Given below are the Class 10 Maths Worksheet with answers for Polynomials

a. cubic polynomials problems

b. quadratic polynomials Problems

c. Word Problems

a. cubic polynomials problems

b. quadratic polynomials Problems

c. Word Problems

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

Find the quadratic polynomial, the sum and product of whose zeroes are 4 and 1, respectively

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

$a^{-1} + b^{-1}-2ab$

$= \frac {a+b-2a^2b^2}{ab}$

$= \frac {-7-98}{7}=-15$

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

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

$p -q = 1$

$(p -q)^2=1$

$(p+q)^2 -4pq=1$

$25-4k=1$

$k=6$

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

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

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,

$k_2 k_3=c$

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

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}$

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 $q -1= -\frac {5}{2}$

So quadratic polynomial will be

$(x + \frac {5}{2})^2$

or

$4x^2 + 20x + 25$

Find the remainder when x

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

p and q are zeroes of the quadratic polynomial $x^2 - (k + 6)x + 2(2k - 1)$. Find the value of k if $2(p+q) =pq$

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 $k=7$

m, n are zeroes of $ax^2 - 5x + c$. Find the value of a and c if $m + n = m \times n = 10$.

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

Find a quadratic polynomial, the sum and product of whose zeroes are √2 and -3/2 respectively. Also, find its zeroes.

$k_1 + k_2 = \sqrt {2}$

$k_1 k_2 = -\frac {3}{2}$

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 $-\frac {1}{\sqrt {2}} , \frac {3}{\sqrt {2}$

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.

$k_1 + k_2 + k_3 = - \frac {Coefficient \; of \; x^2}{Coefficient \; of \; x^3}$

$ 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

Check whether g(x) is a factor of f(x) by dividing f(x) by g(x):

$f(x) = 2x^4 + 4x^3 - 5x^2 - 2x + 2$, $g(x) = x^2 + 2x- 2$.

Here is the division

So $ f(x) = (2x^2 -1) g(x)$

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.

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

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. $m^{-1} + n^{-1}$

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. $m^{-1} + n^{-1} = \frac {m+n}{mn} = \frac {p}{q}$

If the zeroes of the polynomial $f(x) =x^3 -12x^2 + 39x + k$ are in A. P., find the value of k.

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

Using division show that $3y^2 + 5$ is a factor of $6y^5+ 15y^4 + 16y^3 + 4y^2 + 10y - 35$.

If the polynomial f(x) = x

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 x^{2} -2x + k leaves (x + a) as remainder.

So f(x) -(x+a) is divided by x^{2} -2x + k

x^{4} -6x^{3} + 16x^{2} - 25x + 10 -(x+a) =x^{4} -6x^{3} + 16x^{2} - 26x + 10 -a

Now doing the division

Now remainder should be zero

(-10 + 2k)x + (10 - a - 8k + k^{2}) = 0

-10 + 2k = 0 or k=5

Now (10 -a - 8k + k^{2}) = 0

or 10 - a - 8 (5) + 5^{2} = 0

or - a - 5 = 0

a =-5

Find all the zeroes of the polynomial x

√2 , -√2, 2,1

If p and q are he zeroes of the quadratic polynomial f(x) = x

- p + 2, q + 2
- (p-1)/(p+1) , (q-1)/(q+1)

For what value of k, -7 is the zero of the polynomial 2x

k= -3

The other root is 3/2

What must be added to f(x) = 4x

61x - 65

Find k so that x

Find the zeroes of 2x

If (x - 2) and [x - ½ ] are the factors of the polynomials qx

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

Find k so that the polynomial x

For x^{2} + 2x + k is a factor of polynomial 2x^{4} + x^{3} - 14x^{2} + 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 x^{2} + 2x + k becomes x^{2} + 2x -3 = (x-1)(x+3)

Now

2x^{4} + x^{3} - 14x^{2} + 5x + 6= (x^{2} + 2x -3)(2x^{2}-3x-8+2k)

=(x^{2} + 2x -3)(2x^{2}-3x-2)

=(x-1)(x+3)(x-2)(2x+1)

or x= 1,-3,2,=-1/2

On dividing p(x) = x

x^{2} - x +1

a, b, c are zeroes of cubic polynomial x

a,b and c are zeroes of polynomial x

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

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.

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Class 10 Maths Class 10 Science