

Let Î±=0, Î²=0 and Î³ be the zeros of the polynomial

f(x) = ax

^{3}Â + bx^{2}Â + cx + d
Therefore

Î±+Î²+Î³ = -Coefficient of x

^{2}/Coefficient of x^{3}
= -(b/a)

Î±+Î²+Î³ = -b/a

0+0+Î³ = -b/a

Î³ = - b/a

The value of Î³ = -b/a

Hence, the correct choice is c.

22. If two zeros x

^{3}Â + x^{2}Â - 5x - 5 are âˆš5 and -âˆš5, then its third zero is
(a) 1

(b) -1

(c) 2

(d) -2

**Solution**

__Let Î± = âˆš5 and Î² = -âˆš5 be the given zeros and Î³ be the third zero of x__

^{3}Â + x

^{2}Â - 5x - 5 = 0 then

By using Î±+Î²+Î³ = -Coefficient of x

^{2}/Coefficient of x^{3}
Î±+Î²+Î³ = +(+1)/1

Î±+Î²+Î³ = -1

By substituting Î± = âˆš5 and Î² = -âˆš5 in Î±+Î²+Î³ = -1

âˆš5 - âˆš5 + Î³ = -1

Î³ = -1

Hence, the correct choice is bÂ

__.__
23. The product of the zeros of x

^{3}Â + 4x^{2}Â + x âˆ’ 6 is
(a) âˆ’4

(b) 4

(c) 6

(d) âˆ’6

**Solution**

Given Î±,Î²,Î³ be the zeros of the polynomial f(x) = x

^{3}Â + 4x^{2}Â + x -6
Product of the zeros = Constant term/Coefficient of x

^{3}Â = -(-6)/1 =6
The value of product of the zeros is 6.

24. What should be added to the polynomial x

^{2}Â âˆ’ 5x + 4, so that 3 is the zero of the resulting polynomial?
(a) 1

(b) 2

(c) 4

(d) 5

**Solution**

If x = a, is a zero of a polynomial then x - a is a factor of f(x)

Since 3 is the zero of the polynomial f(x) = x

^{2}Â - 5x + 4,
Therefore x -3 is a factor of f(x)

Now we divide f(x) = x

^{2}Â - 5x + 4 by (x-3) we getÂ

Therefore we should add 2 to the given polynomialÂ

Hence, the correct is b.

25. What should be subtracted to the polynomial x

^{2}Â âˆ’ 16x + 30, so that 15 is the zero of the resulting polynomial ?
(a) 30

(b) 14

(c) 15

(d) 16

**Solution**

We know that, if x = a, is zero of a polynomial then x -a is a factor of f(x)

Since 15 is zero of the polynomial f(x) = x

^{2}Â - 16x + 30, therefore (x-15) is a factor of f(x)
Now, we divide f(x) = x

^{2}Â - 16x + 30 by (x-15) we get
Thus we should subtract the remainder 15 from x

^{2}Â - 16x + 30 .
Hence, the correct choice is c.

26. A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is

(a) x

^{2}Â âˆ’ 9
(b) x

^{2}Â + 9
(c) x

^{2}Â + 3
(d) x

^{2}Â âˆ’ 3**Solution**

Since Î± and Î² are the zeros of the quadratic polynomials such that

0 = Î±+Î²

If one of zero is 3 then

Î±+Î² = 0

3+Î² = 0

Î² = 0-3

Î² = -3

Substituting Î² = -3 in Î± + Î² = 0 we get

Î± - 3 = 0

Î± = 3

Let S and P denote the sum and the product of the zeros of the polynomial respectively then

S = Î±+Î²

S = 0

P = Î±Î²

P = 3Ã—-3

P = -9

Hence, the required polynomial is

= (x

^{2}Â -Sx + P)
= (x

^{2}Â - 0x - 9)
= x

^{2}Â - 9
Hence, the correct choice is a .

27. If two zeroes of the polynomial x

^{3}Â + x^{2}Â âˆ’ 9x âˆ’ 9 are 3 and âˆ’3, then its third zero is
(a) âˆ’1

(b) 1

(c) âˆ’9

(d) 9

**Solution**Â

Let Î± = 3 and Î² = -3 be the given zeros and Î³ be the third zero of the polynomial x

^{3}Â + x^{2}Â - 9x -9 then,
By using Î±+Î²+Î³ = -Coefficient of x

^{2}/Coefficient of x^{3}
Î±+Î²+Î³ = -1/1

Î±+Î²+Î³ = -1

Substituting Î± = 3 and Î² = -3 in Î±+Î²+Î³ = -1, we get

3-3+Î³ = -1

Î³ = -1

Hence the correct choice is a .

28. If âˆš5 and -âˆš5 are two zeroes of the polynomial x

^{3}Â + 3x^{2}Â - 5x - 15, then its third zero is
(a) 3

(b) -3

(c) 5

(d) -5

**Solution**

Let Î± = âˆš5 and Î² = -âˆš5 be the given zeros and Î³ be the third zero of the polynomial x

^{3}Â + 3x^{2}Â - 5x - 15. Then,
By using Î±+Î²+Î³ = -Coefficient of x

^{2}/Coefficient of x^{3}
Î±+Î²+Î³ = -3/1

Î±+Î²+Î³ = -3

Substituting Î± = âˆš5 and Î² = -âˆš5 in Î±+Î²+Î³ = -3

We get

âˆš5 - âˆš5 + Î³ = -3

âˆš5 - âˆš5 + Î³ = -3

Î³ = -3

Hence, the correct choice is b.Â

29. If x + 2 is a factor of x

^{2}Â + ax + 2b and a + b = 4, then
(a) a = 1, b = 3

(b) a = 3, b = 1

(c) a = âˆ’1, b = 5

(d) a = 5, b = âˆ’1

**Solution**

Given that x+2 is a factor of x

^{2}Â + ax + 2b and a + b = 4
f(x) = x

^{2}Â + ax + 2b
f(-2) = (-2)

^{2}Â + a(-2) + 2b
0 = 4 - 2a + 2b

-4 = -2a + 2b

By solving -4 and -2a+2b and a+b = 4 by elimination method we get

Multiply a+b = 4 by 2 we get,

2a + 2b = 8. So

-4 = -2a + 2b

+8 = +2a + 2b

4 = 4b

4/4 = b

1 = b

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.Â

30. The polynomial which when divided by - x

^{2}Â + x - 1 gives a quotient x -2 and remainder 3, is
(a) x

^{3}Â - 3x^{2}Â + 3x - 5
(b) -x

^{3}Â - 3x^{2}Â -3x - 5
(c) -x

^{3}Â + 3x^{2}Â -3x + 5
(d) x

^{3}Â -3x^{2}Â -3x + 5**Solution**

We know that,

f(x) = g(x)q(x)+r(x)

= (-x

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

^{3}Â + x^{2}Â - x + 2x^{2}Â - 2x + 2 + 3
= -x

^{3}Â + x^{2}Â + 2x^{2}Â - x - 2x + 2 + 3
= -x

^{3}Â + 3x^{2}Â - 3x + 5
Therefore,

The polynomial which when divide by -x

^{2}Â + x - 1 gives a quotient x-2 and remainder 3, is x^{3}+3x^{2}-3x+5
Hence, the correct choice is c .

31. The number of polynomials having zeros -2 and 5 is

(a) 1

(b) 2

(c) 3

(d) More than 3.

**Solution**

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.

32. If one of the zeroes of the quadratic polynomial (k-1)x

^{2}Â + kx + 1 is -3, then the value of k is
(a) 4/3

(b) -4/3

(c) 2/3

(d) -2/3

**Solution**

The given polynomial is f(x) = (k-1)x

^{2}Â + kx + 1.
Since -3 is one of the zeros of the given polynomial, so f(-3) = 0

(k-1)(-3)

^{2}Â + k(-3) + 1 = 0
â‡’ 9(k-1)-3k + 1 = 0

â‡’ 9k - 9 - 3k + 1 = 0

â‡’ 6k - 8 = 0

â‡’ k = 4/3

Hence, the correct answer is option A.

33. The zeroes of the quadratic polynomial x

^{2}Â + 99x + 127 are
(a) Both positive

(b) Both negative

(c) Both equal

(d) One positive and one negative

**Solution**

Let f(x) = x

^{2}Â + 99x + 127.
Product of the zeroes of f(x) = 127 Ã— 1 = 127 [Product of zeroes = c/a when f(x) = ax

^{2}Â + 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) = ax

^{2}Â + bx + c]
The sum being negative implies that both zeroes are positive is not correct.

So, we conclude that both zeroses are negative.

Hence, the correct answer is option B.

34. If the zeroes of the quadratic polynomialÂ x

^{2}Â +(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

**Solution**

The given quadratic equation is x

^{2}Â + (a+1)x+b = 0.
Since the zeroes of the given equation are 2 and -3.

So,

Î± = 2 and Î² = -3

Now,

Sum of zeroes = -Coefficient of x/Coefficient of x

^{2}
â‡’ 2+(-3)= -(a+1)/1

â‡’ -1 = -a-1

â‡’ a = 0