**Class 10 maths MCQ Chapter 2 Polynomial with detail Solutions for Term-1 CBSE Board Exam**

Class 10 maths MCQ Chapter 2 Polynomial with detail Solutions for Term-1 CBSE Board Exam 2021 are created here by an expert of maths for helping the students in their preparation of Term-1 CBSE Board Exam 2021.Study of these Class 10 maths MCQ Chapter 2 Polynomial with detail Solutions for Term-1 CBSE Board Exam will give you an idea about the type of the questions in Term-1 CBSE Board exam 2021,therefore for achieving excellent percentage in maths every student is required to study MCQ’s of every chapters. Since pattern of question paper is changed so after studying the chapters of NCERT you are required study of these MCQ’s which are addiional questions prepared for Term 1 CBSE board exam 2021.

**Class 10 maths MCQ Chapter 2 Polynomial withย detail Solutions for Term-1 CBSE Board Exam**

**olutions of class 10 last years Science and maths question papers**

**Class 10 Maths MCQ with solutions-Trigonometry for Term-1**

**Class 10 science sample paper for term-1 2021 with Solutions**

**Class 10 maths sample paper for term-1 2021 with Solutions**

**CBSE Class 10 – Question paper of science 2020 with solutions**

**CBSE class 10 -Latest sample paper of science**

**Class 10 maths MCQ Chapter 2 Polynomial with detail Solutions for Term-1 CBSE Board Exam**

**Click for online shopping**

**Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc**

**Q1.If one zero of the polynomial xยฒ +3x +k is 2,then the value of k is**

**(a) 10ย ย ย (b)-10ย ย ย (c) 5ย ย ย ย (d) -5**

Ans. (b) -10

If x is the zero of the polynomial then P(x) =0

โด P(2) = 2ยฒ +3ร2 +k

0 =4 +6 +k

10 +k =0

k =-10

**Q2.Given that two of the zeroes of the cubic polynomial axยณ +bxยฒ + cx +d are 0,the third zero is**

**(a) -b/aย ย ย (b) b/aย ย ย (c) c/aย ย (d) -d/a**

Ans.If two of the zeroes are 0(ฮฑ,ฮฒ ), then applying the relationship between zeroes and coefficients of the polynomial,axยณ +bxยฒ + cx +d

ฮฑ+ฮฒ+ฮณ = -b/a (where ฮฑ,ฮฒ,ฮณ are the zeroes of the polynomial)

0+0+ฮณ =-b/a

ฮณ = -b/a

Therefore, third zero is -b/a

**Q3. If one of the zeroes of the quadratic polynomial (k โ 1) xยฒ + kx + 1 is – 3, then the value of k is**

**(a) 4/3ย ย ย ย (b) -4/3ย ย ย (c) 2/3ย ย ย ย (d) -2/3**

Ans. (a) 4/3

The given polynomial is (k โ 1) xยฒ + kx + 1 is – 3

One of the zero of the given polynomial is -3

So,the given polynomial P(-3) = 0

(k โ 1) (-3)ยฒ + k(-3) + 1 =0

9k -9 -3k +1 =0

6k =8

k = 8/6 = 4/3

**Q4.ย A quadratic polynomial, whose zeroes are -3 and 4, is**

**(a) xยฒ -x +12ย ย ย (b) xยฒ +x +12ย ย ย (c) xยฒ/2ย -x/2ย -6ย ย ย (d) 2xยฒ +2x -24**

Ans.(c) xยฒ/2ย -x/2ย -6

The given zeroes of the quadratic polynomial are -3 and 4

The quadratic polynomial with zeroes ฮฑ and ฮฒ is given as

xยฒ – (ฮฑ + ฮฒ )x + ฮฑฮฒ

Therefore putting the value of ฮฑ =-3 and ฮฒ =4

xยฒ – (-3 + 4 )x + (-3)ร4

xยฒ – x -12

By multiplying a number to the polynomial doesn’t change the polynomial,so dividing it by 2

(1/2)(xยฒ – x -12) = xยฒ/2 – x/2 -6

**Q5.If the zeroes of the quadratic polynomial xยฒย + (a + 1) x + b are 2 and -3, then**

**(a) a =-1,b=-1ย ย (b) a= 5,b = -1ย ย ย (c) a = 2, b=-6ย ย ย (d) a =0, b=-6**

Ans.(d) a = 0,b =-6

The given quadratic polynomial is xยฒ + (a + 1) x + b

Since the zeroes of the polynomial given to us are 2 and -3,theefore value of the polynomial should be zero at x =2 and x =-3

2ยฒ + (a +1)2 +b =0

4 + 2a + 2 +b =0

2a + b =-6……(i)

(-3)ยฒ + (a +1)(-3) + b =0

9 -3a -3 +b =0

-3a + b = -6…..(ii)

Substracting equation (ii) from equation (i)

5aย = 0โ a = 0

Putting the value of a in equation (i)

b = -6

Shortcut Mathod

ฮฑ + ฮฒ = -B/A and ฮฑฮฒ = C/A

2 -3 =-(a +1)/1 and 2ร-3 =b/1

a +1 =1โ a =0 and b = -6

**Q6.The number of polynomials having zeroes as -2 and 5 is**

**(a) 1ย ย ย (b) 2ย ย ย (c) 3ย ย (d) more than 3**

Ans. (d) more than 3

We can creat infinite polynomials by multiplying a polynomial by different numbers but their zeroes remains the same

**Q7.Given that one of the zeroes of the cubic polynomial ax ^{3}ย + bxยฒ + cx + d is zero, the product of the other two zeroes is**

**(a) -c/aย ย ย (b) c/aย ย ย (c) 0ย ย (d) -b/a**

Ans.(a) -c/a

Applying the relationship between the sum of product of two zeroes and coefficient

ฮฑฮฒ + ฮฒฮณ +ฮณฮฑ = -c/a

Putting ฮฑ = 0

ฮฒฮณ = -c/a

**Q8.If one of the zeroes of the cubic polynomial x ^{3}ย + axยฒ + bx + c is -1, then the product of the**

**other two zeroes is**

**(a) b โ a + 1**

**(b) b โ a โ 1**

**(c) a โ b + 1**

**(d) a โ b โ 1**

Ans. (a) b -a + 1

Applying the relationship between product of three zeroes and the coefficients of the polynomial

ฮฑฮฒฮณ = -D/A = -c/1 = -c

Putting ฮฑ = -1

-ฮฒฮณ = -c

ฮฒฮณ = c. but c is not there ,therefore getting the value of c by putting x =-1 since P(-1) =0

(-1)ยณ + a(-1)ยฒ +b(-1) +c = 0

-1 +a -b +c =0

c = b-a +1

ฮฒฮณ = b-a +1

**Q9.The zeroes of the quadratic polynomial xยฒ + 99x +127.**

**(a) both positiveย ย (b) both negativeย ย (c) one positive and one negativeย ย (d) both equal**

Ans.(b) both negative

In the polynomial axยฒ +bx + c, if a >0,b>0 and c>0ย or a <0,b <0 and c<0 then both roots are negative

In the given polynomial xยฒ +99x +127, 1>0,99>0 and 127 >0,therefore given polynomial has both negative roots

**Q10.If zeroes of a quadratic polynomial are -1/2 and 3,then the polynomial is**

**(a) xยฒ -5x -3ย ย ย (b) 2xยฒ -10x + 3ย ย (c) xยฒ -10x + 5ย ย (d)ย 4xยฒ -10x – 6**

Ans.(d)ย 4xยฒ -10x – 6

The quadratic polynomial with zeros ฮฑ and ฮฒ is given as

xยฒ – (ฮฑ +ฮฒ)x + ฮฑฮฒ

Putting the value ฮฑ =-1/2 and ฮฒ = 3

xยฒ – (-1/2 +3)x + (-1/2)3

xยฒ – (-1 +6)x/2 -3/2

xยฒ – (5/2)x -3/2

Multiplying it by 4

4xยฒ -10 x -6

**Q11.ย If the zeroes of the quadratic polynomial axยฒ + bx + c, c # 0 are equal, then**

(a) c and a have opposite signย (b) c and b have opposite sign (c) c and a have the same sign (d) c and b have the same sign

Ans.c and a have the same sign

If two zeros are equal then the polynomial should be a complete square ,as an example (x -2)ยฒ or (x +2)ยฒ

The expansions of them areย xยฒ – 2x + 4 or xยฒ + 2x + 4, here a =1 andย c =4 in both equation,therefore the accurate answer is both a and c should have same sign when both roots are equal

**Q12.The polynomial whose zeros are the reciprocal of the zeros of quadratic polynomial axยฒ +bx +c is**

**(a) xยฒ +x +1ย ย ย (b) xยฒ +bx -aย ย ย ย (c) cxยฒ +bx + aย ย (d) xยฒ -cx + a**

Ans.(c) cxยฒ +bx + a

Let the zeros of the given polynomial axยฒ +bx +c are ฮฑ and ฮฒ

We know

ฮฑ + ฮฒ = -b/a and ฮฑ ฮฒ = c/a

According to question the zeros of the required polynomial are 1/ฮฑ and 1/ฮฒ

Therefore the polynomial is given as

xยฒ – (1/ฮฑ + 1/ฮฒ)x + ฮฑฮฒ

xยฒ – (ฮฑ + ฮฒ)x/ ฮฑฮฒ + 1/ฮฑฮฒ

cxยฒ +bx + a

Q12.If ฮฑ and ฮฒ are the zeros of the polynomial f(x) = xยฒ +x + 1,then 1/ฮฑ + 1/ฮฒย ย = ?

(a) 1ย ย ย (b) -1ย ย ย (c) 0ย ย ย (d) None of these

Ans.(b) -1

The given zeros of the the polynomial f(x ) =xยฒ +x + 1 are ฮฑ and ฮฒ

Therefore

ฮฑ + ฮฒ = -b/a = -1/1 = -1….(i) ,ฮฑฮฒ = c/a = 1/1 =1…..(ii)

1/ฮฑ + 1/ฮฒ =( ฮฑ + ฮฒ)x/ ฮฑฮฒ

Putting the value of (ฮฑ + ฮฒ) and ฮฑฮฒ from equation (i) and equation (ii)

Q13.If zero of the polynomial f(x) = (kยฒ +4)xยฒ +13x +4k is reciprocal of the other,then k=

(a) 2ย ย ย (b) -2ย ย ย (c) 1ย ย ย (d) -1

Ans.(a) 2

Let one zero of the given polynomial f(x) = (kยฒ +4)xยฒ +13x +4kย is ฮฑ then other is 1/ฮฑ

We know the relationship between the coefficients and the zeros of the polynomial axยฒ +bx +c

ฮฑฮฒ = -b/a

Putting ฮฒ =1/ฮฑ and a = (kยฒ +4) and c = 4k

ฮฑ . 1/ฮฑ = 4k/(kยฒ +4)

4k/(kยฒ +4) = 1

kยฒ +4 = 4k

kยฒ -4k +4 =0

(k -2)ยฒ = 0

k – 2 = 0โ k = 2

**Q14. If the sum of the zeros of the polynomial f(x) =2xยณ -3kxยฒ +4x -5 is 6,then value of k is**

**(a) 2ย ย ย ย (b) 4ย ย ย ย (c) -2ย ย ย ย (d) -4**

Ans.(b) 4

Given zero of the given polynomial f(x) =2xยณ -3kxยฒ +4x -5 is 6

The relationship between the coefficients and the zeros of the polynomial axยณ +bxยฒ +cx +d is given as

ฮฑ+ฮฒ+ฮณ = -b/a

6 = -(-3k)/2

3k = 12 โk = 4

**Q15.If ฮฑ,ฮฒ are the zeros of polynomial f(x) =xยฒ -p(x +1) -c,then (ฮฑ +1)(ฮฒ +1) =**

**(a) c -1ย ย ย (b) 1 -cย ย ย (c) cย ย ย (d)ย 1 +c**

Ans.(b) 1 -c

We have to find o

From the polynomial f(xut the value of (ฮฑ +1)(ฮฒ +1)

Expanding the given expression

(ฮฑ +1)(ฮฒ +1)

=ฮฑฮฒ + ฮฑ + ฮฒ + 1) =xยฒ -p(x +1) -c=xยฒ -px -p -c=xยฒ-px -(p +c) wehere a =1,b =-p,c = -(p+c),we have

ฮฑ + ฮฒ = -b/a = p/1 =p and ฮฑฮฒ = c/a = -(p+c)/1 =-(p+c)

โด(ฮฑ +1)(ฮฒ +1)ย = -(p+c) + p +1 = -p-c +p+1 = 1-c

**Q16.If the product of zeros of the polynomial f(x) = axยณ -6xยฒ +11x -6 is 4,then a =**

**(a) 3/2ย ย ย (b) -3/2ย ย ย (c) 2/3ย ย ย (d) -2/3**

Ans.(a) 3/2

Let the zeros of the given polynomial are ฮฑ,ฮฒ and ฮณ

From the given polynomial f(x) = axยณ -6xยฒ +11x -6,we have a =a,b =-6,c = 11 and d = -6

The relationship between the coefficients and the zeros is

ฮฑฮฒฮณ = -d/a = -(-6)/a=6/a

Since ฮฑฮฒฮณ = 4 is given to us

6/a = 4

4a = 6โa =6/4 = 3/2

**Q17.If ฮฑ,ฮฒ,ฮณ are the zeros of the polynomial f(x) =axยณ +bxยฒ +cx +d,then ฮฑยฒ+ฮฒยฒ +ฮณยฒ =**

**(a) (bยฒ-ac)/aยฒย ย ย (b) (bยฒ-2ac)/aย ย (c)ย ย (bยฒ+2ac)/bยฒย ย (d) (bยฒ-2ac)/aยฒ**

Ans.ย (d) (bยฒ-2ac)/aยฒ

From the given polynomial f(x) =axยณ +bxยฒ +cx +d,we have, B =b and A =a , C=c,ฮฑ,ฮฒ,ฮณ are the zeros

ฮฑ+ฮฒ+ฮณ = -B/A =-b/a and ฮฑฮฒ +ฮฒฮณ +ฮณฮฑ = c/a

Squaring both sides,we have

(ฮฑ+ฮฒ+ฮณ)ยฒ = (-b/a)ยฒ

ฮฑยฒ+ฮฒยฒ+ฮณยฒ+2ฮฑฮฒ +2ฮฒฮณ +2ฮณฮฑ = bยฒ/aยฒ

ฮฑยฒ+ฮฒยฒ+ฮณยฒ + 2(ฮฑฮฒ +ฮฒฮณ +ฮณฮฑ) = bยฒ/aยฒ

ฮฑยฒ+ฮฒยฒ+ฮณยฒ + 2c/a = bยฒ/aยฒ

ฮฑยฒ+ฮฒยฒ+ฮณยฒ = bยฒ/aยฒ -2c/a = (bยฒ-2ac)/aยฒ

**Q18**. **If ฮฑ,ฮฒ,ฮณ are the zeros of the polynomial f(x) =xยณ -pxยฒ +qx -r,then 1/ฮฑฮฒ+1/ฮฒฮณ +1/ฮณฮฑ =**

**(a) r/pย ย ย (b) p/rย ย ย (c) -p/rย ย ย (d) -r/p**

Ans.(b) p/r

From the given polynomial f(x) =xยณ -pxยฒ +qx -r,we have, b =-p and a =1 , c=q,d =-r

ฮฑ+ฮฒ+ฮณ = -b/a =-(-p)/1 =p and ฮฑฮฒฮณย = -d/a=-(-r)/1 = r

We have to find out 1/ฮฑฮฒ+1/ฮฒฮณ +1/ฮณฮฑ =

Simplifying the given expression

1/ฮฑฮฒ+1/ฮฒฮณ +1/ฮณฮฑ **= **(ฮฑ +ฮฒ +ฮณ)/ฮฑฮฒฮณ

Substituting the values of (ฮฑ +ฮฒ +ฮณ) and ฮฑฮฒฮณ

1/ฮฑฮฒ+1/ฮฒฮณ +1/ฮณฮฑ = p/r

**Q19.If the polynomial f(x) =axยณ +bx -c is divisible by the polynomial g(x) =xยฒ +bx +c,then ab is**

**(a) 1ย ย ย (b) 1/cย ย ย (c) -1ย ย ย ย (d) -1/c**

Ans.(a) 1

Dividing the given polynomial axยณ +bx -c by another given polynomial xยฒ +bx +c

Since f(x) is divisible by g(x),the remainder should be 0

(b – ac -abยฒ)x + abc -c =0

b – ac -abยฒ = 0 and abc -c =0โabc = cโab =1

**Q20.What should be added to the polynomial xยฒ -5x +4, so that 3 is the zero of the resulting polynomial ?**

**(a) 1ย ย ย (b) 2ย ย ย (c) 4ย ย ย (d) 5**

Ans.(b) 2

The given polynomial isย xยฒ -5x +4

Since 3 is the zero of the resulting polynomial therefore (x -3) should be one of the factor of the resulting polynomial

Therefore dividing the given polynomial by x -3,we can get the remainder

3ยฒ -5ร3 +4 = 9 -15 +4 =-2

Hence -(-2) = 2 should be added to the given polynomial so that the resulting polynomial has 3 as a one of the zeros.

**Q21. If ฮฑ,ฮฒ are the zeros of the polynomial f(x) = axยฒ +bx +c,then 1/ฮฑยฒ + 1/ฮฒยฒ =**

**(a) (bยฒ -2ac)/aยฒย ย ย (b) (bยฒ -2ac)/cยฒย ย (c) (bยฒ +2ac)/aยฒย ย (d) (bยฒ +2ac)/cยฒ**

Ans.(b) (bยฒ -2ac)/cยฒ

From the given polynomial f(x) = axยฒ +bx +c,we have a =a,b =b,c =c and zeros given are ฮฑ and ฮฒ

The relationship between the coefficients and zeros are

ฮฑ + ฮฒ = -b/a and ฮฑ ฮฒ = c/a

โ(ฮฑ + ฮฒ )ยฒ= (-b/a)ยฒ and ฮฑยฒ ฮฒยฒ=cยฒ/aยฒ

ฮฑยฒ +ฮฒยฒ +2ฮฑฮฒ = bยฒ/aยฒ and ฮฑยฒ ฮฒยฒ=cยฒ/aยฒ

ฮฑยฒ +ฮฒยฒ = bยฒ/aยฒ – 2ฮฑฮฒ and ฮฑยฒ ฮฒยฒ=cยฒ/aยฒ

ฮฑยฒ +ฮฒยฒ = bยฒ/aยฒ -2รc/aย and ฮฑยฒ ฮฒยฒ=cยฒ/aยฒ

ฮฑยฒ +ฮฒยฒ =(bยฒ -2ac)/aยฒ….(i) and ฮฑยฒ ฮฒยฒ=cยฒ/aยฒ

Simplifying the given expression

1/ฮฑยฒ + 1/ฮฒยฒ = (ฮฑยฒ +ฮฒยฒ)/ฮฑยฒฮฒยฒ

From equation (i) and equation (ii)

**Q22. If two zeros of xยณ -xยฒ -5x -5 are โ5 and -โ5 then third zero is**

**(a) 1ย ย ย ย (b) -1ย ย ย ย (c) 2ย ย ย ย (d) -2**

Ans.(b) -1

If โ5 and -โ5 are the zeros of the polynomial xยณ + xยฒ -5x -5,then its two factors must be (x -โ5) and (x +โ5) and

their product (x -โ5) (x +โ5)=xยฒ -5 also should be the factor of the given polynomial

Therefore dividing the given polynomial by (xยฒ -5)

Therefore it will have more factor (x +1)

Hence the x+1=0โx =-1

**Q23. If x + 2 is a factor of xยฒ +ax +2b and a + b =4 then**

**(a) a =1,b =3ย ย (b) a =3,b=1ย ย (c) a = -1, b = 5ย ย (d) a =5,b = -1**

Ans.(b) a =3,b=1

If (x +2) is the factor of xยฒ +ax +2b then x = -2 will be its one of the zero

Therefore at x =-2 the value of polynomial is equal to zero

(-2)ยฒ + a(-2) + 2b =0

4ย -2a + 2b =0

-2a + 2b = -4

-a + b = -2…..(i)

The given equation is a + b = 4….(ii)

Adding both equation

2b = 2โ b = 1

Putting the value of b in equation (ii)

a +1 =4

a = 4-1 =3

**Q24.The polynomial which when divided by -xยฒ + x -1 gives a quotient x -2 and remainder 3, is**

**(a) xยณ -3xยฒ +3x -5ย ย ย (b) -xยณ +3xยฒ -3x -5ย ย (c) -xยณ +3xยฒ -3x +5ย (d) -xยณ -3xยฒ -3x +5**

Ans.(c) -xยณ +3xยฒ -3x +5

If a polynomial p(x) is divided by g(x) gives quotient q(x) and remainder r(x) then

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

Here g(x) =-xยฒ + x -1, q(x) = x -2 and r(x) =3

Therefore

p(x) =( -xยฒ + x -1)(x -2) + 3 = -xยณ +2xยฒ +xยฒ -2x -x + 2 +3 =-xยณ +3xยฒย -3xย + 5

**You can compensate us by donating any amount of money for our survival**

**Our Paytm No 9891436286**

**NCERT Solutions ofย Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

**NCERT Solutions of class 9 scienceย **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

**NCERT Solutions of class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT solutions of class 10 science**

**Solutions of class 10 last years Science question papers**

**CBSE Class 10 – Question paper of science 2020 with solutions**

**CBSE class 10 -Latest sample paper of science**

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | ย Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |