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RS Aggarwal (2015) Solutions for Class 10 Math Chapter 2 - Polynomials

RS Aggarwal (2015) Solutions for Class 10 Math Chapter 2 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among class 10 students for Math Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal (2015) Book of class 10 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal (2015) Solutions. All RS Aggarwal (2015) Solutions for class 10 Math are prepared by experts and are 100% accurate.

Page No 43:

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+3x−10=x2+5x−2x−10=x(x+5)−2(x+5)=(x−2)(x+5)∴f(x)=0=>(x−2)(x+5)=0                 =>x−2=0 or x+5=0                  =>x=2 or x=−5So, the zeroes of f(x)are 2 and−5.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)

Page No 43:

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)=6x2−7x−3=6x2−9x+2x−3=3x(2x−3)+1(2x−3)=(3x+1)(2x−3)∴f(x)=0=>(3x+1)(2x−3)=0                 =>3x+1=0 or 2x−3=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)

Page No 43:

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)=4x2−4x−3=4x2−(6x−2x)−3=4x2−6x+2x−3=2x(2x−3)+1(2x−3)=(2x+1)(2x−3)∴f(x)=0=>(2x+1)(2x−3)=0                  =>2x+1=0 or 2x−3=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)

Page No 43:

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)=5x2−4−8x  =5x2−8x-4=5x2−(10x−2x)−4=5x2−10x+2x−4=5x(x−2)+2(x−2)=(5x+2)(x−2)∴f(x)=0=>(5x+2)(x−2)=0                 =>5x+2=0 or x−2=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)

Page No 43:

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)=6x2−3−7x =6x2−7x-3=6x2−(9x−2x)−3=6x2−9x+2x−3=3x(2x−3)+1(2x−3)=(3x+1)(2x−3)∴f(x)=0=>(3x+1)(2x−3)=0                 =>3x+1=0 or 2x−3=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).

Page No 43:

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)=2x2−11x+15=2x2−(6x+5x)+15=2x2−6x−5x+15=2x(x−3)−5(x−3)=(2x−5)(x−3)∴f(x)=0=>(2x−5)(x−3)=0                 =>2x−5=0 or x−3=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)

Page No 43:

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)=x2−5It can be written as x2+0x−5.=(x2−(5)2)=(x+5)(x−5)∴f(x)=0=>(x+5)(x−5)=0                 =>x+5=0 or x−5=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)

Page No 43:

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)=8x2−4It can be written as 8x2+0x−4=4{(2x)2−(1)2}=4(2x+1)(2x−1)∴f(x)=0=>(2x+1)(2x−1)=0                 =>2x+1=0 or 2x−1=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)x−12                                     =x2+4x−12Sum 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=8−312=512Product of the zeroes = αβ=23×−14=−21126=−16∴ Required polynomial = x2−(α+β)x+αβ=x2−512x+−16                                     =x2−512x−16  Sum of the zeroes =512=−(coefficient of x)(coefficient of x2)Product of the zeroes =−16=constant termcoefficient of x2                                   



Page No 44:

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=12∴f(x)=x2−(α+β)x+αβ=>f(x)=x2−8x+12Hence, required polynomial f(x)=x2−8x+12∴f(x)=0=>x2−8x+12=0                 => x2−(6x+2x)+12=0                 => x2−6x−2x+12=0                 => x(x−6)−2(x−6)=0                 => (x−2)(x−6)=0                 => (x−2)=0 or (x−6)=0                 => x=2 or x=6So, the zeroes of f(x) are 2 and 6.

Page No 44:

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=6∴f(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=1∴f(x)=x2−(α+β)x+αβ=>f(x)=x2−52x+1=>f(x)=2x2−5x+2Hence, the required polynomial is f(x)=2x2−5x+2.∴f(x)=0=>2x2−5x+2=0                 => 2x2−(4x+x)+2=0                 => 2x2−4x−x+2=0                 => 2x(x−2)−1(x−2)=0                 => (2x−1)(x−2)=0                 => (2x−1)=0 or (x−2)=0                 => x=12 or x=2So, the zeros of f(x)are 12 and 2.

Page No 44:

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=−1∴f(x)=x2−(α+β)x+αβ=>f(x)=x2−0x+(−1)=>f(x)=x2−1Hence, the required polynomial is f(x)=x2−1.∴f(x)=0=>x2−1=0                 => (x+1)(x−1)=0                 => (x+1)=0 or (x−1)=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=−12∴f(x)=x2−(α+β)x+αβ=>f(x)=x2−2x+(−12)=>f(x)=x2−2x−12Hence, the required polynomial is f(x)=x2−2x−12.∴f(x)=0=>x2−2x−12=0                 => x2−32x+22x−12=0                 => x(x−32)+22(x−32)=0                 => (x−32)(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=4∴f(x)=x2−(α+β)x+αβ=>f(x)=x2−6x+4Hence, the required polynomial is f(x)=x2−6x+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)=(x3−2x2−5x+6)∴p(3)=(33−2×32−5×3+6)=(27−18−15+6)=0p(−2)=[(−23)−2×(−22)−5×(−2)+6]=(−8−8+10+6)=0p(1)=(13−2×12−5×1+6)=(1−2−5+6)=0∴3,−2 and 1 are the zeroes of p(x),Let α=3, β=−2 and γ=1. Then we have:(α+β+γ)=(3−2+1)=2=−(coefficient of x2)(coefficient of x3)(αβ+βγ+γα)=(−6−2+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)=(3x3−10x2−27x+10)p(5)=(3×53−10×52−27×5+10)=(375−250−135+10)=0p(−2)=[3×(−23)−10×(−22)−27×(−2)+10]=(−24−40+54+10)=0p13={3×133)3−10×132−27×13+10}=(3×127−10×19−9+10)=19−109+1=1−10+99=09=0∴5,−2 and 13 are the zeroes of p(x).Let α=5, β=−2 and γ=13. Then we have:(α+β+γ)=5−2+13=103=−(coefficient of x2)(coefficient of x3)(αβ+βγ+γα)=−10−23+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=−2, β=−3 and γ=−1Now, (α+β+γ)=(−2−3−1)=−6and (αβ+βγ+γα)=(6+3+2)=11Also, αβγ= {(−2)×(−3)×(−1)}=(−6)∴Required polynomial = x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγi.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=3, β=12 and γ=−1Now, (α+β+γ)={3+12+(−1)}=6+1−22=52and (αβ+βγ+γα)={(3×12)+12×(−1)+(−1)×3}=32−12−3=−42Also, αβγ={3×12×(−1)}=−32∴Required polynomial = x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγi.e., x3−52x2+−42x−−32=x3−52x2−42x+32=2x3−5x2−4x+3

Page No 52:

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)=(4x3−8x2+8x+1), q(x)=(2x−1) 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)]=(4x3−8x2+8x+1)−(x+3)                           =(4x3−8x2+7x−2)On dividing (4x3−8x2+7x−2) by (2x−1), we get g(x).


∴g(x)=2x2−3x+2

Page No 52:

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=(2x−5) and Remainder=0=>(Quotient×divisor + remainder)=(2x−5)×(x+3)+0=2x2−5x+6x−15=2x2+x−15=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.

Page No 52:

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=(3x−1) and remainder=0⇒(Quotient×divisor)+remainder=(3x−1)×(x2−x+2)+0=3x3−x2−3x2+x+6x−2=3x3−4x2+7x−2=dividendThus, (quotient×divisor)+remainder=dividendHence, the division algorithm is verified.

Page No 52:

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+5x−3x2)+0=4x+6+10x2+15x−6x3−9x2+0=−6x3+x2+19x+6=dividendThus, (quotient×divisor)+remainder=dividendHence, the division algorithm is verified.

Page No 52:

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)=x3−4x2+x+6Since 2 is a zero of f(x), (x−2) is a factor of f(x).On dividing f(x) by (x−2), we get:



f(x)=(x3−4x2+x+6)=(x−2)(x2−2x−3)=(x−2)(x2−3x+x−3)=(x−2){x(x−3)+1(x−3)}=(x−2)(x−3)(x+1)∴f(x)=0=>(x−2)=0 or (x−3)=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+2x2−11x−12Since -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+2x2−11x−12=(x+1)(x2+x−12)=(x+1){x2+4x−3x−12}=(x+1){x(x+4)−3(x+4)}=(x+1)(x−3)(x+4)∴f(x)=0=>(x+1)(x−3)(x+4)=0                  =>(x+1)=0 or (x−3)=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)=x3−4x2−7x+10Since 1 and −2 are the zeroes of f(x), it follows that each one of (x−1) and (x+2) is a factor of f(x).Consequently, (x−1)(x+2)=(x2+x−2) is a factor of f(x).On dividing f(x) by (x2+x−2), we get:



f(x)=0=>(x2+x−2)(x−5)=0                 =>(x−1)(x+2)(x−5)=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+x3−11x2−9x+18Since 3 and −3 are the zeroes  of f(x), it follows that each one of (x+3) and (x−3) is a factor of f(x).Consequently, (x−3)(x+3)=(x2−9) is a factor of f(x).On dividing f(x) by (x2−9), we get:



f(x)=0 =>(x2+x−2)(x2−9)=0                    =>(x2+2x-x-2)(x−3)(x+3)                   =>(x−1)(x+2)(x−3)(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+x3−34x2−4x+120Since 2 and −2 are the zeroes of f(x), it follows that each one of (x−2) and (x+2) is a factor of f(x).Consequently, (x−2)(x+2)=(x2−4) is a factor of f(x).On dividing f(x) by (x2−4),we get:



 ∴f(x)=0=>(x2+x−30)(x2−4)=0=>(x2+6x-5x-30)(x−2)(x+2)=>[x(x+6)-5(x+6)](x−2)(x+2)=>(x−5)(x+6)(x−2)(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+x3−23x2−3x+60Since 3 and −3 are the zeroes  of f(x), it follows that each one of (x−3) and (x+3) is a factor of f(x).Consequently, (x−3)(x+3)=(x2−3) is a factor of f(x).On dividing f(x) by (x2−3), we get:


 ∴f(x)=0 =>(x2+x−20)(x2−3)=0=>(x2+5x-4x-20)(x2−3)=>[x(x+5)-4(x+5)](x2−3)=>(x−4)(x+5)(x−3)(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)=2x4−3x3−5x2+9x−3.Since 3 and −3 are the zeroes of f(x), it follows that each one of (x−3) and (x+3) is a factor of f(x).Consequently, (x−3)(x+3)=(x2−3) is a factor of f(x).On dividing f(x) by (x2−3), we get:

 
 ∴f(x)=0=>2x4−3x3−5x2+9x−3=0=>(x2−3)(2x2−3x+1)=0=>(x2−3)(2x2-2x-x+1)=0=>(x−3)(x+3)(2x−1)(x−1)=0=>x=3 or x=−3 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)=2x4−11x3+7x2+13x−7.Since (3+2) and (3−2) are the zeroes of f(x), it follows that each one of (x+3+2) and (x+3−2) is a factor of f(x).Consequently, [x−(3+2)] [x−(3−2)]=[(x−3)−2][(x−3)+2]=[(x−3)2−2]=x2−6x+7, which is a factor of f(x)On dividing f(x) by(x2-6x+7), we get:


∴f(x)=0⇒2x4−11x3+7x2+13x−7=0=>(x2−6x+7)(2x2+x−1)=0=>(x+3+2)(x+3−2)(2x−1)(x+1)=0=>x=−3−2 or x=−3+2or x=12 or x=−1Hence, all the zeros are (−3−2), (−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+4x3−2x2−20x−15.Since (x−5) and (x+5) are the zeroes of f(x), it follows that each one of (x−5) and (x+5) is a factor of f(x).Consequently, (x−5)(x+5)=(x2−5) is a factor of f(x).On dividing f(x) by(x2−5), we get:


∴f(x)=0=>x4+4x3−7x2−20x−15=0=>(x2−5)(x2+4x+3)=0=>(x−5)(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)=x2−2x−3=0         =x2−3x+x−3=0         =x(x−3)+1(x−3)=0         =(x−3)(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)=x2−2x−12=0           =>x2−32x+22x−12=0           =>x(x−32)+22(x−32)=0            =>(x−32)(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+52x−3=0      =>4x2+62x−2x−3=0      =>22x(2x+3)−1(2x+3)=0      =>(2x+3)(22x−1)=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+16x−2=0      =>6x2+x−12=0      =>6x2+9x−8x−12=0      =>3x(2x+3)−4(2x+3)=0      =>(2x+3)(3x−4)=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)=7x2−113x−23=0⇒21x2−11x−2=0⇒21x2−14x+3x−2=0⇒7x(3x−2)+1(3x−2)=0⇒(3x−2)(7x+1)=0⇒x=23 or x=−17

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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) x2−2x−15 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 x2−2x−15.

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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) 10x2 − x + 3
(d) x2-110x-310

Answer:

(d)  x2−110x−310 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 x2−110x−310.

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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)  x2−3x−10Given: Sum of zeroes, α+β = 3 Also, product of zeroes, αβ=-10∴Required polynomial=x2-(α+β)+αβ=x2−3x−10

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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(x−4)(x+2) or n(x2−2x−8), 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.



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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

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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+5x−9.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 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

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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 (k−1)x2+kx+1, we have: (k−1)×(−4)2+k×(−4)+1=0  =>16k−16−4k+1=0=>12k−15=0=>k=155124 =>k=54

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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=>4−2a−2+b=0=>b−2a=−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=−2−4=−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>0∴b=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)  c 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 3x2−8x+k.Then product of zeroes=k3     =>α×1α=k3     =>1=k3      =>k=3

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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+6x−6. Then α+β=−62=−3 and αβ=−62=−3 ∴α+β=αβ

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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 x2−5x+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 x2−5x+c, we have: 32−5×3+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

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Question 26:

If α, β, γ are the zeros of the polynomial x3 − 6x2 − x + 30, then (αβ + βγ + γα) = ?

(a) −1
(b) 1
(c) −5
(d) 30

Answer:

 (a) -1It is given that α, β and γ are the zeroes of x3−6x2−x+30.   ∴(αβ+βγ+γα)=co-efficient of x co-efficient of x3=−11=−1

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Question 27:

If α, β, γ are the zeros of the polynomial 2x3 + x2 − 13x + 6, then αβγ = ?

(a) −3
(b) 3
(c) -12
(d) -132

Answer:

 (a) -3 Since α, β and γ are the zeroes of 2x3+x2−13x+6, we have:      αβγ=−(constant term)co-efficient of x3=−62=−3

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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) x3−3x2-10x+24 Given: α, β and γ are the zeroes of polynomial p(x).Also, (α+β+γ)=3, (αβ+βγ+γα)=−10 and αβγ=−24∴p(x)=x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ           =x3−3x2-10x+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) a − b − 1
(b) b − a − 1
(c) 1 − a + b
(d) 1 + a − b

Answer:

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

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Question 32:

If the zeros of the polynomial x3 − 3x2 + x + 1 are a −d, 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 a−d, a and a+d are the zeroes of x3−3x2+x+1, we have:  Sum of zeroes=3  =>a−d+a+a+d=3 =>3a=3=>a=1 So, the zeroes are (1−d),1 and (1+d).Now, product of the zeroes=−1  =>(1−d)×1×(1+d)=−1 =>1−d2=−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 x2−8x+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)2−αβ=214=>−522−k2=214=>254−k2=214=>k2=254−214=44=1=>k=2 

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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).

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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) b2−2aca2Here, α+β=−ba and αβ=ca∴(α2+β2)=(α+β)2−2αβ=b2a2−2ca=b2−2aca2

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Question 38:

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

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

Answer:

(c)


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

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Question 39:

Read the statements given below:
I. If α, β are the zeros of the polynomial x2 − p(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 x2−px−(p−c).∴α+β=p and αβ=−(p−c)=>(α+1)(β+1)=(α+β)+αβ+1=p−(p−c)+1=1+c∴I 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=1q∴Required polynomial=x2−pqx+1q, i.e., qx2−px+1∴II is true.(III)Let p(x)=x3+3x2−5x+4 be divided by (x+1). Then remainder=p(−1)=(−1)3+3×(−1)2−5×(−1)+4=−1+3+5+4=11∴III is false.



Page No 61:

Question 40:

Read the statements given below:

I. If the polynomial p(x) = 2x3 − kx2 + 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)3−k(−2)2+5×(−2)+2=0=>−16−4k−10+2=0=>4k=−24=>k=−6∴I is true.(II) p(1)=2=>13−7×1−k=2=>k=−8∴II 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 )=5∴α+(−α)+β=5=>β=5So the third zero is 5.∴III is true.

Page No 61:

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=>k2−4k+4=0=>(k−2)2=0=>k=2∴Assertion (A) is true, but reason (R) is not its correct explanation.

Page No 61:

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+1≠0, 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).



Page No 62:

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)2−3a×(−1)+3a−7=6=>6a=12=>a=2∴Assertion (A) is false.

Page No 62:

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.

Page No 62:

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. 

Page No 62:

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 (a − d), a and (a + d), then
m = ......... .		 (r) 4
(d) If the zeros of x3 + 9x2 + 23x + 15
are (a − d), a and (a + d), then
a = ......... .		 (s) 6

Answer:

(a) α−β=1 and α+β=5=>2α=6=>α=3As α is a zero of the given polynomial, it will satisfy the given polynomial.∴32−5×3+k=0=>k=6(b) If one zero is α, then the other zero is 1α. Product of the roots =p4⇒α×1α=p4=>p=4(c) (a−d)+a+(a+d)=6=>3a=6=>a=2As a is a zero of the given polynomial, it will satisfy the given equation. ∴23−6×22+3×2+m=0=>8−24+6+m=0=>m=10(d) (a−d)+a+(a+d)=−9=>3a=−9=>a=−3∴The correct matches are (a)-(s),(b)-(r),(c)-(p) and (d)-(q).



Page No 63:

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+x−6.(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= x2−4x+1.(c) α=32 and β=-12α+β=32−12=1αβ=−34∴Required polynomial = x2-(α+β)+αβ= x2−x−34, i.e., 4x2−4x−3(d)  α=3+1 and β=3-1α+β=23αβ=3-1=2∴The required polynomial is x2-(α+β)+α= x2−23x+2.Thus, the correct matches  are (a)-(r),(b)-(p),(c)-(s) and (d)-(q).



Page No 69:

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

Page No 69:

Question 2:

If α, β, γ are the zeros of the polynomial x3 − 6x2 − x + 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

Page No 69:

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)

Page No 69:

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.

Page No 69:

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 (a − b), 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=±√2∴a=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=0∴2 is a zero of p(x).

Page No 69:

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.

Page No 70:

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=-1∴1α+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=k1∴k=6

Page No 70:

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±-2⇒x=-2±-2, which is not a real number.The zeroes of a polynomial should be real numbers.∴The given f(x) has no zeroes.

Page No 70:

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-2∴The 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-1∴The 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 3x2−4x+2

Page No 70:

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 21∴3k+48=21⇒3k=-27⇒k=-9



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