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Question 1:

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 2:

Find the zeros of the quadratic polynomial (6x2 − 7x − 3) and verify the relation between its zeros an coefficients.

Answer:

 We have:f(x)=6x27x3=6x29x+2x3=3x(2x3)+1(2x3)=(3x+1)(2x3)f(x)=0=>(3x+1)(2x3)=0                 =>3x+1=0 or 2x3=0                 =>x=13 or x=32So, the zeroes of f(x) are 13 and 32.Sum of the zeroes =13+32=2+96=76=(coefficient of x)(coefficient of x2)Product of the zeros= 13×32 36=constant term(coefficient of x2)

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Question 3:

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 4:

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 5:

Find the zeros of the quadratic polynomial 6x2 − 3 − 7x and verify the relationship between the zeros and the coefficients of the given polynomial.

Answer:

We have:f(x)=6x237x =6x27x-3=6x2(9x2x)3=6x29x+2x3=3x(2x3)+1(2x3)=(3x+1)(2x3)f(x)=0=>(3x+1)(2x3)=0                 =>3x+1=0 or 2x3=0                 =>x=13 or x=32So, the zeroes of f(x) are 13 and 32.Sum of zeroes =13+32= 2+96=76=(coefficient of x)(coefficient of x2)Product of zeroes= 13× 32 =3162=12=constant term(coefficient of x2).

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Question 6:

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 7:

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 8:

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 9:

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 10:

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 11:

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 12:

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 13:

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

Answer:

Let α and β be the zeros of the required polynomial f(x).Then (α+β)=5 and αβ=6f(x)=x2(α+β)x+αβ=>f(x)=x2(5)x+6=>f(x)=x2+5x+6Hence, the required polynomial is f(x)=x2+5x+6.f(x)=0=>x2+5x+6=0                 => x2+3x+2x+6=0                 => x(x+3)+2(x+3)=0                 => (x+2)(x+3)=0                 => (x+2)=0 or (x+3)=0                 => x=2 or x=3So, the zeros of f(x) are 2 and 3.

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Question 14:

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 15:

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 16:

Find the quadratic polynomial, the sum of whose zeros is 2 and their product is −12. Hence, find the zeros of the polynomial.

Answer:

 Let α and β be the zeros of the required polynomial f(x).Then (α+β)=2 and αβ=12f(x)=x2(α+β)x+αβ=>f(x)=x22x+(12)=>f(x)=x22x12Hence, the required polynomial is f(x)=x22x12.f(x)=0=>x22x12=0                 => x232x+22x12=0                 => x(x32)+22(x32)=0                 => (x32)(x+22)=0                 => x=32 or x=22So, the zeros of f(x) are 32 and 22.

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Question 17:

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

Answer:

Given: (α+β)=6 and αβ=4f(x)=x2(α+β)x+αβ=>f(x)=x26x+4Hence, the required polynomial is f(x)=x26x+4.



Page No 52:

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 zeros are −2, −3 and −1.

Answer:

 Let α=2β=3 and γ=1Now, (α+β+γ)=(231)=6and (αβ+βγ+γα)=(6+3+2)=11Also, αβγ{(2)×(3)×(1)}=(6)Required polynomial = x3(α+β+γ)x2+(αβ+βγ+γα)xαβγi.e., x3+6x2+11x+6

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Question 4:

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

Answer:

 Let α=3β=12 and γ=1Now, (α+β+γ)={3+12+(1)}=6+122=52and (αβ+βγ+γα)={(3×12)+12×(1)+(1)×3}=32123=42Also, αβγ={3×12×(1)}=32Required polynomial = x3(α+β+γ)x2+(αβ+βγ+γα)xαβγi.e., x352x2+42x32=x352x242x+32=2x35x24x+3

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Question 5:

When f(x) = 4x3 − 8x2 + 8x + 1 is divided by a polynomial g(x), we get (2x − 1) as quotient and (x + 3) as remainder. Find g(x).

Answer:

 Let f(x)=(4x38x2+8x+1), q(x)=(2x1) and r(x)=(x+3).Then, f(x)=g(x)×q(x)+r(x)=>g(x)=f(x)r(x)q(x)Now, [f(x)r(x)]=(4x38x2+8x+1)(x+3)                           =(4x38x2+7x2)On dividing (4x38x2+7x2) by (2x1), we get g(x).


g(x)=2x23x+2

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Question 6:

Divide (2x2 + x − 15) by (x + 3) and verify the division algorithm.

Answer:

 First, we will write the given polynomials in their standard forms in decreasing order of degrees and then perform the division as shown below:

Quotient=(2x5) and Remainder=0=>(Quotient×divisor + remainder)=(2x5)×(x+3)+0=2x25x+6x15=2x2+x15=dividendThus, quotient×divisor+remainder=dividendHence, the division algorithm is verified.

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Question 7:

Divide (12 − 17x − 5x2) by (3 − 5x) and verify the division algorithm.

Answer:

 First, we will write the given polynomials in their standard forms in decreasing order of degrees and then perform the division as shown below:

   

Quotient=(x+4) and remainder=0⇒ (Quotient×divisor)+remainder=(x+4)×(-5x+3)+0=-5x2-20x+3x+12=5x2-17x+12=dividendThus, (quotient×divisor)+remainder=dividendHence, the division algorithm is verified.

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Question 8:

Divide (3x3 − 4x2 + 7x − 2) by (x2 x + 2) and verify the division algorithm.

Answer:

 First, we will write the given polynomials in their standard forms in decreasing order of degrees and then perform the division as shown below:



Quotient=(3x1) and remainder=0(Quotient×divisor)+remainder=(3x1)×(x2x+2)+0=3x3x23x2+x+6x2=3x34x2+7x2=dividendThus, (quotient×divisor)+remainder=dividendHence, the division algorithm is verified.

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Question 9:

Divide (6 + 19x + x2 − 6x3) by (2+ 5x − 3x2) and verify the division algorithm.

Answer:

 First, we will write the given polynomials in their standard forms in decreasing order of degrees and then perform the division as shown below:



Quotient=(2x+3) and remainder=0(Quotient×divisor)+remainder = (2x+3)×(2+5x3x2)+0=4x+6+10x2+15x6x39x2+0=6x3+x2+19x+6=dividendThus, (quotient×divisor)+remainder=dividendHence, the division algorithm is verified.

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Question 10:

It being given that 2 is one of the zeros of the polynomial x3 − 4x2 + x + 6. Find its other zeros.

Answer:

Let f(x)=x34x2+x+6Since 2 is a zero of f(x), (x2) is a factor of f(x).On dividing f(x) by (x2), we get:



f(x)=(x34x2+x+6)=(x2)(x22x3)=(x2)(x23x+x3)=(x2){x(x3)+1(x3)}=(x2)(x3)(x+1)f(x)=0=>(x2)=0 or (x3)=0 or (x+1)=0                 =>x=2 or x=3 or x=1Thus, the other two zeroes are 3 and 1.

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Question 11:

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 12:

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 13:

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 14:

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 15:

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 16:

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 17:

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 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 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

Page No 57:

Question 8:

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 9:

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 10:

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 11:

How many polynomials are there having 4 and −2 as zeros?

(a) One
(b) Two
(c) Three
(d) More than three

Answer:

(d)  More than threeAn infinite number of polynomials can be formed having zeroes 4 and -2, which will be in the form n(x4)(x+2) or n(x22x8), for any arbitrary real number n.

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Question 12:

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.



Page No 58:

Question 13:

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

Page No 58:

Question 14:

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

Page No 58:

Question 15:

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

Page No 58:

Question 16:

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

Page No 58:

Question 17:

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 18:

If one of the zeroes of the quadratic polynomial x2 + bx + c is negative of the other, then

(a) b = 0 and c is positive
(b) b = 0 and c is negative
(c) b ≠ 0 and c is positive
(d) b ≠ 0 and c is negative

Answer:

 (b)  b=0 and c is negative Let α and α be the zeroes of x2+bx+c.Then  α+(α)=b =>b=0 =>b=0 Also, α×(α)=c =>-α2=c  =>c is negative, as α2>0b=0 and c is negative.

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Question 19:

If the zeros of the quadratic polynomial ax2 + bx + c, where a ≠ 0 and c ≠ 0, are equal then

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

Answer:

 (a)  and a have the same sign Let α and α be the zeroes of ax2+bx+c.Then α+α=ba   >2α=ba   =>α=b2a    Also, α×α=ca      =>α2=ca>0  c and a must have the same sign.

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Question 20:

The zeros of the quadratic polynomial x2 + kx + k, where k > 0

(a) are both positive
(b) are both negative
(c) are always equal
(d) are always unequal

Answer:

(b)   are both negative  Let α and β be the zeroes of x2+kx+k.Then α+β=k and α×β=kThis is possible only when α and β are both negative.

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Question 21:

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

Page No 58:

Question 22:

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



Page No 59:

Question 23:

If α, β are the zeros of f (x) = 2x2 + 6x − 6, then

(a) α + β = αβ
(b) α + β > αβ
(c) α + β < αβ
(d) α + β + αβ = 0

Answer:

 (a)  α+β=αβ  Given: α and β are zeroes of 2x2+6x6. Then α+β=62=3 and αβ=62=3 α+β=αβ

Page No 59:

Question 24:

If α, β are the zeros of the polynomial x2 − 5x + c and α − β = 1, then c = ?

(a) 0
(b) 1
(c) 4
(d) 6

Answer:

(d)  6 Given: α and β are the zeroes of x25x+c.  Also, α-β=1   α+β=5    Solving  α+β = 5  and  α-β=1, we  get  α=3 and  β=2. Since 3 is a zero of the polynomial x25x+c, we have: 325×3+c=0 =>c=6

Page No 59:

Question 25:

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

Page No 59:

Question 26:

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

Page No 59:

Question 27:

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

Page No 59:

Question 28:

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 29:

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 30:

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 31:

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

Page No 59:

Question 32:

If the zeros of the polynomial x3 − 3x2 + x + 1 are ad, a and a + d, then a + d is

(a) a natural number
(b) an integer
(c) a rational number
(d) an irrational number

Answer:

 (d) an irrational number    Since ad, a and a+d are the zeroes of x33x2+x+1, we have:  Sum of zeroes=3  =>ad+a+a+d=3 =>3a=3=>a=1 So, the zeroes are (1d),1 and (1+d).Now, product of the zeroes=1  =>(1d)×1×(1+d)=1 =>1d2=1 =>d2=2=>d=2    a+d=(1±2), which is irrational.  

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Question 33:

If α, β be the zeros of the polynomial x2 − 8x + k such that α2 + β2 = 40, then k = ?

(a) 6
(b) 9
(c) 12
(d) −12

Answer:

(c) 12   Since α and β are the zeroes of x28x+k, we have: α+β=8 and αβ=k Also, it is given that α2+β2=40. Now, (α+β)2=α2+β2+2αβ    =>82=40+2k    =>2k=24    =>k=12         



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Question 34:

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 

Page No 60:

Question 35:

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).

Page No 60:

Question 36:

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 37:

If α, β are the zeros of the polynomial ax2 + bx + c, then (α2 + β2) = ?

(a) a2-2bcb2
(b) b2-2aca2
(c) a2+2bcb2
(d) b2+2aca2

Answer:

(b) b22aca2Here, α+β=ba and αβ=ca(α2+β2)=(α+β)22αβ=b2a22ca=b22aca2

Page No 60:

Question 38:

Which of the following is not a graph of a quadratic polynomial?

(a)
(b)
(c)
(d)

Answer:

(c)


 The graph cuts the xaxis at three distinct points. So, it represents a cubic polynomial.

Page No 60:

Question 39:

Read the statements given below:
I. If α, β are the zeros of the polynomial x2p(x + 1) −c, then (a + 1)(β + 1) = 1 − c.

II. If α, β are the zeros of the polynomial x2 + px + q, then the polynomial having 1α,1β as zeros is qx2 + px + 1.
III. When x3 + 3x2 − 5x + 4 is divided by (x + 1), then the remainder is 9.

Which of the above statements is false?

(a) I only
(b) II only
(c) III only
(d) I and III both

Answer:

(c)  III only The given polynomial is x2px(pc).α+β=p and αβ=(pc)=>(α+1)(β+1)=(α+β)+αβ+1=p(pc)+1=1+cI is true.(II)α and β are the zeroes of x2+px+q.α+β=p and αβ=q1α+1β=α+βαβ=pq and 1α×1β=1qRequired polynomial=x2pqx+1q, i.e., qx2px+1II is true.(III)Let p(x)=x3+3x25x+4 be divided by (x+1)Then remainder=p(1)=(1)3+3×(1)25×(1)+4=1+3+5+4=11III is false.



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Question 40:

Read the statements given below:

I. If the polynomial p(x) = 2x3kx2 + 5x + 2 is exactly divisible by (x + 2), then k = −6.

II. If the polynomial q(x) = x3 − 7x + k when divided by (x − 1) leaves the remainder 2, then k = 6.

III. If two zeros of the polynomial f(x) = x3 − 5x2 − 16x + 80 are equal in magnitude and opposite in sigh, then the third zero is 5.

Which of the above statements is not true?

(a) I only
(b) II only
(c) III only
(d) I as well as II

Answer:

(b) II only (I) p(2)=0=>2×(2)3k(2)2+5×(2)+2=0=>164k10+2=0=>4k=24=>k=6I is true.(II) p(1)=2=>137×1k=2=>k=8II is false.(III) Let the zeroes of f(x) be α,α and β.Then sum of zeroes=(co-efficient of x2 )(co-efficient of x3 )=5α+(α)+β=5=>β=5So the third zero is 5.III is true.

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Question 41:

Assertion (A)
If one zero of the polynomial p(x) = (k2 + 4) x2 + 9x + 4k is the reciprocal of the other zero, then k = 2.

Reason (R)
If (x − α) is a factor of the polynomial p(x), then α is a zero of p(x).

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(b) The Reason (R) is the factor theorem.Let α and 1α be the zeroes of p(x).Then product of the zeroes=α×1α=constant termco-efficient of x2 =4kk2+4=>4kk2+4=1=>k24k+4=0=>(k2)2=0=>k=2Assertion (A) is true, but reason (R) is not its correct explanation.

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Question 42:

Assertion (A)
The polynomial p(x) = x3 + x has one real zero.

Reason (R)
A polynomial of nth degree has at most n zeros.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(b) Both Assertion (A) and Reason (R) are true but  Reason (R) is not a correct explanation of Assertion (A).p(x)=0=>(x3+x)=0=>x(x2+1)=0=>x=0 or x2+1=0But x2+10, for any real value of x.      [x2+1>0]p(x) has one real zero, i.e., 0.Asseration(A) is true and reason(R) does not clearly explain Assertion(A).



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Question 43:

Assertion (A)
If on dividing the polynomial p(x) = x2 − 3ax + 3a − 7 by (x + 1), we get 6 as remainder, then a= 3.

Reason (R)
When a polynomial p(x) is divided by (x − α), then the remainder is p(α).

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(d) Assertion (A) is false. By Remainder Theorem, when p(x) is divided by (x+1), then the remainder is p(1).So, p(1)=6=>(1)23a×(1)+3a7=6=>6a=12=>a=2Assertion (A) is false.

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Question 44:

Assertion (A)
A monic quadratic polynomial having 4 and −2 as zeroes is x2 − 2x − 8.

Reason (R)
The monic quadratic polynomial having α and β as zeroes is given by p(x) = x2 − (α + β) x + αβ.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Both the assertion (A) and reason (R) are true and reason (R) is a correct explanation of (A). Let α=4 and β=2.Then α+β=2 and αβ=8 Required polynomial = x2-(α+β)x+αβ=x2-2x-8i.e., assertion (A) is true.

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Question 45:

If the zeros of a quadratic polynomial ax2 + bx + c are both negative, then a, b, c will have the same sign.

(a) True
(b) False

Answer:

(a) True Let α and β be the negative zeroes of ax2+bx+c.Then α+β=ba<0 and αβ=ca=>ba>0 and ca>0=>a,b,c have the same sign.The given statement is true. 

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Question 46:

Match the following columns:

Column I Column II
(a) If α and β be the zeros of the
polynomial x2 − 5x + k such that
(α − β) = 1, then k = ......... .
(p) 10
(b) If one zero of 4x2 + 17x + p is
the reciprocal of the other, then
p = ......... .
(q) −3
(c) If the zeros of x3 − 6x2 + 3x + m
are (ad), a and (a + d), then
m = ......... .
(r) 4
(d) If the zeros of x3 + 9x2 + 23x + 15
are (ad), a and (a + d), then
a = ......... .
(s) 6

Answer:

(a) αβ=1 and α+β=5=>2α=6=>α=3As α is a zero of the given polynomial, it will satisfy the given polynomial.325×3+k=0=>k=6(b) If one zero is α, then the other zero is 1α. Product of the roots =p4α×1α=p4=>p=4(c) (ad)+a+(a+d)=6=>3a=6=>a=2As a is a zero of the given polynomial, it will satisfy the given equation. 236×22+3×2+m=0=>824+6+m=0=>m=10(d) (ad)+a+(a+d)=9=>3a=9=>a=3The correct matches are (a)-(s),(b)-(r),(c)-(p) and (d)-(q).



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Question 47:

Match the following columns:

Column I Column II
(a) The polynomial whose zeros
are 2 and −3 is ......... .
(p) x2 − 4x + 1
(b) The polynomial whose zeros are 2+3 and 2-3 is ......... . (q) x2-23x+2
(c) The polynomial whose zeros
are 32 and -12 is ......... .
(r) x2 + x − 6
(d) The polynomial whose zeros
are 3+1 and 3-1
(s) 4x2 − 4x − 3

Answer:

(a) α=2 and β=-3α+β=2+(-3)=-1αβ=2×(-3)=-6 The required polynomial will be x2-(α+β)+αβ=x2+x6.(b) α=2+3 and β=2-3α+β=2+3 +2-3=4αβ=(2+3)(2-3)=4-3=1 The required polynomial will be x2-(α+β)+αβ= x24x+1.(c) α=32 and β=-12α+β=3212=1αβ=34Required polynomial = x2-(α+β)+αβ= x2x34, i.e., 4x24x3(d)  α=3+1 and β=3-1α+β=23αβ=3-1=2The required polynomial is x2-(α+β)+α= x223x+2.Thus, the correct matches  are (a)-(r),(b)-(p),(c)-(s) and (d)-(q).



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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

Page No 69:

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

Page No 69:

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

Page No 69:

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

Page No 69:

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



Page No 70:

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.

Page No 70:

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

Page No 70:

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

Page No 70:

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.

Page No 70:

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.

Page No 70:

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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