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Question 1:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
x2+7x+12

Answer:

x2+7x+12=0x2+4x+3x+12=0xx+4+3x+4=0

x+4x+3=0x+4=0 or x+3=0x=-4 or x=-3

Sum of zeroes = -4+-3=-71=-coefficient of xcoefficient of x2
Product of zeroes =  -4-3=121=constant termcoefficient of x2

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Question 2:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
x2-2x-8

Answer:

x2-2x-8=0x2-4x+2x-8=0xx-4+2x-4=0
x-4x+2=0x-4 =0 or x+2=0x=4 or x=-2

Sum of zeroes = 4+-2=2=21=-coefficient of xcoefficient of x2 4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes = 4-2=-8=-81=constant termcoefficient of x2 (4)(3)=121=constant termcoefficient of x2

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Question 3:

Find the zeros of the quadratic polynomial (x2 + 3x − 10) and verify the relation between its zeros and coefficients.

Answer:

We have:f(x)=x2+3x10=x2+5x2x10=x(x+5)2(x+5)=(x2)(x+5)f(x)=0=>(x2)(x+5)=0                 =>x2=0 or x+5=0                  =>x=2 or x=5So, the zeroes of f(x)are 2 and5.Sum of the zeroes = 2+(5)= 3=31=(coefficient of x)(coefficient of x2)Product of the zeroes= 2× (5) = 10 =101=constant term(coefficient of x2)

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Question 4:

Find the zeros of the quadratic polynomial 4x2 − 4x − 3 and verify the relation between the zeros and the coefficients.

Answer:

 We have:f(x)=4x24x3=4x2(6x2x)3=4x26x+2x3=2x(2x3)+1(2x3)=(2x+1)(2x3)f(x)=0=>(2x+1)(2x3)=0                  =>2x+1=0 or 2x3=0                  =>x=12 or x=32So, the zeros of f(x) are 12 and 32.Sum of the zeros =12+32=1+32=22=1=(coefficient of x)(coefficient of x2)Product of the zeros= 12×32 =34=constant term(coefficient of x2)

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Question 5:

Find the zeros of the quadratic polynomial 5x2 − 4 − 8x and verify the relationship between the zeros and the coefficients of the given polynomial.

Answer:

 We have:f(x)=5x248x  =5x28x-4=5x2(10x2x)4=5x210x+2x4=5x(x2)+2(x2)=(5x+2)(x2)f(x)=0=>(5x+2)(x2)=0                 =>5x+2=0 or x2=0                 =>x=25 or x=2So, the zeros of f(x) are 25 and 2.Sum of the zeros = 25+2= 2+105=85=(coefficient of x)(coefficient of x2)Product of the zeros= 25 × 2 =45=constant term(coefficient of x2)

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Question 6:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
23x2-5x+3

Answer:


23x2-5x+323x2-2x-3x+32x3x-1-33x-1=0

3x-1 or 2x-3=03x-1=0 or 2x-3=0x=13 or x=32x=13×33=33 or x=32

Sum of zeroes = 33 +32=536=-coefficient of xcoefficient of x2 4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes = 33 ×32=12=constant termcoefficient of x2 (4)(3)=121=constant termcoefficient of x2

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Question 7:

Find the zeros of the quadratic polynomial 2x2 − 11x + 15 and verify the relation between the zeros and the coefficients.

Answer:

We have:f(x)=2x211x+15=2x2(6x+5x)+15=2x26x5x+15=2x(x3)5(x3)=(2x5)(x3)f(x)=0=>(2x5)(x3)=0                 =>2x5=0 or x3=0                  =>x=52 or x=3So, the zeroes of f(x) are 52 and 3.Sum of the zeroes =52+3=5+62=112=(coefficient of x)(coefficient of x2)Product of the zeroes = 52× 3=152=constant term(coefficient of x2)

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Question 8:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
4x2-4x+1

Answer:

4x2-4x+1=02x2-22x1+12=02x-12=0            a2-2ab+b2=a-b2

2x-12=0x=12 or x=12

Sum of zeroes = 12 +12=1=11=-coefficient of xcoefficient of x2 4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes = 12 ×12=14=constant termcoefficient of x2 (4)(3)=121=constant termcoefficient of x2



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Question 9:

Find the zeros of the quadratic polynomial (x2 − 5) and verify the relation between the zeros and the coefficients.

Answer:

We have:f(x)=x25It can be written as x2+0x5.=(x2(5)2)=(x+5)(x5)f(x)=0=>(x+5)(x5)=0                 =>x+5=0 or x5=0                 =>x=5 or x=5So, the zeroes of f(x) are 5 and 5.Here, the coefficient of x is 0 and the coefficient of x2 is 1.Sum of the zeroes = 5+5=01=(coefficient of x)(coefficient of x2)Product of the zeroes= 5×5=51=constant term(coefficient of x2)

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Question 10:

Find the zeros of the quadratic polynomial (8 x2 − 4) and verify the relation between the zeros and the coefficients.

Answer:

 We have:f(x)=8x24It can be written as 8x2+0x4=4{(2x)2(1)2}=4(2x+1)(2x1)f(x)=0=>(2x+1)(2x1)=0                 =>2x+1=0 or 2x1=0                 =>x=12 or x=12So, the zeroes of f(x) are 12  and 12Here the coefficient of x is 0 and the coefficient of x2 is 2Sum of the zeroes =12+12=1+12=02=(coefficient of x)(coefficient of x2)Product of the zeroes=  12×12=1×42×4=-48=constant term(coefficient of x2)

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Question 11:

Find the zeros of the quadratic polynomial (5u2 + 10u) and verify the relation between the zeros and the coefficients.

Answer:

 We have:f(u)=5u2+10uIt can be written as 5u(u+2)f(u)=0=>5u=0    or u+2=0                 =>u=0 or u=-2                 So, the zeroes of f(u) are 2  and 0.Sum of the zeroes = 2+ 0=2=2×51×5=-105=(coefficient of u)(coefficient of u2)Product of the zeroes=  2×0=0=0×51×5=05=constant term(coefficient of u2)

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Question 12:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
3x2-x-4

Answer:

3x2-x-4=03x2-4x+3x-4=0x3x-4+13x-4=0
3x-4x+1=03x-4 or x+1=0x=43 or x=-1

Sum of zeroes = 43 +-1=13=-coefficient of xcoefficient of x2 4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes = 43 ×-1=-43=constant termcoefficient of x2 (4)(3)=121=constant termcoefficient of x2

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Question 13:

Find the quadratic polynomial whose zeros are 2 and −6. Verify the relation between the coefficients and the zeros of the polynomial.

Answer:

Let α=2 and β=6Sum of the zeroes, (α+β)=2+(6)=4Product of the zeroes, αβ=2×(6)=12∴ Required polynomial =x2(α+β)x+αβ=x2(4)x12                                     =x2+4x12Sum of the zeroes =4=41=(coefficient of x)(coefficient of x2)Product of the zeroes =12=121=constant termcoefficient of x2

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Question 14:

Find the quadratic polynomial whose zeros are 23 and -14. Verify the relation between the coefficients and the zeros of the polynomial.

Answer:

 Let α=23 and β=14.Sum of the zeroes = (α+β)=23+14=8312=512Product of the zeroes = αβ=23×14=21126=16∴ Required polynomial = x2(α+β)x+αβ=x2512x+16                                     =x2512x16  Sum of the zeroes =512=(coefficient of x)(coefficient of x2)Product of the zeroes =16=constant termcoefficient of x2                                   

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Question 15:

Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12. Hence, find the zeros of the polynomial.

Answer:

Let α and β be the zeroes of the required polynomial f(x).Then (α+β)=8 and αβ=12f(x)=x2(α+β)x+αβ=>f(x)=x28x+12Hence, required polynomial f(x)=x28x+12f(x)=0=>x28x+12=0                 => x2(6x+2x)+12=0                 => x26x2x+12=0                 => x(x6)2(x6)=0                 => (x2)(x6)=0                 => (x2)=0 or (x6)=0                 => x=2 or x=6So, the zeroes of f(x) are 2 and 6.

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Question 16:

Find the quadratic polynomial, the sum of whose zeros is 0 and their product is −1. Hence, find the zeros of the polynomial.

Answer:

Let α and β be the zeros of the required polynomial f(x).Then (α+β)=0 and αβ=1f(x)=x2(α+β)x+αβ=>f(x)=x20x+(1)=>f(x)=x21Hence, the required polynomial is f(x)=x21.f(x)=0=>x21=0                 => (x+1)(x1)=0                 => (x+1)=0 or (x1)=0                 => x=1 or x=1So, the zeros of f(x) are 1 and 1.

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Question 17:

Find the quadratic polynomial, the sum of whose zeros is 52 and their product is 1. Hence, find the zeros of the polynomial.

Answer:

 Let α and β be the zeros of the required polynomial f(x).Then (α+β)=52 and αβ=1f(x)=x2(α+β)x+αβ=>f(x)=x252x+1=>f(x)=2x25x+2Hence, the required polynomial is f(x)=2x25x+2.f(x)=0=>2x25x+2=0                 => 2x2(4x+x)+2=0                 => 2x24xx+2=0                 => 2x(x2)1(x2)=0                 => (2x1)(x2)=0                 => (2x1)=0 or (x2)=0                 => x=12 or x=2So, the zeros of f(x)are 12 and 2.

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Question 18:

Find the quadratic polynomial, the sum of whose root is 2 and their product is 13

Answer:

We can find the quadratic equation if we know the sum of the roots and product of the roots by using the formula
x2 − (Sum of the roots)x + Product of roots = 0
x2-2x+13=03x2-32x+1=0

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Question 19:

If x=23 and x=-3 are the roots of the quadratic equation ax2+7x+b=0 then find the values of a and b.

Answer:

Given: ax2+7x+b=0
Since, x=23 is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
a232+723+b=049a+143+b=04a+42+9b=04a+9b=-42                   ...1

Since, x=-3 is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
a-32+7-3+b=09a-21+b=09a+b=21                      ...2
From (1) and (2), we get
a=3, b=-6

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Question 20:

If x+a is a factor of the polynomial 2x2+2ax+5x+10, find the value of a.

Answer:

Given: x+a is a factor of 2x2+2ax+5x+10
So, we have
x+a=0x=-a
Now, It will satisfy  the above polynomial.
Therefore, we will get
2-a2+2a-a+5-a+10=02a2-2a2-5a+10=0-5a=-10a=2

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Question 21:

One zero of the polynomial 3x3+16x2+15x-18 is 23. Find the other zeroes of the polynomial.

Answer:

Given: x=23 is one of the zero of 3x3+16x2+15x-18
Now, we have
x=23x-23=0
Now, we divide 3x3+16x2+15x-18 by x-23 to find the quotient



So, the quotient is 3x2+18x+27
Now,
3x2+18x+27=03x2+9x+9x+27=03xx+3+9x+3=0

x+33x+9=0x+3=0 or 3x+9=0x=-3 or x=-3



Page No 60:

Question 1:

Verity that 3, −2, 1 are the zeros of the cubic polynomial p(x) = x3 − 2x2 − 5x + 6 and verify the relation between its zeros and coefficients.

Answer:

 The given polynomial is p(x)=(x32x25x+6)p(3)=(332×325×3+6)=(271815+6)=0p(2)=[(23)2×(22)5×(2)+6]=(88+10+6)=0p(1)=(132×125×1+6)=(125+6)=03,2 and 1 are the zeroes of p(x),Let α=3, β=2 and γ=1. Then we have:(α+β+γ)=(32+1)=2=(coefficient of x2)(coefficient of x3)(αβ+βγ+γα)=(62+3)=51=coefficient of xcoefficient of x3 αβγ={3×(2)×1}=61=(constant term)(coefficient of x3)

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Question 2:

Verify that 5, −2 and 13 are the zeros of the cubic polynomial p(x) = 3x3 − 10x2 − 27x + 10 and verify the relation between its zeros and coefficients.

Answer:

 p(x)=(3x310x227x+10)p(5)=(3×5310×5227×5+10)=(375250135+10)=0p(2)=[3×(23)10×(22)27×(2)+10]=(2440+54+10)=0p13={3×133)310×13227×13+10}=(3×12710×199+10)=19109+1=110+99=09=05,2 and 13 are the zeroes of p(x).Let α=5, β=2 and γ=13. Then we have:(α+β+γ)=52+13=103=(coefficient of x2)(coefficient of x3)(αβ+βγ+γα)=1023+53=273=coefficient of xcoefficient of x3 αβγ={5×(2)×13}=103=(constant term)(coefficient of x3)

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Question 3:

Find a cubic polynomial whose zeroes are 2, −3 and 4

Answer:

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x3-a+b+cx2+ab+bc+cax-abc                                  ...(1)
Let a=2, b=-3 and c=4
Substituting the values in (1), we get
x3-2-3+4x2+-6-12+8x--24x3-3x2-10x+24

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Question 4:

Find a cubic polynomial whose zeroes are 12, 1 and −3.

Answer:

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x3-a+b+cx2+ab+bc+cax-abc                                ...(1)
Let a=12, b=1 and c=-3
Substituting the values in (1), we get
x3-12+1-3x2+12-3-32x--32x3--32x2-4x+322x3+3x2-8x+3

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Question 5:

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes are 5, −2 and −24 respectively.

Answer:

We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as
x3 −(Sum of the zeroes)x2 + (sum of the product of the zeroes taking two at a time)x − Product of zeroes
Therefore, the required polynomial is
x3-5x2-2x+24

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Question 6:

Find the quotient and the remainder when:
fx=x3-3x2+5x-3 is divided by gx=x2-2

Answer:



Quotient  qx=x-3
Remainder  rx=7x-9

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Question 7:

Find the quotient and remainder when:
fx=x4-3x2+4x+5 is divided by gx=x2+1-x

Answer:



Quotient  qx=x2+x-3
Remainder  rx=8

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Question 8:

Find the quotient and remainder when
fx=x4-5x+6 is divided by gx=2-x2

Answer:

We can write

fx as x4+0x3+0x2-5x+6  and gx as-x2+2



Quotient  qx=-x2-2
Remainder  rx=-5x+10

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Question 9:

By actual division, show that x2-3 is a factor of 2x4+3x3-2x2-9x-12

Answer:

Let fx=2x4+3x3-2x2-9x-12 and gx=x2-3



Quotient  qx=2x2+3x+4
Remainder  rx=0
Since, the remainder is 0.
Hence, x2-3 is a factor of 2x4+3x3-2x2-9x-12

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Question 10:

On dividing, 3x3+x2+2x+5 by a polynomial g(x), the quotient and remainder are 3x-5 and 9x+10 respectively. Find g(x)

Answer:

By using division rule, we have
Divided = Quotient × Divisor + Remainder
3x3+x2+2x+5=3x-5gx+9x+103x3+x2+2x+5-9x-10=3x-5gx3x3+x2-7x-5=3x-5gxgx=3x3+x2-7x-53x-5



gx=x2+2x+1

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Question 11:

Verify division algorithm for the polynomials fx=8+20x+x2-6x3 and gx=2+5x-3x2

Answer:

We can write fx as-6x3+x2+20x+ 8 and gx as -3x2+5x+2



Quotient = 2x+3
Remainder = x+2

By using division rule, we have

Divided = Quotient × Divisor + Remainder

-6x3+x2+20x+ 8=-3x2+5x+22x+3+x+2-6x3+x2+20x+ 8=-6x3+10x2+4x-9x2+15x+6+x+2-6x3+x2+20x+ 8=-6x3+x2+20x+ 8

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Question 12:

It is given that −1 is one of the zeros of the polynomial x3 + 2x2 − 11x − 12. Find all the given zeros of the given polynomial.

Answer:

Let f(x)=x3+2x211x12Since -1 is a zero of f(x), (x+1) is a factor of f(x).On dividing f(x) by (x+1), we get:



f(x)=x3+2x211x12=(x+1)(x2+x12)=(x+1){x2+4x3x12}=(x+1){x(x+4)3(x+4)}=(x+1)(x3)(x+4)f(x)=0=>(x+1)(x3)(x+4)=0                  =>(x+1)=0 or (x3)=0 or (x+4)=0                  =>x=1 or x=3 or x=4Thus, all the zeroes are 1, 3 and 4.



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Question 13:

If 1 and −2 are two zeros of the polynomial (x3 − 4x2 − 7x + 10), find its third zero.

Answer:

 Let f(x)=x34x27x+10Since 1 and 2 are the zeroes of f(x), it follows that each one of (x1) and (x+2) is a factor of f(x).Consequently, (x1)(x+2)=(x2+x2) is a factor of f(x).On dividing f(x) by (x2+x2), we get:



f(x)=0=>(x2+x2)(x5)=0                 =>(x1)(x+2)(x5)=0                 =>x=1 or x=2 or x=5Hence, the third zero is 5.

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Question 14:

If 3 and −3 are two zeros of the polynomial (x4 + x3 − 11x2 − 9x + 18), find all the zeros of the given polynomial.

Answer:

 Let f(x)=x4+x311x29x+18Since 3 and 3 are the zeroes  of f(x), it follows that each one of (x+3) and (x3) is a factor of f(x).Consequently, (x3)(x+3)=(x29) is a factor of f(x).On dividing f(x) by (x29), we get:



f(x)=0 =>(x2+x2)(x29)=0                    =>(x2+2x-x-2)(x3)(x+3)                   =>(x1)(x+2)(x3)(x+3)=0                    =>x=1 or x=2 or x=3 or x=3Hence, all the zeroes are 1, 2, 3 and 3.

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Question 15:

If 2 and −2 are two zeros of the polynomial (x4 + x3 − 34x2 − 4x + 120), find all the zeros of given polynomial.

Answer:

Let f(x)=x4+x334x24x+120Since 2 and 2 are the zeroes of f(x), it follows that each one of (x2) and (x+2) is a factor of f(x).Consequently, (x2)(x+2)=(x24) is a factor of f(x).On dividing f(x) by (x24),we get:



 f(x)=0=>(x2+x30)(x24)=0=>(x2+6x-5x-30)(x2)(x+2)=>[x(x+6)-5(x+6)](x2)(x+2)=>(x5)(x+6)(x2)(x+2)=0=>x=5 or x=6 or x=2 or x=2Hence, all the zeros are 2, 2, 5 and 6.

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Question 16:

Find all the zeros of (x4 + x3 − 23x2 − 3x + 60), if it is given that two of its zeros are 3 and -3.

Answer:

Let f(x)=x4+x323x23x+60Since 3 and 3 are the zeroes  of f(x), it follows that each one of (x3) and (x+3) is a factor of f(x).Consequently, (x3)(x+3)=(x23) is a factor of f(x).On dividing f(x) by (x23), we get:


 f(x)=0 =>(x2+x20)(x23)=0=>(x2+5x-4x-20)(x23)=>[x(x+5)-4(x+5)](x23)=>(x4)(x+5)(x3)(x+3)=0=>x=4 or x=5 or x=3 or x=3Hence, all the zeroes are 3,3, 4 and 5.

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Question 17:

Find all the zeros of (2x4 − 3x3 − 5x2 + 9x − 3), it being given that two of its zeros are 3 and -3.

Answer:

 The given polynomial is f(x)=2x43x35x2+9x3.Since 3 and 3 are the zeroes of f(x), it follows that each one of (x3) and (x+3) is a factor of f(x).Consequently, (x3)(x+3)=(x23) is a factor of f(x).On dividing f(x) by (x23), we get:

 
 f(x)=0=>2x43x35x2+9x3=0=>(x23)(2x23x+1)=0=>(x23)(2x2-2x-x+1)=0=>(x3)(x+3)(2x1)(x1)=0=>x=or x=3 or x=12or x=1

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Question 18:

Obtain all other zeros of (x4 + 4x3 − 2x2 − 20x − 15) if two of its zeros are 5 and -5.

Answer:


 The given polynomial is f(x)=x4+4x32x220x15.Since (x5) and (x+5) are the zeroes of f(x), it follows that each one of (x5) and (x+5) is a factor of f(x).Consequently, (x5)(x+5)=(x25) is a factor of f(x).On dividing f(x) by(x25), we get:


f(x)=0=>x4+4x37x220x15=0=>(x25)(x2+4x+3)=0=>(x5)(x+5)(x+1)(x+3)=0=>x=5 or x=5 or x=1 or x=3Hence, all the zeros are 5,5,1 and 3.

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Question 19:

Find all the zeros of the polynomial (2x4 − 11x3 + 7x2 + 13x), it being given that two if its zeros are 3+2 and 3-2.

Answer:

 The given polynomial is f(x)=2x411x3+7x2+13x7.Since (3+2) and (32) are the zeroes of f(x), it follows that each one of (x+3+2) and (x+32) is a factor of f(x).Consequently, [x(3+2)] [x(32)]=[(x3)2][(x3)+2]=[(x3)22]=x26x+7, which is a factor of f(x)On dividing f(x) by(x2-6x+7), we get:


f(x)=02x411x3+7x2+13x7=0=>(x26x+7)(2x2+x1)=0=>(x+3+2)(x+32)(2x1)(x+1)=0=>x=32 or x=3+2or x=12 or x=1Hence, all the zeros are (32), (3+2), 12 and 1.

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Question 1:

If one zero of the polynomial x2-4x+1  is 2+3 . Write the other zero.  [CBSE 2010]

Answer:

Let the other zeroes of x2-4x+1 be a.
By using the relationship between the zeroes of the quadratic ploynomial.
We have, Sum of zeroes = -coefficient of xcoefficent of x2
2+3+a=--41a=2-3
Hence, the other zeroes of x2-4x+1 is 2-3.

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Question 2:

Find the zeros of the polynomial x2 + x − p(p + 1).   [CBSE 2011]

Answer:

fx=x2+x-p(p+1)
By adding and subtracting px, we get
fx=x2+px+x-px-p(p+1)=x2+p+1x-px-p(p+1)=xx+p+1-px+(p+1)

=x+p+1x-pfx=0x+p+1x-p=0x+p+1=0 or x-p=0x=-p+1 or x=p
So, the zeros of f(x) are −(p + 1) and p.

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Question 3:

Find the zeros of the polynomial x2 − 3x − m(m + 3).   [CBSE 2011]

Answer:

fx=x2-3x-m(m+3)
By adding and subtracting mx, we get
fx=x2-mx-3x+mx-m(m+3)=xx-m+3+mx-(m+3)=x-m+3x+m

fx=0x-m+3x+m=0x-m+3=0 or x+m=0x=m+3 or x=-m
So, the zeros of f(x) are −m and m + 3.

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Question 4:

If α, β are the zeros of a polynomial such that α + β = 6 and αβ = 4 the write the polynomial.   [CBSE 2010]

Answer:

If the zeroes of the quadratic polynomial are α and β then the quadratic polynomial can be found as
x2 − (α + β)x + αβ                .....(1)
Substituting the values in (1), we get
x2 − 6x + 4

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Question 5:

If one zero of the quadratic polynomial kx2 + 3x + k is 2 then find the value of k.

Answer:

Given: x = 2 is one zero of the quadratic polynomial kx2 + 3x + k
Therefore, It will satisfy the above polynomial.
Now, we have
k22+32+k=04k+6+k=05k+6=0k=-65

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Question 6:

If 3 is a zero of the polynomial 2x2 + x + k, find the value of k.  [CBSE 2010]

Answer:

Given: x = 3 is one zero of the polynomial 2x2 + x + k
Therefore, It will satisfy the above polynomial.
Now, we have
232+3+k=021+k=0k=-21



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Question 7:

If −4 is a zero of the quadratic polynomial x2x − (2k + 2) then find the value of k.

Answer:

Given: x = −4 is one zero of the polynomial x2x −(2k + 2)
Therefore, It will satisfy the above polynomial.
Now, we have
-42--4-2k+2=016+4-2k-2=0-2k=-18k=9

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Question 8:

If 1 is a zero of the polynomial ax2 − 3(a − 1) x − 1, then find the value of a.

Answer:

Given: x = 1 is one zero of the polynomial ax2 − 3(a − 1) x − 1
Therefore, It will satisfy the above polynomial.
Now, we have
a12-3a-11-1=0a-3a+3-1=0-2a=-2a=1

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Question 9:

If −2 is a zero of the polynomial 3x2 + 4x + 2k then find the value of k.

Answer:

Given: x = −2 is one zero of the polynomial 3x2 + 4x + 2k
Therefore, It will satisfy the above polynomial.
Now, we have
3-22+4-2+2k=012-8+2k=0k=-2

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Question 10:

Write the zeros of the polynomial x2 −− 6

Answer:

fx=x2-x-6=x2-3x+2x-6=xx-3+2x-3

=x-3x+2fx=0x-3x+2=0x-3=0 or x+2=0x=3 or x=-2
So, the zeros of f(x) are 3 and −2.

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Question 11:

If the sum of the zeros of the quadratic polynomial kx2 − 3x + 5 is 1, write the value of k.

Answer:

By using the relationship between the zeros of the quadratic ploynomial.
We have
Sum of zeroes = -coefficient of xcoefficent of x2
1=--3kk=3

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Question 12:

If the product of the zeros of the quadratic polynomial x2 − 4x + k is 3 then write the value of k.

Answer:

By using the relationship between the zeros of the quadratic ploynomial.
We have
Product of zeroes = constant termcoefficent of x2
3=k1k=3

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Question 13:

If (x + a) is a factor of (2x2 + 2ax + 5x + 10), find the value of a.   [CBSE 2010]

Answer:

Given: (x + a) is a factor of 2x2 + 2ax + 5x + 10
We have
x+a=0x=-a
Since, (x + a) is a factor of 2x2 + 2ax + 5x + 10
Hence, It will satisfy the above polynomial
2-a2+2a-a+5-a+10=0-5a+10=0a=2

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Question 14:

If (a − b), a and (a + b) are zeros of the polynomial 2x3 − 6x2 + 5x − 7, write the value of a.

Answer:

By using the relationship between the zeroes of the cubic ploynomial.
We have
Sum of zeroes = -coefficient of x2coefficent of x3
a-b+a+a+b=--623a=3a=1

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Question 15:

If x3 + x2ax + b is divisible by (x2x), write the values of a and b.

Answer:

Equating x2x to 0 to find the zeros, we will get
xx-1=0x=0 or x-1=0x=0 or x=1

Since,  x3 + x2ax + b is divisible by x2x.
Hence, the zeros of x2x will satisfy x3 + x2ax + b
03+02-a0+b=0b=0

and
13+12-a1+0=0       b=0a=2

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Question 16:

If α and β are the zeroes of a polynomial 2x2 + 7x + 5, write the value of α + β + αβ.   [CBSE 2010]

Answer:

By using the relationship between the zeros of the quadratic ploynomial.
We have,
Sum of zeroes = -coefficient of xcoefficent of x2 and Product of zeroes = constant termcoefficent of x2
α+β=-72 and αβ=52Now, α+β+αβ=-72+52=-1

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Question 17:

State division algorithm for polynomials.

Answer:

“If f(x) and g(x) are two polynomials such that degree of f(x) is greater than degree of g(x) where g(x) ≠ 0, then there exists unique polynomials q(x) and r(x) such that

f(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).

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Question 18:

The sum of the zero and the product of zero of a quadratic polynomial are -12 and −3 respectively, write the polynomial.

Answer:

We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula
x2 − (Sum of the zeros)x + Product of zeros
x2--12x+-3x2+12x-3
Hence, the required polynomial is x2+12x-3.

x22x+13=03x232x+1=0

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Question 19:

Write the zeros of the quadratic polynomial f(x) = 6x2 − 3

Answer:

To find the zeros of the quadratic polynomial we will equate f(x) to 0
fx=06x2-3=032x2-1=02x2-1=0
2x2=1x2=12x=±12
Hence, the zeros of the quadratic polynomial f(x) = 6x2 − 3 are 12,-12.

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Question 20:

Find the zeros of the quadratic polynomial fx=43x2+5x-23

Answer:

To find the zeros of the quadratic polynomial we will equate f(x) to 0
fx=043x2+5x-23=043x2+8x-3x-23=04x3x+2-33x+2=0
3x+24x-3=03x+2=0 or 4x-3=0x=-23 or x=34
Hence, the zeros of the quadratic polynomial fx=43x2+5x-23 are -23 or 34.

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Question 21:

If α and β are the zeroes of a polynomial f(x) = x2 − 5x + k, such that αβ = 1, find the value of k.

Answer:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = -coefficient of xcoefficent of x2 and Product of zeroes = constant termcoefficent of x2
α+β=--51 and αβ=k1α+β=5 and αβ=k1
Solving αβ = 1 and α + β = 5, we will get
α = 3 and β = 2
Substituting these values in αβ=k1, we will get
k = 6

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Question 22:

If α and β are the zeroes of a polynomial f(x) = 6x2 + x − 2, find the value of αβ+βα

Answer:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = -coefficient of xcoefficent of x2 and Product of zeroes = constant termcoefficent of x2
α+β=-16 and αβ=-13Now,αβ+βα=α2+β2αβ                        =α2+β2+2αβ-2αβαβ
                        =α+β2-2αβαβ                        =-162-2-13-13                        =136+23-13                        =-2512

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Question 23:

If α and β are the zeroes of a polynomial f(x) = 5x2 − 7x +1, find the value of 1α+1β

Answer:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = -coefficient of xcoefficent of x2 and Product of zeroes = constant termcoefficent of x2
α+β=--75 and αβ=15α+β=75 and αβ=15Now,1α+1β=α+βαβ

                        =7515                        =7

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Question 24:

If α and β are the zeroes of a polynomial f(x) = x2 + x − 2, find the value of 1α-1β

Answer:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = -coefficient of xcoefficent of x2 and Product of zeroes = constant termcoefficent of x2

α+β=-11 and αβ=-21α+β=-1 and αβ=-2Now,1α-1β2=β-ααβ2

=α+β2-4αβαβ2          β-α2=α+β2-4αβ=-12-4-2-22          α+β=-1 and αβ=-2=-12-4-24=94

1α-1β2=941α-1β=±32



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Question 25:

If the zeros of the polynomial f(x) = x3 − 3x2 + x + 1 are (a − b), a and (a + b), Find a and b.

Answer:

By using the relationship between the zeroes of the cubic ploynomial.
We have, Sum of zeroes = -coefficient of x2coefficent of x3
a-b+a+a+b=--313a=3a=1

Now, Product of zeros = -constant termcoefficent of x3
a-baa+b=-111-b11+b=-1     a=11-b2=-1b2=2b=±2



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Question 1:

Which of the following is a polynomial?

(a) x2-5x+6x+2
(b) x3/2-x+x1/2+1
(c) x+1x
(d) None of these

Answer:

(d) none of these

A polynomial in x of degree n is an expression of the form p(x) =ao +a1x+a2x2 +...+an xn, where an 0.

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Question 2:

Which of the following is not a polynomial?

(a) 3x2-23x+5
(b) 9x2-4x+2
(c) 32x3+6x2-12x-8
(d) x+3x

Answer:

(d) x+3x is not a polynomial.
It is because in the second term, the degree of x is −1 and an expression with a negative degree is not a polynomial.

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Question 3:

The zeros of the polynomial x2 − 2x − 3 are

(a) −3, 1
(b) −3, −1
(c) 3, −1
(d) 3, 1

Answer:

 (c)3,-1Let f(x)=x22x3=0         =x23x+x3=0         =x(x3)+1(x3)=0         =(x3)(x+1)=0        =x=3 or x=1

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Question 4:

The zeros of the polynomial x2-2x-12 are
(a) 2,-2
(b) 32, -22
(c) -32, 22
(d) 32, 22

Answer:

 (b) 32,22Let f(x)=x22x12=0           =>x232x+22x12=0           =>x(x32)+22(x32)=0            =>(x32)(x+22)=0           =>x=32 or  x=22

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Question 5:

The zeros of the polynomial 4x2+52x-3 are
(a) -32,2
(b) -32,22
(c) -322,24
(d) none of these

Answer:

(c) 32,24 Let f(x)=4x2+52x3=0      =>4x2+62x2x3=0      =>22x(2x+3)1(2x+3)=0      =>(2x+3)(22x1)=0      =>x=32 or x=122      =>x=32 or  x=122×22=24



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Question 6:

The zeros of the polynomial x2+16x-2 are
(a) −3, 4
(b) -32,43
(c) -43,32
(d) none of these

Answer:

(b)  32,43  Let f(x) =x2+16x2=0      =>6x2+x12=0      =>6x2+9x8x12=0      =>3x(2x+3)4(2x+3)=0      =>(2x+3)(3x4)=0         x=32 or x=43              

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Question 7:

The zeros of the polynomial 7x2-113x-23 are
(a) 23,-17
(b) 27,-13
(c) -23,17
(d) none of these

Answer:

(a) 23,-17Let f(x)=7x2113x23=021x211x2=021x214x+3x2=07x(3x2)+1(3x2)=0(3x2)(7x+1)=0x=23 or x=17

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Question 8:

The sum and product of the zeros of a quadratic polynomial are 3 and −10 respectively. The quadratic polynomial is

(a) x2 − 3x + 10
(b) x2 + 3x −10
(c) x2 − 3x −10
(d) x2 + 3x + 10

Answer:

 (c)  x23x10Given: Sum of zeroes, α+β = 3 Also, product of zeroes, αβ=-10Required polynomial=x2-(α+β)+αβ=x23x10

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Question 9:

A quadratic polynomial whose zeros are 5 and −3, is

(a) x2 + 2x − 15
(b) x2 − 2x + 15
(c) x2 − 2x − 15
(d) none of these

Answer:

(c) x22x15 Here, the zeroes are 5 and 3.Let α=5 and β So, sum of the zeroes, α+β = 5+(3)=2 Also, product of the zeroes, αβ = 5×(3)=15The polynomial will be x2-(α+β)x+αβ. The required polynomial is x22x15.

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Question 10:

A quadratic polynomial whose zeros are 35 and -12,is
(a) 10x2 + x + 3
(b) 10x2 + x − 3
(c) 10x2x + 3
(d) x2-110x-310

Answer:

(d)  x2110x310 Here, the zeroes are 35 and 12.Let α=35 and β=12  So, sum of the zeroes, α+β=35+12=110 Also, product of the zeroes, αβ=35×12=310The polynomial will be x2-(α+β)x+αβ.   The required polynomial is x2110x310.

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Question 11:

The zeros of the quadratic polynomial x2 + 88x + 125 are

(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal

Answer:

(b) both negative Let α and β be the zeroes of x2+88x+125.Then α+β=88 and α×β=125This can only happen when both the zeroes are negative.

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Question 12:

If α and β are the zero of x2 + 5x + 8, then the value of (α + β) is

(a) 5
(b) −5
(c) 8
(d) −8

Answer:

(b)  -5Given: α and β are the zeroes of x2+5x+8.If α+β is the sum of the roots and αβ is the product, then the required polynimial will be x2-(α+β)+αβ. α+β=5

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Question 13:

If α and β are the zero of 2x2 + 5x − 8, then the value of (αβ) is

(a) -52
(b) 52
(c) -92
(d) 92

Answer:

(c) 92   Given: α and β are the zeroes of 2x2+5x9.If α and β are the zeroes, then x2-(α+β)x+αβ is the required polynomial.The polynomial will be x2-52x-92.     ∴ αβ=92

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Question 14:

If one zero of the quadratic polynomial kx2 + 3x + k is 2, then the value of k is

(a) 56
(b) -56
(c) 65
(d) -65

Answer:

(d) 65 Since 2 is a zero of kx2+3x+k, we have:k×(2)2+3×2+k=0=>4k+k+6=0=>5k=6=>k=65

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Question 15:

If one zero of the quadratic polynomial (k − 1) x2 + kx + 1 is −4, then the value of k is

(a) -54
(b) 54
(c) -43
(d) 43

Answer:

 (b)  54 Since 4 is a zero of (k1)x2+kx+1, we have: (k1)×(4)2+k×(4)+1=0  =>16k164k+1=0=>12k15=0=>k=155124 =>k=54

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Question 16:

If −2 and 3 are the zeros of the quadratic polynomial x2 + (a + 1) x + b, then

(a) a = −2, b = 6
(b) a = 2, b = −6
(c) a = −2, b = −6
(d) a = 2, b = 6

Answer:

 (c)  a=-2, b=-6 Given: 2 and 3 are the zeroes of x2+(a+1)x+b.  Now, (2)2+(a+1)×(2)+b=0=>42a2+b=0=>b2a=2    ...(1)Also, 32+(a+1)×3+b=0=>9+3a+3+b=0    =>b+3a=12      ...(2)On subtracting (1) from (2),  we get a=2∴ b=24=6      [From (1)]

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Question 17:

If one zero of 3x2 + 8x + k be the reciprocal of the other, then k = ?

(a) 3
(b) −3
(c) 13
(d) -13

Answer:

(a) k=3 Let α and 1α be the zeroes of 3x28x+k.Then product of zeroes=k3     =>α×1α=k3     =>1=k3      =>k=3



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Question 18:

If the sum of the zeros of the quadratic polynomial kx2 + 2x + 3k is equal to the product of its zeros, then k = ?

(a) 13
(b) -13
(c) 23
(d) -23

Answer:

(d) 23 Let α and β be the zeroes of kx2+2x+3k. Then α+β=2k and αβ=3kk=3     =>α+β=αβ     =>2k=3     =>k=23

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Question 19:

If α, β are the zeros of the polynomial x2 + 6x + 2, then 1α+1β=?
(a) 3
(b) −3
(c) 12
(d) −12

Answer:

 (b)  -3Since α and β are the zeroes of x2+6x+2,we have:      α+β=6 and αβ=2      (1α+1β)=(α+βαβ)=62=3

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Question 20:

If α, β, γ are the zeros of the polynomial x3 − 6x2x + 30, then (αβ + βγ + γα) = ?

(a) −1
(b) 1
(c) −5
(d) 30

Answer:

 (a) -1It is given that αβ and γ are the zeroes of x36x2x+30.   (αβ+βγ+γα)=co-efficient of x co-efficient of x3=11=1

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Question 21:

If α, β, γ are the zeros of the polynomial 2x3x2 − 13x + 6, then αβγ = ?

(a) −3
(b) 3
(c) -12
(d) -132

Answer:

 (a) -3 Since αβ and γ are the zeroes of 2x3+x213x+6, we have:      αβγ=(constant term)co-efficient of x3=62=3

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Question 22:

If α, β, γ be the zeros of the polynomial p(x) such that (α + β + γ) = 3, (αβ + βγ + γα) = −
10 and αβγ = −24, then p(x) = ?

(a) x3 + 3x2 − 10x + 24
(b) x3 + 3x2 + 10x −24
(c) x3 − 3x2 −10x + 24
(d) None of these

Answer:

 (c) x33x2-10x+24 Given: αβ and γ are the zeroes of polynomial p(x).Also, (α+β+γ)=3, (αβ+βγ+γα)=10 and αβγ=24p(x)=x3(α+β+γ)x2+(αβ+βγ+γα)xαβγ           =x33x2-10x+24

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Question 23:

If two of the zeros of the cubic polynomial ax3 + bx2 + cx + d is 0, then the third zeros is

(a) -ba
(b) ba
(c) ca
(d) -da

Answer:

 (a) ba  Let α, 0 and 0 be the zeroes of ax3+bx2+cx+d=0.Then sum of the zeroes=ba        =>α+0+0=ba        =>α=ba  Hence, the third zero is ba.

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Question 24:

If one of the zeros of the cubic polynomial ax3 + bx2 + cx + d is 0, then the product of the other two zeros is

(a) -ca
(b) ca
(c) 0
(d) -ba

Answer:

(b)  ca Let α, β and 0 be the zeroes of ax3+bx2+cx+d.Then, sum of the products of zeroes taking two at at a time is given by   (αβ+β×0+α×0)=ca   =>αβ=ca   The product of the other two zeroes is ca.

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Question 25:

If one of the zeros of the cubic polynomial x3 + ax2 + bx + c is −1, then the product of the other two zeros is

(a) ab − 1
(b) b a − 1
(c) 1 − a + b
(d) 1 + a b

Answer:

(c) 1a+b      Since 1 is a zero of x3+ax2+bx+c, we have:    (1)3+a×(1)2+b×(1)+c=0   =>ab+c1=0   =>c=1a+b Also, product of all zeroes is given by  αβ×(1)=c=>αβ=c=>αβ=1a+b

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Question 26:

If α, β be the zero of the polynomial 2x2 + 5x + k such that α2 + β2 + αβ = 214, then k = ?
(a) 3
(b) −3
(c) −2
(d) 2

Answer:

(d) 2Since α and β are the zeroes of 2x2+5x+k, we have: α+β=52 and αβ=k2Also, it is given that α2+β2+αβ=214.=>(α+β)2αβ=214=>522k2=214=>254k2=214=>k2=254214=44=1=>k=2 

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Question 27:

On dividing a polynomial p(x) by a non-zero polynomial q(x), let g(x) be the quotient and r(x) be the remainder, than p(x) = q(x)⋅g(x) + r(x), where

(a) r(x) = 0 always
(b) deg r (x) <deg g(x) always
(c) either r(x) = 0 or deg r(x) <deg g(x)
(d) r(x) = g(x)

Answer:

(c) either r(x) =0 or deg r(x)<deg g(x)By division algorithm on polynomials, either r(x)=0 or deg r(x)<deg g(x).



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Question 28:

Which of the following is a true statement?

(a) x2 + 5x − 3 is a linear polynomial.
(b) x2 + 4x − 1 is a binomial.
(c) x + 1 is a monomial.
(d) 5x3 is a monomial.

Answer:

(d) 5x2 is a monomial. 5x2 consists of one term only. So, it is a monomial.   



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Question 1:

Zeros of p(x) = x2 − 2x − 3 are

(a) 1, −3
(b) 3, −1
(c) −3, −1
(d) 1, 3

Answer:

(b) 3,-1
Here, p(x)=x2-2x-3

Let x2-2x-3=0=>x2-(3-1)x-3=0=>x2-3x+x-3=0=>xx-3+1x-3=0=>x-3x+1=0=>x=3,-1

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Question 2:

If α, β, γ are the zeros of the polynomial x3 − 6x2x + 30, then the value of (αβ + βγ + γα) is

(a) −1
(b) 1
(c) −5
(d) 30

Answer:

(a) −1
Here, p(x) =x3-6x2-x+3

Comparing the given polynomial with x3-α+β+γx2+αβ+βγ+γαx -αβγ, we get:
 αβ+βγ+γα=-1

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Question 3:

If α, β are the zeroes of kx2 − 2x + 3k such that α + β = αβ, then k = ?

(a) 13
(b) -13
(c) 23
(d) -23

Answer:

 (c) 23
Here, p(x)=x2-2x+3k
Comparing the given polynomial with ax2+bx+c, we get:
a=1, b=-2 and c=3k
It is given that α and β
are the roots of the polynomial.
α+ β=-ba=>α+β=--21=>α+β=2     ...(i)

Also,  αβ=ca
 =>αβ=3k1=> αβ=3k       ...(ii)Now, α+ β= αβ=>2=3k      [Using (i) and (ii)]=>k=23

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Question 4:

It is given that the difference between the zeroes of 4x2 − 8kx + 9 is 4 and k > 0. Then, k = ?

(a) 12
(b) 32
(c) 52
(d) 72

Answer:

 (c) 52
Let the zeroes of the polynomial be α and α+4.
Here,
p(x) =4x2-8kx+9
Comparing the given polynomial with ax2+bx+c, we get:
a = 4, b = −8k and c = 9
Now, sum of the roots=-ba
=>α+α+4=-(-8k)4=>2α+4=2k=>α+2=k=>α=(k-2)      ...(i)Also, product of the roots, αβ=ca=> α(α+4)=94 =>(k-2)(k-2+4)=94=>k-2k+2=94=>k2-4=94=>4k2-16=9=>4k2=25=>k2=254=>k=52        (k>0)

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Question 5:

Find the zeros of the polynomial x2 + 2x − 195.

Answer:

Here, p(x)= x2+2x-195

Let p(x) =0 =>x2+(15-13)x-195=0=>x2+15x-13x-195=0=>xx+15-13(x+15)=0=>x+15x-13=0=>x=-15,13Hence, the zeroes are -15 and 13.

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Question 6:

If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is the reciprocal of the other, find the value of a.

Answer:

a+9x2-13x+6a=0Here, A=a2+9, B=13 and C=6aLet α and 1α be the two zeroes.Then, product of the zeroes=CA=>α.1α=6aa2+9=>1=6aa2+9=>a2+9=6a=>a2-6a+9=0=>a2-2×a×3+32=0=>a-32=0=>a-3=0=>a=3

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Question 7:

Find a quadratic polynomial whose zeros are 2 and −5.

Answer:

It is given that the two roots of the polynomial are 2 and −5.
Let α=2 and β=-5
Now, sum of the zeroes, α+β = 2 + (5) = 3
Product of the zeroes, αβ = 2×5 = 10
∴ Required polynomial = x2-(α+β)x+αβ
=x2(-3)x+(-10)=x2+3x-10

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Question 8:

If the zeroes of the polynomial x3 − 3x2 + x + 1 are (ab), a and (a + b), find the values of a and b.

Answer:

 The given polynomial =x3-3x2+x+1  and its roots are (a-b), a and (a+b).

Comparing the given polynomial with Ax3+Bx2+Cx+D, we have:A=1, B=-3, C=1 and D=1Now, (a-b)+a+(a+b)=-BA=>3 a=--31=> a=1Also, (a-b)×a×(a+b)=-DA=>aa2-b2=-11=>1(12-b2)=-1=>1-b2=-1=>b2=2=>b=±2a=1 and b=±2

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Question 9:

Verify that 2 is a zero of the polynomial x3 + 4x2 − 3x − 18.

Answer:

Let p(x)=x3+4x2-3x-18

Now, p2=23+4×22-3×2-18=02 is a zero of p(x).

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Question 10:

Find the quadratic polynomial, the sum of whose zeroes is −5 and their product is 6.

Answer:

Given:
Sum of the zeroes = −5
Product of the zeroes = 6
∴ Required polynomial = x2-(sum of the zeroes)x+product of the zeroes
=x2--5x+6=x2+5x+6



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Question 11:

Find a cubic polynomial whose zeros are 3, 5 and −2.

Answer:

Let α, β and γ be the zeroes of the required polynomial. Then we have:α+β+γ=3+5+(-2)=6αβ+βγ+γα=3×5+5×(-2)+(-2)×3=-1 and αβγ=3×5×-2=-30Now, px=x3-x2α+β+γ+xαβ+βγ+γα-αβγ         =x3-x2×6+x×-1--30          =x3-6x2-x+30So, the required polynomial is px=x3-6x2-x+30.

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Question 12:

Using remainder theorem, find the remainder when p(x) = x3 + 3x2 − 5x + 4 is divided by (x − 2).

Answer:

Given: px=x3+3x2-5x+4Now, p2 =23+3(22)-52+4               =8+12-10+4               =14

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Question 13:

Show that (x + 2) is a factor of f(x) = x3 + 4x2 + x − 6.

Answer:

Given: fx=x3+4x2+x-6Now, f-2=-23+4-22+-2-6                  =-8+16-2-6                   =0 x+2 is a factor of fx=x3+4x2+x-6.

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Question 14:

If α, β, γ are the zeroes of the polynomial p(x) = 6x3 + 3x2 − 5x + 1, find the value of 1α+1β+1γ

Answer:

Given: px=6x3+3x2-5x+1                   =6x2--3x2+-5x-(-1)

Comparing the polynomial with x3-x2α+β+γ+xαβ+βγ+γα-αβγ, we get:
αβ+βγ+γα=-5 and αβγ=-11α+1β+1γ=βγ+αγ+αβαβγ=-5-1=5

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Question 15:

If α, β are the zeros of the polynomial f(x) = x2 − 5x + k such that α − β = 1, find the value of k.

Answer:

Given: fx=x2-5x+kThe co-efficients are a=1, b=-5 and c=k.α+β=-ba=>α+β=-(-5)1=>α+β=5         1Also, α-β=1       2From 1 & 2, we get:2α=6=>α=3Putting the value of α in (1), we get β=2.Now, αβ=ca=>3×2=k1k=6

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Question 16:

Show that the polynomial f(x) = x4 + 4x2 + 6 has no zeroes.

Answer:

Let t=x2So, f(t)= t2+4t+6Now, to find the zeroes, we will equate f(t)=0.t2+4t+6=0Now, t=-4±16-242           =-4±-82           =-2±-2i.e., x2=-2±-2x=-2±-2, which is not a real number.The zeroes of a polynomial should be real numbers.The given f(x) has no zeroes.

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Question 17:

If one zero of the polynomial p(x) = x3 − 6x2 + 11x − 6 is 3, find the other two zeroes.

Answer:

Given: px=x3-6x2+11x-6 and its factor, x+3Let us divide px by (x-3).

Here, x3-6x2+11x-6=x-3x2-3x+2                                  =x-3 x2-2+1x+2                                  =x-3(x2-2x-x+2)                                  =x-3[xx-2-1(x-2)]                                  =x-3x-1x-2The other two zeroes are 1 and 2.

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Question 18:

If two zeroes of the polynomial p(x) = 2x4 − 3x3 − 3x2 + 6x − 2 are 2 and -2, find its other two zeroes.

Answer:

Given: px=2x4-3x3-3x2+6x-2 and the two zeroes, 2 and -2So, the polynomial is x+2x-2=x2-2.Let us divide px by x2-2.

Here, 2x4-3x3-3x2+6x-2=x2-22x2-3x+1                                           =x2-22x2-2+1x+1                                           =x2-2(2x2-2x-x+1)                                           =(x2-2)[(2x(x-1)-1(x-1)]                                           =x2-22x-1x-1The other two zeroes are 12 and 1.

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Question 19:

Find the quotient when p(x) = 3x4 + 5x3 − 7x2 + 2x + 2 is divided by (x2 + 3x + 1).

Answer:

Given: px=3x4+5x3-7x2+2x+2Dividing px by x2+3x+1, we have:             



The quotient  is 3x24x+2

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Question 20:

Use remainder theorem to find the value of k, it being given that when x3 + 2x2 + kx + 3 is divided by (x − 3), then the remainder is 21.

Answer:

Let px=x3+2x2+kx+3Now, p3=33+232+3k+3                 =27+18+3k+3                 =48+3kIt is given that the remainder is 213k+48=213k=-27k=-9



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