Polynomials Class 10th Mathematics CBSE Solution

Class 10th Mathematics CBSE Solution

Exercise 2.1
Question 1.

The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials, p(x). Find the number of zeroes of p(x) in each case.




Answer:

The number of zeroes for any graph is the number of values of x for which y is equal to zero. And y is equal to zero at the point where a graph cuts x axis.
So, to find the number of zeroes of a polynomial, watch the number of times it cuts the x axis.
(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.


(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.


(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.


(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.


(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.


(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.



Exercise 2.2
Question 1.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x- 2x - 8
(ii) 4s- 4s + 1
(iii) 6x- 3 - 7x
(iv) 4u+ 8u

(v) t- 15
(vi) 3x- x - 4


Answer:

Zeroes of the polynomial are the values of the variable of the polynomial when the polynomial is put equal to zero.
Let p(x) be a polynomial with any number of terms any number of degree.
Now, zeroes of the polynomial will be the values of x at which p(x) = 0.
If p(x) = ax+ bx + c is a quadratic polynomial (highest power is equal to 2) and its roots are α and β, then
Sum of the roots = α + β = - b/a
Product of roots = αβ = c/a

(i) p(x) = x- 2x - 8
So, the zeroes will be the values of x at which p(x) = 0
Therefore,

⇒ x2 - 4x + 2x - 8 = 0

(We will factorize 2 such that the product of the factors is equal to 8 and difference is equal to 2)
⇒ x(x - 4) + 2(x - 4) = 0
= (x - 4)(x + 2)

The value of x- 2x - 8 is zero when x − 4 = 0 or x + 2 = 0,

i.e, x = 4 or x = −2

Therefore, The zeroes of x- 2x - 8 are 4 and −2.

Sum of zeroes = 4 + (-2) = 2 


Hence, it is verified that, 

Product of zeroes = 

Hence, it is verified that, 


(ii) 4s- 4s + 1
= (2s)2 - 2(2s)1 + 12
As, we know (a - b)2 = a2 - 2ab + b2, the above equation can be written as
= (2s - 1)2

The value of 4s2 − 4s + 1 is zero when 2s − 1 = 0, when, s = 1/2 , 1/2.

Therefore, the zeroes of 4s2 − 4s + 1 are  and .

Sum of zeroes = 

Product of zeroes = 

Hence Verified.


(iii)6x- 3 - 7x
= 6x2 - 7x - 3
(We will factorize 7 such that the product of the factors is equal to 18 and the difference is equal to - 7)
= 6x2 + 2x - 9x - 3
= 2x(3x + 1) - 3(3x + 1)
= (3x + 1)(2x - 3)

The value of 6x2 − 3 − 7x is zero when 3x + 1 = 0 or 2x − 3 = 0,

i.e. 

Therefore, the zeroes of 6x2 − 3 − 7x are .

Sum of zeroes = 

Product of zeroes =

Hence, verified.


(iv) 4u+ 8u 
= 4u+ 8u + 0

= 4u (u + 2)

The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0,

i.e., u = 0 or u = −2

Therefore, the zeroes of 4u2 + 8u are 0 and −2.

Sum of zeroes = 

Product of zeroes = 



(v) t2 – 15

= t2 – (√15)2

= (t - √15)(t + √15) [As, x- y2 = (x - y)(x + y)]

The value of t2 − 15 is zero when (t - √15) = 0 or (t + √15) = 0,

i.e., when t = √15 or t = -√15

Therefore, the zeroes of t2 − 15 are √15 and -√15.

Sum of zeroes = 

Product of zeroes = 

Hence verified.


(vi) 3x2 – x – 4
(We will factorize 1 in such a way that the product of factors is equal to 12 and the difference is equal to 1)
= 3x2 - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x – 4 )(x + 1)

The value of 3x2 − x − 4 is zero when 3x − 4 = 0 or x + 1 = 0,

when 

Therefore, the zeroes of 3x2 − x − 4 are and -1

Sum of zeroes = 

Product of zeroes =

Hence, verified.


Question 2.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.


Answer:


(i) , -1

we know that for a quadratic equation in the form ax2 + bx + c = 0, and its zerors are α and β, then
sum of zeroes is

and product of zeroes is

Let the polynomial be , then




Let a = 4, then b = -1, c = -4


Therefore, the quadratic polynomial is 4x2 − x − 4.


(ii) 

we know that for a quadratic equation in the form ax2 + bx + c = 0, and its zerors are α and β, then
sum of zeroes is

and product of zeroes is

Let the polynomial be , then

and


If a = 3, then b = , and c = 1


Therefore, the quadratic polynomial is 


(iii) 0, 

we know that for a quadratic equation in the form ax2 + bx + c = 0, and its zerors are α and β, then
sum of zeroes is

and product of zeroes is

Let the polynomial be , then




If a = 1, then b = 0, c = 


Therefore, the quadratic polynomial is .


(iv) 1, 1

we know that for a quadratic equation in the form ax2 + bx + c = 0, and its zerors are α and β, then
sum of zeroes is

and product of zeroes is

Let the polynomial be , then




If a = 1, then b = -1, c = 1


Therefore, the quadratic polynomial is .


(v) 

we know that for a quadratic equation in the form ax2 + bx + c = 0, and its zerors are α and β, then
sum of zeroes is

and product of zeroes is

Let the polynomial be , then




If a = 4, then b = 1, c = 1


Therefore, the quadratic polynomial is .


(vi) 4, 1

we know that for a quadratic equation in the form ax2 + bx + c = 0, and its zerors are α and β, then
sum of zeroes is

and product of zeroes is

Let the polynomial be , then




If a = 1, then b = -4, c = 1


Therefore, the quadratic polynomial is .



Exercise 2.3
Question 1.

Divide the polynomial by the polynomial and find the quotient and remainder in each of the following :

(i) 

(ii) 

(iii) 


Answer: 

(i)By long division method we have,


Quotient = x − 3

Remainder = 7x − 9


(ii) By long division method we have,


Quotient = x2 + x − 3


Remainder = 8


(iii) By long division method we have,


Quotient = −x2 − 2


Remainder = −5x +10


Question 2.

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) 

(ii) 

(iii) 


Answer: 

(i) t2-3 = t2+0t-3



Since the remainder is 0,


Hence, t2 – 3 is a factor of 2t4+3t3-2t2-9t-12.


(ii)



Since the remainder is 0,


Hence, x2+3x+1 is a factor of 3x4+5x3-7x2+2x+2.


(iii)



Since the remainder ≠0,


Hence, x3-3x+1 is not a factor of x5-4x3+ x2+3x+1.


Question 3.

Obtain all other zeroes of 3x+ 6x- 2x- 10x - 5, if two of its zeroes areand 


Answer:

p(x) = 3x+ 6x- 2x- 10x - 5

Since the two zeroes are  .

 is a factor of 3x+ 6x- 2x- 10x - 5.

Therefore, we divide the given polynomial by .


We know,
Dividend = (Divisor × quotient) + remainder

3 x+ 6 x- 2x- 10 x - 5 = 

3 x+ 6 x- 2 x- 10 x - 5 = 
As (a+b)2 = a2 + b2 + 2ab

So, x2 + 2x + 1 = (x+1)2

3 x+ 6 x- 2 x- 10 x - 5 = 3 () (x + 1)2

Therefore, its zero is given by x + 1 = 0.
⇒ x = −1,-1

Hence, the zeroes of the given polynomial are  and - 1 , -1.


Question 4.

On dividing x3 - 3x2 + x + 2 by a polynomial g(x) the quotient and remainder were (x - 2) and (-2x + 4), respectively. Find g(x).


Answer:

Given, 
Polynomial, p(x) = x3 - 3x2 + x + 2 (dividend)

Quotient = (x − 2)

Remainder = (− 2x + 4)


To find : divisor = g(x)
we know, 
Dividend = Divisor × Quotient + Remainder
⇒ x3 - 3x2 + x +2 = g(x) × (x - 2) + (-2x + 4)

⇒ x3 - 3x2 + x + 2 + 2x - 4 = g(x)(x - 2)

⇒ x3 - 3x2 + 3x - 2 = g(x)(x - 2)

g(x) is the quotient when we divide (x3 - 3x2 + 3x - 2) by (x - 2)



Question 5.

Give examples of polynomials and which satisfy the division algorithm and

(i) 

(ii) 

(iii) 


Answer:

Degree of a polynomial is the highest power of the variable in the polynomial. For example if f(x) = x3 - 2x+ 1, then the degree of this polynomial will be 3.


(i) By division Algorithm : p(x) = g(x) x q(x) + r(x)
It means when P(x) is divided by g(x) then quotient is q(x) and remainder is r(x)
We need to start with p(x) = q(x)
This means that the degree of polynomial p(x) and quotient q(x) is same. This can only happen if the degree of g(x) = 0 i.e p(x) is divided by a constant
Let p(x) = x2 + 1 and g(x) = 2
The,

Clearly, Degree of p(x) = Degree of q(x)
2. Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)




Thus, the division algorithm is satisfied.


(ii) Let us assume the division of x3+ x by x2,


Here,


p(x) = x3+ x


g(x) = x2


q(x) = x and r(x) = x


Clearly, the degree of q(x) and r(x) is the same i.e., 
Checking for division algorithm,

p(x) = g(x) × q(x) + r(x) x3+ x


= (x2 ) × x + x x3+ x = x3+ x


Thus, the division algorithm is satisfied.


(iii) Degree of the remainder will be 0 when the remainder comes to a constant.


Let us assume the division of x3+ 1by x2.


Here,


p(x) = x3+ 1 g(x) = x2


q(x) = x and r(x) = 1


Clearly, the degree of r(x) is 0. Checking for division algorithm,


p(x) = g(x) × q(x) + r(x)x3+ 1


= (x2 ) × x + 1 x3+ 1 = x3+1


Thus, the division algorithm is satisfied.



Exercise 2.4
Question 1.

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 

(ii) 


Answer:

(i) P(x) = 


Now for zeroes, putting the given values in x.


P(1/2) = 2(1/2)3 + (1/2)2 - 5(1/2) + 2
= (1/4) + (1/4) - (5/2) + 2
= (1 + 1 - 10 + 8)/2
= 0/2 = 0

P(1) = 


P(-2) = 


Thus, 1/2, 1 and -2 are zeroes of given polynomial.

Comparing given polynomial with ax3 + bx2 + cx + d and Taking zeroes as α, β, and γ, we have

Now, We know the relation between zeroes and the coefficient of a standard cubic polynomial as


Substituting value, we have

Since, LHS = RHS (Relation Verified)


Since LHS = RHS, Relation verified.


Since LHS = RHS, Relation verified.

Thus, all three relationships between zeroes and the coefficient is verified.


(ii) p(x) = x3 – 4x2 + 5x – 2

Now for zeroes , put the given value in x.

P(2) =  = 

P(1) = 

P(1) = 


Thus, 2, 1 , 1 are the zeroes of the given polynomial.

Now,

Comparing the given polynomial with ax3 + bx2 + cx + d, we get

Now, 

4 = 4



5 = 5


αβγ = 

2 × 1 × 1 = 2

2 = 2


Thus, all three relationships between zeroes and the coefficient is verified.


Question 2.

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.


Answer:

For a cubic polynomial equation, ax3 + bx2 + cx + d, and zeroes α, β and γ

we know that

Let the polynomial be ax3 + bx2 + cx + d, and zeroes α, β and γ.
A cubic polynomial with respect to its zeroes is given by,
x3 - (sum of zeroes) x2 + (Sum of the product of roots taken two at a time) x - Product of Roots = 0
x3 - (2) x2 + (- 7) x - (- 14) = 0
x3 - (2) x2 + (- 7) x + 14 = 0

Hence, the polynomial is x3 - 2x2 - 7x + 14.


Question 3.

If the zeroes of the polynomial are (a - b), a and (a + b). Find a and b.


Answer:

Given,

p(x) = 


zeroes are = a – b , a + b , a


C



Sum of zeroes = 


a = 


The zeroes are = (1 - b), 1 and (1 + b)


Product of zeroes = (1 - b)(1 + b)


(1 - b)(1 + b) = -q/m





So,


We get, 


Question 4.

If two zeroes of the polynomial x4 - 6x3 - 26x2 + 138x - 35 are find other zeroes.


Answer:

Given:

2+√3 and 2-√3 are zeroes of given equation, 

Therefore, 

(x - 2 + √3)(x - 2 - √3) should be a factor of given equation.

Also, (x - 2 + √3)(x - 2 - √3) = x2 - 2x - √3x -2x + 4 + 2√3 + √3x - 2√3 - 3
= x2 - 4x + 1

To find other zeroes, we divide given equation by x2 - 4x + 1


We get ,

x4 - 6x3 - 26x2 + 138x - 35 = (x2 - 4x + 1)(x2 - 2x - 35)


Now factorizing x2 - 2 x - 35 we get,


The value of polynomial is also zero when ,

x - 7 = 0

or

x = 7

And, x + 5 = 0

or

x = -5


Hence, 7 and -5 are also zeroes of this polynomial.


Question 5.

If the polynomial x4 - 6x3 + 16x2 - 25x + 10 is divided by another polynomial x2 - 2k + k the remainder comes out to be x + a, find k and a. 


Answer:

To solve this question divide x4 - 6 x3 + 16 x2 - 25 x + 10 by x2 - 2 x + k by long division method
Let us divide,  by 


So, remainder = (2k - 9)x + (10 - 8k + k2)
But given remainder = x + a
⇒ (2k - 9)x + (10 - 8k + k2) = x + a
Comparing coefficient of x, we have
2k - 9 = 1
⇒ 2k = 10
⇒ k = 5
and
Comparing constant term,
10 - 8k + k2 = a
⇒ a = 10 - 8(5) + 52

⇒ a = 10 - 40 + 25
⇒ a = -5
So, the value of k is 5 and a is -5.


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