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Class 10 Maths Worksheet for Polynomials




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

Short Answer type

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

Answer

$x^2 + x -12$
$= x^2 + 4x -3x -12$
$=x(x+4) -3(x+4) = (x-3)(x+4)$
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 $a^{-1} + b^{-1}-2ab$

Answer

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).

Answer

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.

Answer

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.

Answer

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.

Answer

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.

Answer

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

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

Answer

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 $a^2 + b^2$.

Answer

$f(x) = 2x^2 - 7x + 3$
$=2x^2 -x -6x+ 3= x(2x-1) -3(2x -1) = (x-3)(2x-1)$
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 $q-1$

Answer

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


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

Long Answer type

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.

Answer

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,
$p+q = k+6$
$pq = 2(2k-1)$
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
$k_1 k_2 k_3 = - \frac {Constant term}{Coefficient \; of \; x^3}$
$a(a+b)(a+2b) =-10$
$2(2-b)(2+b)=-10$
$(4-b^2) =-5$
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
worksheet
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
$m+n=p$
$mn=q$
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
$k_1 k_2 + k_2 k_3 + k_1 k_3 = \frac {Coefficient \; of \; x}{Coefficient \; of \; x^3}$
$(a - d)a + a(a + d) + (a + d)(a - d) = 39$
$3a^2 - d^2 = 39$
Substituting the value of a from above
$3(4)^2 - d^ = 39$
$d = \pm 9$
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

Class 10 Maths Worksheet  for Polynomials


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
polynomials class 10 practice questions
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

polynomials class 10 practice questions
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
Class 10 Maths Class 10 Science

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



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