Polynomials - Test Papers

 CBSE Test Paper 01

Chapter 2 Polynomials

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1. The zeroes of a polynomial x2+5x−24$x^2+5x-24$ are (1)
   1. one positive and one negative
   2. both positive
   3. both negative
   4. both equal
2. If ‘α$\alpha$’ and ‘β$\beta$’ are the zeroes of a quadratic polynomial x2− 5x + b$x^2-5x+b$ andα − β = 1$\alpha -\beta =1$, then the value of ‘b’ is (1)
   1. -6
   2. -5
   3. 5
   4. 6
3. Degree of the polynomial 2x4+3x3-5x2+9x+1 is (1)
   1. 3
   2. 1
   3. 2
   4. 4
4. If α$\alpha$ and β$\beta$ are zeros of x2 + 5x + 8, then the value of (α+β)$(\alpha +\beta )$is (1)
   1. -8
   2. 8
   3. 5
   4. -5
5. Which of the following expressions is not a polynomial? (1)
   1. 5x3−3x2−x−−√+2$5x^3-3x^2-\sqrt{x}+2$
   2. 5x3−3x2−x+2–√$5x^3-3x^2-x+\sqrt{2}$
   3. 5x2−23x+25–√$5x^2-\frac{2}{3}x+2\sqrt{5}$
   4. 5–√x3−35x+17$\sqrt{5}{x}^3-\frac{3}{5}x+\frac{1}{7}$
6. Find the zeroes of the polynomial 3–√$\sqrt{3}$x2 - 8x + 43–√$\sqrt{3}$. (1)
7. If the product of the zeros of the polynomial (ax2 - 6x - 6) is 4. Find the value of a. (1)
8. If x3 + x2 - ax + b is divisible by (x2 - x), write the values of a and b. (1)
9. Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients: 3x2 - x - 4. (1)
10. Find all the zeroes of f(x) = x2 - 2x. (1)
11. If α and β are the zeroes of the polynomial 4x2 - 2x + ( k - 4) and α=1β,$\alpha =\frac{1}{\beta },$ find the value of k. (2)
12. If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is the reciprocal of the other, find the value of a. (2)
13. Find a cubic polynomial whose zeros are 3, 12$\frac{1}{2}$ and -1. (2)
14. If α$\alpha$ and β$\beta$ are zeroes of the polynomial x2−p(x+1)+c$x^2-p(x+1)+c$ such that (α+1)(β+1)=0$(\alpha +1)(\beta +1)=0$, then find the value of c. (3)
15. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a. (3)
16. Find the zeroes of the given quadratic polynomials and verify the relationship between the zeroes and the coefficients. x2 - 2x - 8 (3)
17. A polynomial g(x) of degree zero is added to polynomial 2x3+5x2−14x+10$2x^3+5x^2-14x+10$, so that it becomes exactly divisible by 2x−3$2x-3$. Find g(x). (3)
18. A village of the North-East India is suffering from flood. A group of students decide to help them with food items, clothes etc, So the student collects some amount of rupees, which is represented by x4+x3+8x2+ax+b$x^4+x^3+8x^2+ax+b$

    1. If the number of students is represented by x2+1$x^2+1$, find the values of a and b.
    2. What values have been depicted by the group of students? (4)
19. If α and β$\alpha \text{ and }\beta$ are the zeroes of the polynomial p(x) = 6x2 + 5x - k satisfying the relation, α−β=16$\alpha -\beta =\frac{1}{6}$ , then find the value of k. (4)
20. If α and β$\alpha \text{ and }\beta$ are the zeroes of polynomial p(x) = 3x2 + 2x + 1, find the polynomial whose zeroes are 1−α1+α and 1−β1+β$\frac{1-\alpha }{1+\alpha }\text{ and }\frac{1-\beta }{1+\beta }$. (4)

CBSE Test Paper 01
Chapter 2 Polynomials

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Solution

1. 1. one positive and one negative
      Explanation: x2+5x−24$x^2+5x-24$
      = x2+8x−3x−24$x^2+8x-3x-24$
      = x(x+8)−3(x+8)$x\left(x+8\right)-3\left(x+8\right)$ =0
      (x+8)(x−3)$\left(x+8\right)\left(x-3\right)$=0
      ∴$\therefore$x+8=0$x+8=0$ or x−3=0$x-3=0$
      ⇒$\Rightarrow$x=−8$x=-8$orx=3$x=3$
2. 4. 6
      Explanation: Here α+β=−ba$\alpha +\beta =\frac{-b}{a}$ = −(−5)1$\frac{-\left(-5\right)}{1}$α+β=5$\alpha +\beta =5$ ……….(i)
      And it is given that α−β=1$\alpha -\beta =1$ ……….(ii)
      On solving eq. (i) and eq. (ii), we get
      α+β=5$\alpha +\beta =5$
      α−β=1$\alpha -\beta =1$
      --------------
      2a=6$2a=6$ (β$\beta$ is cancelled)
      α=$\alpha =$62$\frac{6}{2}$
      α=​​3$\alpha =3$ Put the value of α$\alpha$ in eq. (i)
      α+β=5$\alpha +\beta =5$
      ⇒$\Rightarrow$3+β=5$3+\beta =5$
      ⇒β=5−3$\Rightarrow \beta =5-3$
      ⇒β=2$\Rightarrow \beta =2$
      ∴$\therefore$ αβ=ca$\alpha \beta =\frac{c}{a}$
      ⇒$\Rightarrow$ 3×2=b1$3\times 2=\frac{b}{1}$⇒$\Rightarrow$b=6$b=6$
3. 4. 4
      Explanation: The highest power of the variable is 4. So, the degree of the polynomial is 4.
4. 4. -5
      Explanation: x2+5x+8$x^2+5x+8$
      α+β=−Coefficient of xCoefficient of x2$\alpha +\beta =\frac{-\text{Coefficient of x}}{\text{Coefficient of }{x}^2}$
      =−51$=\frac{-5}{1}$
      = -5
5. 1. 5x3−3x2−x−−√+2$5x^3-3x^2-\sqrt{x}+2$
      Explanation: 5x3−3x2−x−−√+2$5x^3-3x^2-\sqrt{x}+2$ is not a polynomial because each term of a polynomial should be a product of a constant and one or more variable raised to a positive, zero or integral power. Here x−−√$\sqrt{x}$does not satisfy the condition of being a polynomial.
6. We have to find the zeroes of the polynomial 3–√$\sqrt{3}$x2 - 8x + 43–√$\sqrt{3}$.
   p(x) =3–√$\sqrt{3}$x2 - 8x + 43–√$\sqrt{3}$
   = 3–√$\sqrt{3}$x2 - 6x - 2x + 43–√$\sqrt{3}$ = 0
   = 3–√(x−23–√)−2(x−23–√)$\sqrt{3}(x-2\sqrt{3})-2(x-2\sqrt{3})$
   = (3–√x−2)(x−23–√)$(\sqrt{3}x-2)(x-2\sqrt{3})$= 0
   ∴$\therefore$ Zeroes = 23√,23–√$\frac{2}{\sqrt{3}},2\sqrt{3}$
7. According to the question,we have to find the value of a such that the product of the zeros of the polynomial (ax2 - 6x - 6) is 4.
   Let α$\alpha$ and β$\beta$ be the zeros of the polynomial (ax2 - 6x - 6)
   Then, α$\alpha$β$\beta$ =  constant term  coefficient of x2=−6a$\frac{\text{ constant term }}{\text{ coefficient of }{x}^2}=\frac{-6}{a}$
   But, α$\alpha$β$\beta$ = 4 (given).
   ∴−6a=4⇒4a=−6⇒a=−64=−32$\therefore \frac{-6}{a}=4\Rightarrow 4a=-6\Rightarrow a=\frac{-6}{4}=\frac{-3}{2}$
   Hence, a = −32$\frac{-3}{2}$
8. Since f(x) = x3 + x2 - ax + b is divisible by (x2 - x), we have
   x2 - x = 0
   ⇒$\Rightarrow$ x(x - 1) = 0
   ⇒$\Rightarrow$ x = 0 or x = 1
   Hence,
   f(0) = 0
   ⇒$\Rightarrow$ x3 + x2 - ax + b = 0
   ⇒$\Rightarrow$ 03 + 02 - a(0) + b = 0
   ⇒$\Rightarrow$ b = 0
   Also,
   f(1) = 0
   ⇒$\Rightarrow$ x3 + x2 - ax + b = 0
   ⇒$\Rightarrow$ 13 + 12 - a(1) + 0 = 0
   ⇒$\Rightarrow$ 1 + 1 - a = 0
   ⇒$\Rightarrow$ 2 - a = 0
   ⇒$\Rightarrow$ a = 2
   Hence , the value of a and b in given polynomial are a = 2 and b = 0.
9. We have,f(x) = 3x2 - x - 4
   = 3x2 - 4x + 3x - 4
   = x(3x - 4) + 1(3x - 4)
   = (3x - 4) )(x + 1)
   ∴$\therefore$ f(x) = 0
   ⇒$\Rightarrow$ (3x - 4)(x +1) = 0
   ⇒$\Rightarrow$ 3x - 4 = 0 or x + 1 = 0
   ⇒x=43$\Rightarrow x=\frac{4}{3}$ or x = -1
   So, the zeros of f(x) are 43 and −1$\frac{4}{3}\text{ and }-1$
   Now sum of zeros =43+(−1)=13=−( coefficient of x)( coefficient of x2)$=\frac{4}{3}+(-1)=\frac{1}{3}=\frac{-(\text{ coefficient of }x)}{\left(\text{ coefficient of }{x}^2\right)}$.
   And product of zeros =43×(−1)=−43= constant term ( coefficient of x2)$=\frac{4}{3}\times (-1)=\frac{-4}{3}=\frac{\text{ constant term }}{\left(\text{ coefficient of }{x}^2\right)}$
10. f(x) = x2 - 2x
    = x (x -2)
    f(x) = 0 ⇒$\Rightarrow$ x = 0 or x = 2
    Hence, zeroes are 0 and 2.
11. Here p(x) = 4x2 − 2x +k − 4
    Here a=4,b=-2,c=k-4
    Given α and β are zeros of the given polynomial
    α=1β,$\alpha =\frac{1}{\beta },$
    ⇒$\Rightarrow$β=1α$\frac{1}{\alpha }$
    αβ=1
    also αβ= ca$\frac{c}{a}$=k−44$\frac{k-4}{4}$
    So k−44$\frac{k-4}{4}$ =1
    k - 4 = 4
    k = 4 + 4 = 8
12. Let α$\alpha$ and 1α$\frac{1}{\alpha }$ be the zeros of (a2 + 9)x2 + 13x + 6a.
    Then, we have
    α×1α=6aa2+9$\alpha \times \frac{1}{\alpha }=\frac{6a}{a^2+9}$
    ⇒ 1 = 6aa2+9$\frac{6a}{a^2+9}$
    ⇒ a2 + 9 = 6a
    ⇒ a2 - 6a + 9 = 0
    ⇒ a2 - 3a - 3a + 9 = 0
    ⇒ a(a - 3) - 3(a - 3) = 0
    ⇒ (a - 3) (a - 3) = 0
    ⇒ (a - 3)2 = 0
    ⇒ a - 3 = 0
    ⇒ a = 3
    So, the value of a in given polynomial is 3.
13. Let α$\alpha$ = 3, β$\beta$ = 12$\frac{1}{2}$ and γ$\gamma$ = -1. Then,
    (α+β+γ)=(3+12−1)=52$(\alpha +\beta +\gamma )=\left(3+\frac{1}{2}-1\right)=\frac{5}{2}$,
    (αβ+βγ+γα)=(32−12−3)=−42$(\alpha \beta +\beta \gamma +\gamma \alpha )=\left(\frac{3}{2}-\frac{1}{2}-3\right)=\frac{-4}{2}$ = -2
    and αβy={3×12×(−1)}=−32$\alpha \beta y=\left\{3\times \frac{1}{2}\times (-1)\right\}=\frac{-3}{2}$
    The polynomial with zeros α,β and γ$\gamma$ is:
    x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ${\mathrm{x}}^3-(\alpha +\beta +\gamma ){\mathrm{x}}^2+(\alpha \beta +\beta \gamma +\gamma \alpha )\mathrm{x}-\alpha \beta \gamma$
    =x3−52x2−2x+32$={\mathrm{x}}^3-\frac{5}{2}{\mathrm{x}}^2-2\mathrm{x}+\frac{3}{2}$
    Thus, 2x3- 5x2- 4x + 3 is the desired polynomial.
14. Given, α$\alpha$ and β$\beta$ are the zeroes of polynomial x2−p(x+1)+c$x^2-p(x+1)+c$
    which can be written as x2−px+c−p$x^2-px+c-p$
    So, sum of zeroes, α+β=p$\alpha +\beta =p$ [∵$[\because$ sum of coefficients = −(coefficient(x))coefficient(x2)$\frac{-(coefficient(x))}{\mathrm{co}efficient(x^2)}$]
    and product of zeroes αβ=c−p$\alpha \beta =c-p$ [∵$[\because$ product of cofficients= constant_termcoefficient(x2)$\frac{constant\mathrm{\_}term}{coefficient(x^2)}$,]
    Also, (α+1)(β+1)=0$(\alpha +1)(\beta +1)=0$
    αβ+α+β+1=0$\alpha \beta +\alpha +\beta +1=0$
    ⇒c−p+p+1=0$\Rightarrow c-p+p+1=0$
    ⇒c=−1$\Rightarrow c=-1$
15. On dividing x4 - 6x3 - 16x2 - 25x + 10 by x2 - 2x + k
    
    ∴$\therefore$ Remainder = (2k - 9)x - (8 - k)k + 10
    But the remainder is given as x+a.
    On comparing their coefficients,
    2k - 9 = 1
    ⇒$\Rightarrow$ k = 10
    ⇒$\Rightarrow$ k = 5 and,
    -(8 - k)k + 10 = a
    ⇒$\Rightarrow$ a = -(8 - 5)5 + 10 = -15 + 10 = -5
    Hence, k = 5 and a = -5
16. Let p(x) = x2 - 2x - 8
    By the method of splitting the middle term,
    x2−2x−8=x2−4x+2x−8$x^2-2x-8=x^2-4x+2x-8$
    =x(x−4)+2(x−4)=(x−4)(x+2)$=x(x-4)+2(x-4)=(x-4)(x+2)$
    For zeroes of p(x),
    p(x) = 0
    ⇒(x−4)(x+2)=0$\Rightarrow (x-4)(x+2)=0$
    ⇒x−4=0 or x+2=0$\Rightarrow x-4=0\text{ or }x+2=0$
    ⇒x=4 or x=−2$\Rightarrow x=4\text{ or }x=-2$
    ⇒x=4,−2$\Rightarrow x=4,-2$
    So, the zeroes of p(x) are 4 and -2.
    We observe that, Sum of its zeroes
    = 4 + (-2) = 2
    =−(−2)1=− (Coefficient of x) Coefficient of x2$=\frac{-(-2)}{1}=\frac{-\text{ (Coefficient of }x)}{\text{ Coefficient of }{x}^2}$
    Product of its zeroes
    =4x(−2)=−8=−81= Constant term  Coefficient of x2$=4x(-2)=-8=\frac{-8}{1}=\frac{\text{ Constant term }}{\text{ Coefficient of }{x}^2}$
    Hence, relation between zeroes and coefficients is verified.
17. According to the question, g(x) of degree degree zero is added to the polynomial (2x3+2x2−14x+10)$(2x^3+2x^2-14x+10)$
    such that it becomes completely divisible by 2x−3$2x-3$.
    Let g(x)=k, then 2x3+5x2−14x+10$2x^3+5x^2-14x+10$+ k will be exactly divisible by 2x−3$2x-3$.
    P(x)= 2x3+5x2−14x+10+k$2x^3+5x^2-14x+10+k$
    We know dividend = quotient x divisor + remainder
    On dividing 2x3+5x2−14x+10+k$2x^3+5x^2-14x+10+k$ by 2x−3$2x-3$, we get quotient x2+4x−1$x^2+4x-1$ and remainder = k+7
    The degree of g(x) is zero then g(x) = 0
    ⇒k+7=0$\Rightarrow k+7=0$ ⇒k=−7$\Rightarrow k=-7$
    ∴$\therefore$ g(x) = -7
18. 1. First we divide x4+x3+8x2+ax+b$x^4+x^3+8x^2+ax+b$ by x2+1$x^2+1$ as follows:
       
       Since x4+x3+8x2+ax+b$x^4+x^3+8x^2+ax+b$ is divisible by x2+1$x^2+1$, therefore remainder = 0
       i.e. (a–1)x+(b–7)=0$(a–1)x+(b–7)=0$
       or (a–1)x+(b–7)=0x+0$(a–1)x+(b–7)=0x+0$
       Equating the corresponding terms, We have
       a–1=0$a–1=0$ and b–7=0$b–7=0$
       i.e. a=1$a=1$ and b=7$b=7$
    2. Common good, Social responsibility
19. According to the question, α and β$\alpha \text{ and }\beta$ are zeroes of p(x) = 6x2 - 5x + k
    So, Sum of zeroes = α+β=−(−56)=56$\alpha +\beta =-\left(\frac{-5}{6}\right)=\frac{5}{6}$.......(i)
    α−β=16$\alpha -\beta =\frac{1}{6}$(Given) ..........(ii)
    Adding equations (i) and (ii) , we get
    2α=1$2\alpha =1$
    or, α=12$\alpha =\frac{1}{2}$
    On putting the value of α$\alpha$ in equation (ii), we get
    12−β=16$\frac{1}{2}-\beta =\frac{1}{6}$
    β=12−16$\beta =\frac{1}{2}-\frac{1}{6}$
    β=26=13$\beta =\frac{2}{6}=\frac{1}{3}$
    ∴αβ=k6=12×13$\therefore \alpha \beta =\frac{k}{6}=\frac{1}{2}\times \frac{1}{3}$=16$=\frac{1}{6}$
    Hence, k = 1
20. Since α and β$\alpha \text{ and }\beta$ are the zeroes of polynomial 3x2 + 2x + 1.
    Hence, α+β=−23$\alpha +\beta =-\frac{2}{3}$
    and αβ=13$\alpha \beta =\frac{1}{3}$
    Now, for the new polynomial,
    Sum of zeroes = 1−α1+α+1−β1+β$\frac{1-\alpha }{1+\alpha }+\frac{1-\beta }{1+\beta }$
    =(1−α+β−αβ)+(1+α−β−αβ)(1+α)(1+β)$=\frac{(1-\alpha +\beta -\alpha \beta )+(1+\alpha -\beta -\alpha \beta )}{(1+\alpha )(1+\beta )}$
    =2−2αβ1+α+β+αβ=2−231−23+13$=\frac{2-2\alpha \beta }{1+\alpha +\beta +\alpha \beta }=\frac{2-\frac{2}{3}}{1-\frac{2}{3}+\frac{1}{3}}$
    ∴$\therefore$ Sum of zeroes = 4/32/3=2$\frac{4/3}{2/3}=2$
    Product of zeroes = [1−α1+α][1−β1+β]$\left[\frac{1-\alpha }{1+\alpha }\right]\left[\frac{1-\beta }{1+\beta }\right]$
    =(1−α)(1−β)(1+α)(1+β)$=\frac{(1-\alpha )(1-\beta )}{(1+\alpha )(1+\beta )}$
    =1−(α+β)+αβ1+(α+β)+αβ$=\frac{1-(\alpha +\beta )+\alpha \beta }{1+(\alpha +\beta )+\alpha \beta }$
    =$=$ 1+23+131−23+13=6333=3$\frac{1+\frac{2}{3}+\frac{1}{3}}{1-\frac{2}{3}+\frac{1}{3}}=\frac{\frac{6}{3}}{\frac{3}{3}}=3$
    Hence, Required polynomial = x2 - (Sum of zeroes)x + Product of zeroes
    = x2 - 2x + 3