NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Class 10 Maths Chapter 2 – Polynomials

Exercise 2.1 Page: 28

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.

Ncert solutions class 10 chapter 2-1

Solutions:

Graphical method to find zeroes:-

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.

(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.

(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.

(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.

(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.

(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.


Exercise 2.2 Page: 33

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Solutions:

(i) x2–2x –8

x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2)

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x2)

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)

(ii) 4s2–4s+1

⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s2)

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )

(iii) 6x2–3–7x

⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2)

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2 )

(iv) 4u2+8u

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2)

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )

(v) t2–15

⇒ t2 = 15 or t = ±√15

Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2)

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )

(vi) 3x2–x–4

⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2)

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x2 )

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

(i) 1/4 , -1

Solution:

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α β

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x +αβ = 0

x2–(1/4)x +(-1) = 0

4x2–x-4 = 0

Thus,4x2–x–4 is the quadratic polynomial.

(ii)√2, 1/3

Solution:

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x +αβ = 0

x2 –(√2)x + (1/3) = 0

3x2-3√2x+1 = 0

Thus, 3x2-3√2x+1 is the quadratic polynomial.

(iii) 0, √5

Solution:

Given,

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

as:-

x2–(α+β)x +αβ = 0

x2–(0)x +√5= 0

Thus, x2+√5 is the quadratic polynomial.

(iv) 1, 1

Solution:

Given,

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x +αβ = 0

x2–x+1 = 0

Thus , x2–x+1is the quadratic polynomial.

(v) -1/4, 1/4

Solution:

Given,

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x +αβ = 0

x2–(-1/4)x +(1/4) = 0

4x2+x+1 = 0

Thus,4x2+x+1 is the quadratic polynomial.

(vi) 4, 1

Solution:

Given,

Sum of zeroes = α+β =4

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x+αβ = 0

x2–4x+1 = 0

Thus, x2–4x+1 is the quadratic polynomial.


Exercise 2.3 Page: 36

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x3-3x2+5x–3 , g(x) = x2–2

Solution:

Given,

Dividend = p(x) = x3-3x2+5x–3

Divisor = g(x) = x2– 2

Ncert solutions class 10 chapter 2-2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

(ii) p(x) = x4-3x2+4x+5 , g(x) = x2+1-x

Solution:

Given,

Dividend = p(x) = x4 – 3x2 + 4x +5

Divisor = g(x) = x2 +1-x

Ncert solutions class 10 chapter 2-3

Therefore, upon division we get,

Quotient = x2 + x–3

Remainder = 8

(iii) p(x) =x4–5x+6, g(x) = 2–x2

Solution:

Given,

Dividend = p(x) =x4 – 5x + 6 = x4 +0x2–5x+6

Divisor = g(x) = 2–x2 = –x2+2

Ncert solutions class 10 chapter 2-4

Therefore, upon division we get,

Quotient = -x2-2

Remainder = -5x + 10

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

(i) t2-3, 2t4 +3t3-2t2-9t-12

Solutions:

Given,

First polynomial = t2-3

Second polynomial = 2t4 +3t3-2t2 -9t-12

Ncert solutions class 10 chapter 2-5

As we can see, the remainder is left as 0. Therefore, we say that, t2-3 is a factor of 2t4 +3t3-2t2 -9t-12.

(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2

Solutions:

Given,

First polynomial = x2+3x+1

Second polynomial = 3x4+5x3-7x2+2x+2

Ncert solutions class 10 chapter 2-6

As we can see, the remainder is left as 0. Therefore, we say that, x2 + 3x + 1 is a factor of 3x4+5x3-7x2+2x+2.

(iii) x3-3x+1, x5-4x3+x2+3x+1

Solutions:

Given,

First polynomial = x3-3x+1

Second polynomial = x5-4x3+x2+3x+1

Ncert solutions class 10 chapter 2-7

As we can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1 .

3. Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Solutions:

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

√(5/3) and – √(5/3) are zeroes of polynomial f(x).

(x –√(5/3)) (x+√(5/3) = x2-(5/3) = 0

(3x2−5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x2−5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Ncert solutions class 10 chapter 2-8

Therefore, 3x+6x−2x−10x–5 = (3x–5)(x2+2x+1)

Now, on further factorizing (x2+2x+1) we get,

x2+2x+1 = x2+x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

So, its zeroes are given by: x= −1 and x = −1.

Therefore, all four zeroes of given polynomial equation are:

√(5/3),- √(5/3) , −1 and −1.

Hence, is the answer.

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).

Solution:

Given,

Dividend, p(x) = x3-3x2+x+2

Quotient = x-2

Remainder = –2x+4

We have to find the value of Divisor, g(x) =?

As 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)

Therefore, g(x) × (x-2) = x3-3x2+3x-2

Now, for finding g(x) we will divide x3-3x2+3x-2 with (x-2)

Ncert solutions class 10 chapter 2-9

Therefore, g(x) = (x2–x+1)

5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

Solutions:

According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor × Quotient + Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

(i) deg p(x) = deg q(x)

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, p(x) = 3x2+3x+3 is a polynomial to be divided by g(x) = 3.

So, (3x2+3x+3)/3 = x2+x+1 = q(x)

Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).

Hence, division algorithm is satisfied here.

(ii) deg q(x) = deg r(x)

Let us take an example, p(x) = x+ 3 is a polynomial to be divided by g(x) = x – 1.

So, x+ 3 = (x – 1)×(x) + (x + 3)

Hence, quotient q(x) = x

Also, remainder r(x) = x + 3

Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).

Hence, division algorithm is satisfied here.

(iii) deg r(x) = 0

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x+ 1 is a polynomial to be divided by g(x) = x.

So, x+ 1 = (x)×(x) + 1

Hence, quotient q(x) = x

And, remainder r(x) = 1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here.


Exercise 2.4 Page: 36

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) 2x3+x2-5x+2; -1/2, 1, -2

Solution:

Given, p(x) = 2x3+x2-5x+2

And zeroes for p(x) are = 1/2, 1, -2

 

∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)3+(1)2-5(1)+2 = 0

p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax3+bx2+cx+d = 2x3+x2-5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x3-4x2+5x-2 ;2, 1, 1

Solution:

Given, p(x) = x3-4x2+5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 23-4(2)2+5(2)-2 = 0

p(1) = 13-(4×12 )+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2

Now, comparing the given polynomial with general expression, we get;

∴ ax3+bx2+cx+d = x3-4x2+5x-2

a = 1, b = -4, c = 5 and d = -2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

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.

Solution:

Let us consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

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

3. If the zeroes of the polynomial x3-3x2+x+1 are a – b, a, a + b, find a and b.

Solution:

We are given with the polynomial here,

p(x) = x3-3x2+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴px3+qx2+rx+s = x3-3x2+x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b2

-1/1 = 1-b2

b2 = 1+1 = 2

b = ±√2

Hence,1-√2, 1 ,1+√2 are the zeroes of x3-3x2+x+1.

4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±3, find other zeroes.

Solution:

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let f(x) = x4-6x3-26x2+138x-35

Since 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

∴ [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

On multiplying the above equation we get,

x2-4x+1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

Ncert solutions class 10 chapter 2-10

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

Now, on further factorizing (x2–2x−35) we get,

x2–(7−5)x −35 = x2– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, its zeroes are given by:

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3 , 2-√3, −5 and 7.

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