# Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

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## 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.

## 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 )

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

(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 equation 3x2 – 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 )

## 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:-

–(α+β)x +αβ = 0

–(1/4)x +(-1) = 0

4x2 –x-4 = 0

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

## 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.

## 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, x

2+√5 is the quadratic polynomial.

## 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.

## 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.

## 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.

## Solution:

Given,

Dividend = p(x) = x3 -3x2 +5x–3

Divisor = g(x) = x2 – 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

## Solution:

Given,

Dividend = p(x) = x4 – 3x2 +4x+5

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

Therefore, upon division we get,

Quotient = x2 + x–3

Remainder = 8

## Solution:

Given,

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

Divisor = g(x) = 2– x2 = –x2 +2

## Solutions:

Given,

First polynomial = t2 -3

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

As we can see, the remainder is left as 0. Therefore, we say that,

t2 -3 is a factor of 2t2 +3t +4.

## Solutions:

Given,

First polynomial = x2+3x+1

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

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.

## Solutions:

Given,

First polynomial = x2 -3x+1

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

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.

√(5/3),- √(5/3) , −1 and −1.

Hence, is the answer

## Solutions:

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

## 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, 3x2+3x+3 is a polynomial to be divided by 3.

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

Thus, you can see, the degree of quotient is equal to the degree of dividend.

Hence, division algorithm is satisfied here.

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

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

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

Also, remainder, r(x) = 0

Thus, you can see, the degree of quotient is equal to the degree of remainder.

Hence, division algorithm is satisfied here.

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

(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) = x2 + 1 is a polynomial to be divided by g(x)=x.

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

And r(x)=1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here