Class 10 Maths Worksheet for Polynomials

Given below are the Class 10 Maths Worksheet with answers for Polynomials
a. cubic polynomials problems
c. Word Problems

Question 1
Find the zeroes of the quadratic polynomial $x^2 + x -12$ and verify the relationship between the zeroes and the coefficients.

$x^2 + x -12$
$= x^2 + 4x -3x -12$
$=x(x+4) -3(x+4) = (x-3)(x+4)$
So zeroes are 3 and -4

Question 2.
Find the quadratic polynomial, the sum and product of whose zeroes are 4 and 1, respectively
Question 3
If a and b are zeroes of the $x^2 + 7x + 7$, find the value of $a^{-1} + b^{-1}-2ab$

for $f(x)=x^2 + 7x + 7$
we get
$a+b=-7$
$ab=7$
Now
$a^{-1} + b^{-1}-2ab$
$= \frac {a+b-2a^2b^2}{ab}$
$= \frac {-7-98}{7}=-15$

Question 4
Find remainder when $x^3 - ax^2 + 6 - a$ is divided by (x - a).

Given $p(x) =x^3 - ax^2 + 6 - a$
By remainder theorem
p(a) = 6-a

Question 5
If p and q are zeroes of $f(x) = x^2 - 5x + k$, such that $p -q = 1$, find the value of k.

for $f(x)=x^2 - 5x + k$
we get
$p+q= 5$
$pq=k$
Now
$p -q = 1$
$(p -q)^2=1$
$(p+q)^2 -4pq=1$
$25-4k=1$
$k=6$

Question 6
Given that two of the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$ are 0, then find the third zero.

Two zeroes = 0, 0
Let the third zero be k.
The, using relation between zeroes and coefficient of polynomial, we have:
$k + 0 + 0 = -\frac {b}{a}$
Third zero = k = -b/a

Question 7
If the zeroes of the quadratic polynomial $x^2 + (a + 1) x + b$ are 2 and -3, then find the value of a and b.

Let $f(x)=x^2 + (a + 1) x + b$
Then
$2-3= - (a+1) or a=0$
$-6 = b$
So a=0 and b=6

Question 8
If one of the zeroes of the cubic polynomial $x^3 + ax^2+ bx + c$ is -1, then find the product of the other two zeroes.

$k_1 k_2 k_3 = - \frac {Constant term}{Coefficient \; of \; x^3}$

Hence
$k_1 k_2 k_3 =-c$
Now $k_1=-1$
So,
$k_2 k_3=c$

Question 9
If a-b, a a+b , are zeroes of $x^3-6x^2 + 8x$, then find the value of b

Let $f(x)=x^3-6x^2 + 8x$
Method -1
$=x(x^2 -6x +8)=x(x-2)(x-4)$
So 0,2,4 are zeroes of the polynomial. or a=2 and b=2 or -2
Method -2
$k_1 + k_2 + k_3 = - \frac {Coefficient \; of \; x^2}{Coefficient \; of \; x^3}$

$a-b + a + a+b=6$ or a=2
Now $k_1 k_2 k_3 = - \frac {Constant term}{Coefficient \; of \; x^3}$
$(2-b)2(2+b) =0$ or b=+2 or -2

Question 10
If a and b are zeroes of the polynomial $f(x) = 2x^2 - 7x + 3$, find the value of $a^2 + b^2$.

$f(x) = 2x^2 - 7x + 3$
$=2x^2 -x -6x+ 3= x(2x-1) -3(2x -1) = (x-3)(2x-1)$
So zeroes are 3 and 1/2
Now
$a^2 + b^2$
$=9 + \frac {1}{4}= \frac {37}{4}$

Question 11
Quadratic polynomial $4x^2 + 12x + 9$ has zeroes as p and q . Now form a quadratic polynomial whose zeroes are $p -1$ and $q-1$

$4x^2 + 12x + 9$
$=4x^2 + 6x + 6x + 9$
$= 2x(2x+ 3) + 3(2x + 3)=(2x+3)^2$
So p=-3/2 and q=-3/2
So, $p -1= -\frac {5}{2}$ and $q -1= -\frac {5}{2}$
$(x + \frac {5}{2})^2$
or
$4x^2 + 20x + 25$

Question 12.
Find the remainder when x51 +51 is divided by (x+1).

Question 13
Verify the 1/2 ,1, -2 are zeroes of cubic polynomial $2x^3 + x^2 -5x + 2$. Also verify the relationship between the zeroes and their coefficients.

Let $f(x) =2x^3 + x^2 -5x + 2$
For Verification of zeroes, we can simply substitute in the polynomial and verify
$f(1/2) = 2 (\frac {1}{2}}^3 + (\frac {1}{2})^2 -5 \frac {1}{2} + 2 = 0$
$f(1) = 2(1)^3 + (1)^2 -5(1) +2 =0$
$f(2) = 2(2)^3 + (2)^2 -5(2) +2 =0$
The relationship between the zeroes and coefficient is given by
i. $k_1 + k_2 + k_3 = - \frac {Coefficient \; of \; x^2}{Coefficient \; of \; x^3}$
or
$\frac {1}{2} + 1 -2 = -\frac {1}{2}$
$-\frac {1}{2} = -\frac {1}{2}$
Verified

ii. $k_1 k_2 k_3 = - \frac {Constant term}{Coefficient \; of \; x^3}$
$\frac {1}{2} \times 1 \times (-2) = -1$
$-1=-1} Verified iii.$k_1 k_2 + k_2 k_3 + k_1 k_3 = \frac {Coefficient \; of \; x}{Coefficient \; of \; x^3} \frac {1}{2} -2 -1 = \frac {-5}{2} \frac {-5}{2}= \frac {-5}{2}$Verified Question 14 p and q are zeroes of the quadratic polynomial$x^2 - (k + 6)x + 2(2k - 1)$. Find the value of k if$2(p+q) =pq$Answer for$f(x)=x^2 - (k + 6)x + 2(2k - 1)$We get,$p+q = k+6pq = 2(2k-1)$Now$2(p+q) =pq$Therefore,$2(k+6) = 2(2k-1)$or$k+6=2k-1$or$k=7$Question 15 m, n are zeroes of$ax^2 - 5x + c$. Find the value of a and c if$m + n = m \times n = 10$. Answer for$f(x)=ax^2 - 5x + c$we get$m+n=\frac {5}{a}mn=\frac {c}{a}$Given$m + n = m \times n = 10$Therefore, 5/a=10 a=1/2 c/a=10 or c=5 Question 16 Find a quadratic polynomial, the sum and product of whose zeroes are √2 and -3/2 respectively. Also, find its zeroes. Answer$k_1 + k_2 = \sqrt {2}k_1 k_2 = -\frac {3}{2}$Now the quadratic polynomial is given by$x^2 -(k_1 + k_2) x + k_1 k_2=x^2 -\sqrt {2} x -\frac {3}{2}$or$=2x^2 -2 \sqrt {2} x -3$Now we need to find the zeroes of this polynomial$2x^2 -2 \sqrt {2} x -3=2x^2 -3 \sqrt {2} x + \sqrt {2}x -3= \sqrt {2}x( \sqrt {2}x -3) + 1( \sqrt {2}x -3)=(\sqrt {2}x +1)(\sqrt {2}x -3)$or roots are$-\frac {1}{\sqrt {2}} , \frac {3}{\sqrt {2}$Question 17 Given that the zeroes of the cubic polynomial$x^3 - 6x^2 + 3x + 10$are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial. Answer$k_1 + k_2 + k_3 = - \frac {Coefficient \; of \; x^2}{Coefficient \; of \; x^3} a+a+b +a +2b = 6$or$a+b= 2$or$a =2-b$Now$k_1 k_2 k_3 = - \frac {Constant term}{Coefficient \; of \; x^3}a(a+b)(a+2b) =-102(2-b)(2+b)=-10(4-b^2) =-5$or b= -3 or 3 So a= 5 or -1 The zeroes with a=5 ,b=-3 can be expressed as 5 ,2,-1 The zeroes with a=-1 ,b=3 can be expressed as -1 ,2,5 Question 18 Check whether g(x) is a factor of f(x) by dividing f(x) by g(x):$f(x) = 2x^4 + 4x^3 - 5x^2 - 2x + 2$,$g(x) = x^2 + 2x- 2$. Answer Here is the division So$ f(x) = (2x^2 -1) g(x)$Question 19 If one zero of the polynomial$2x^2 - 5x - (2k + 1)$is twice the other, find both the zeroes of the polynomial and the value of k. Answer Let a be one zero ,then another will be 2a Now$a + 2a= \frac {5}{2}$or a= 5/6 Also$a \times 2a = \frac {-(2k+1)}{2}$or$\frac {25}{9} = -(2k+1)2k = -\frac {34}{9}$or k= -17/9 Question 20 If m and n are the zeroes of the quadratic polynomial$f(x) = x^2 - px + q$, then find the values of: a.$m^2+ n^2$b.$m^{-1} + n^{-1}$Answer for$f(x) = x^2 - px + q$We get$m+n=pmn=q$a.$m^2+ n^2=(m+n)^2 -2mn= p^2-q$b.$m^{-1} + n^{-1} = \frac {m+n}{mn} = \frac {p}{q}$Question 21 If the zeroes of the polynomial$f(x) =x^3 -12x^2 + 39x + k$are in A. P., find the value of k. Answer Let a-b,a ,a+b are the zeroes Now$k_1 + k_2 + k_3 = - \frac {Coefficient \; of \; x^2}{Coefficient \; of \; x^3}a -b +a +a +b= 12$or a=4 Also$k_1 k_2 + k_2 k_3 + k_1 k_3 = \frac {Coefficient \; of \; x}{Coefficient \; of \; x^3}(a - d)a + a(a + d) + (a + d)(a - d) = 393a^2 - d^2 = 39$Substituting the value of a from above$3(4)^2 - d^ = 39d = \pm 9$hence, series are 1 , 4 , 7 or 7, 4 , 1 now,$k_1 k_2 k_3 = - \frac {Constant term}{Coefficient \; of \; x^3}7 \times 4 \times 1 = -k$k=-28 Question 22 Using division show that$3y^2 + 5$is a factor of$6y^5+ 15y^4 + 16y^3 + 4y^2 + 10y - 35$. Answer Question 23. If the polynomial f(x) = 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. Answer Dividend = Divisor X Quotient + Remainder Dividend - Remainder = Divisor X Quotient Dividend - Remainder is always divisible by the divisor. Now, it is given that f(x) when divided by x2 -2x + k leaves (x + a) as remainder. So f(x) -(x+a) is divided by x2 -2x + k x4 -6x3 + 16x2 - 25x + 10 -(x+a) =x4 -6x3 + 16x2 - 26x + 10 -a Now doing the division Now remainder should be zero (-10 + 2k)x + (10 - a - 8k + k2) = 0 -10 + 2k = 0 or k=5 Now (10 -a - 8k + k2) = 0 or 10 - a - 8 (5) + 52 = 0 or - a - 5 = 0 a =-5 Question 24. Find all the zeroes of the polynomial x4 - 3x3 + 6x - 4, if two of its zeroes are √2 and -√2 Answer √2 , -√2, 2,1 Question 25. If p and q are he zeroes of the quadratic polynomial f(x) = x2 - 2x + 3, find a polynomial whose roots are: 1. p + 2, q + 2 2. (p-1)/(p+1) , (q-1)/(q+1) Question 26. For what value of k, -7 is the zero of the polynomial 2x2 + 11x + (6k - 3)? Also find the other zero of the polynomial Answer k= -3 The other root is 3/2 Question 27. What must be added to f(x) = 4x4 + 2x3 - 2x2 + x - 1 so that the resulting polynomial is divisible by g(x) = x2 + 2x -3? Answer 61x - 65 Question 28. Find k so that x2 + 2x + k is a factor of 2x4 + x3 - 14 x2 + 5x + 6. Also find all the zeroes of the two polynomials. Question 29. Find the zeroes of 2x3 - 11x2 + 17x - 6. Question 30. If (x - 2) and [x - ½ ] are the factors of the polynomials qx2 + 5x + r prove that q = r Answer 4q+10+r=0 -(1) q/4 +5/2 +r=0 or q+10+4r=0 -(2) Subtracting 1 from 2 3q-3r=0 q=r Question 31. Find k so that the polynomial x2 + 2x + k is a factor of polynomial 2x4 + x3 - 14x2 + 5x + 6. Also, find all the zeroes of the two polynomials. Answer For x2 + 2x + k is a factor of polynomial 2x4 + x3 - 14x2 + 5x + 6, it should be able to divide the polynomial without any remainder Comparing the coefficient of x we get. 21+7k=0 or k=-3 So x2 + 2x + k becomes x2 + 2x -3 = (x-1)(x+3) Now 2x4 + x3 - 14x2 + 5x + 6= (x2 + 2x -3)(2x2-3x-8+2k) =(x2 + 2x -3)(2x2-3x-2) =(x-1)(x+3)(x-2)(2x+1) or x= 1,-3,2,=-1/2 Question 32. On dividing p(x) = x3 - 3x2 + x + 2 by a polynomial q(x), the quotient and remainder were x - 2 and -2x + 4, respectively. Find g(x). Answer x2 - x +1 Question 33. a, b, c are zeroes of cubic polynomial x3 - 2x2 + qx - r. If a + b = 0 then show that 2q = r. Question 34. a,b and c are zeroes of polynomial x3 + px2 + qx + 2 such that a b + 1 = 0. Find the value of 2p + q + 5. Answer By cubic polynomial equation, we have a+b+c=-p ab+bc+ac=q abc=-2 Now given ab+1=0 or ab=-1 So abc=-2 (-1)c=-2 or c=2 2p + q + 5 =-2(a+b+c) +(ab+bc+ac) +5 =-2(a+b+2) +[-1+2(a+b)] +5 =-4 -1+5=0 Summary This Class 10 Maths Worksheet for Polynomials with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail. Also Read Go back to Class 10 Main Page using below links Practice Question Question 1 What is$1 - \sqrt {3}\$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20