The document NCERT Solutions - Polynomials, Class 10, Maths Class 10 Notes | EduRev is a part of Class 10 category.

All you need of Class 10 at this link: Class 10

**Page No: 28 Exercises 2.1 1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case**

**Answer**

(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.

**Page No: 33 Exercise 2.2**

(i)

(ii) 4

(iii) 6

(iv) 4

(v)

(vi) 3

**Answer**

(i) *x*^{2} – 2*x* – 8

= (*x* - 4) (*x* 2)

The value of *x*^{2} – 2*x* – 8 is zero when *x* - 4 = 0 or *x* 2 = 0, i.e., when *x* = 4 or *x* = -2

Therefore, the zeroes of *x*^{2} – 2*x* – 8 are 4 and -2.

Sum of zeroes = 4 (-2) = 2 = -(-2)/1 = -(Coefficient of *x*)/Coefficient of *x*^{2}

Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of *x*^{2}

(ii) 4*s*^{2} – 4*s* + 1= (2*s*-1)^{2}

The value of 4*s*^{2} - 4*s +*1 is zero when 2*s* - 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s^{2} - 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 1/2 = 1 = -(-4)/4 = -(Coefficient of *s)*/Coefficient of *s*^{2}

Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of *s*^{2}.

(iii) 6*x*^{2} – 3 – 7*x**= *6*x*^{2 }– 7*x *– 3

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

The value of 6*x*^{2} - 3 - 7*x* is zero when 3*x* + 1 = 0 or 2*x* - 3 = 0, i.e., *x* = -1/3 or *x* = 3/2

Therefore, the zeroes of 6*x*^{2} - 3 - 7*x* are -1/3 and 3/2.

Sum of zeroes = -1/3 3/2 = 7/6 = -(-7)/6 = -(Coefficient of *x*)/Coefficient of *x*^{2}

Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of *x*^{2}.

(iv) 4*u*^{2} 8*u**= *4*u*^{2} 8*u *0

= 4*u*(*u* 2)

The value of 4*u*^{2} 8*u* is zero when 4*u* = 0 or *u* 2 = 0, i.e., *u* = 0 or *u* = - 2

Therefore, the zeroes of 4*u*^{2} 8*u* are 0 and - 2.

Sum of zeroes = 0 (-2) = -2 = -(8)/4 = -(Coefficient of *u*)/Coefficient of *u*^{2}

Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of *u*^{2}.

(v) *t*^{2} – 15

= *t*^{2 }- 0.*t* - 15

= (*t *- √15) (*t* √15)

The value of *t*^{2} - 15 is zero when *t* - √15 = 0 or *t* √15 = 0, i.e., when *t* = √15 or *t *= -√15

Therefore, the zeroes of *t*^{2} - 15 are √15 and -√15.Sum of zeroes = √15 -√15 = 0 = -0/1 = -(Coefficient of *t*)/Coefficient of *t*^{2}

Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of *u*^{2}.

(vi) 3*x*^{2} –* x* – 4

= (3*x* - 4) (*x* 1)

The value of 3*x*^{2} –* x* – 4 is zero when 3*x* - 4 = 0 and *x* 1 = 0,i.e., when *x* = 4/3 or *x* = -1

Therefore, the zeroes of 3*x*^{2} –* x* – 4 are 4/3 and -1.

Sum of zeroes = 4/3 (-1) = 1/3 = -(-1)/3 = -(Coefficient of *x*)/Coefficient of *x*^{2}

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of *x*^{2}.**2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.**

**(i) 1/4 , -1**

**(ii) √2 , 1/3 **

**(iii) 0, √5**

**(iv) 1,1 **

**(v) -1/4 ,1/4 **

**(vi) 4,1**

**Answer**

(i) 1/4 , -1

Let the polynomial be *ax*^{2} *bx* *c*, and its zeroes be α and ß

α ß = 1/4 = -*b*/*a*

αß = -1 = -4/4 = *c*/*a*

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

Therefore, the quadratic polynomial is 4*x*^{2} - *x* -4.

(ii) √2 , 1/3

Let the polynomial be *ax*^{2} *bx* *c*, and its zeroes be α and ß

α ß = √2 = 3√2/3 = -*b*/*a*

αß = 1/3 = *c*/*a*

If *a* = 3, then *b* = -3√2, *c* = 1

Therefore, the quadratic polynomial is 3*x*^{2} -3√2*x* 1.

(iii) 0, √5

Let the polynomial be *ax*^{2} *bx* *c*, and its zeroes be α and ß

α ß = 0 = 0/1 = -*b*/*a*

αß = √5 = √5/1 = *c*/*a*

If *a* = 1, then *b* = 0, *c* = √5

Therefore, the quadratic polynomial is *x*^{2} √5.

(iv) 1, 1

Let the polynomial be *ax*^{2} *bx* *c*, and its zeroes be α and ß

α ß = 1 = 1/1 = -*b*/*a*

αß = 1 = 1/1 = *c*/*a*

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

Therefore, the quadratic polynomial is *x*^{2} - *x* 1.

(v) -1/4 ,1/4

Let the polynomial be *ax*^{2} *bx* *c*, and its zeroes be α and ß

α ß = -1/4 = -*b*/*a*

αß = 1/4 = *c*/*a*

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

Therefore, the quadratic polynomial is 4*x*^{2} *x* 1.

(vi) 4,1

Let the polynomial be *ax*^{2} *bx* *c*, and its zeroes be α and ß

α ß = 4 = 4/1 = -*b*/*a*

αß = 1 = 1/1 = *c*/*a*

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

Therefore, the quadratic polynomial is *x*^{2} - 4*x* 1.

**Page No: 36**

**Exercise 2.3**

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

**Answer**

(i) *p*(*x*) = *x*^{3} – 3*x*^{2} 5*x* – 3, *g*(*x*) = *x*^{2} – 2

Quotient = *x*-3 and remainder 7*x* - 9

(ii) *p*(*x*) = *x*^{4} – 3*x*^{2} 4x 5, *g*(*x*) = *x*^{2} 1 – *x*

Quotient = *x*^{2} *x *- 3 and remainder 8

(iii) *p*(*x*) = *x*^{4} – 5*x* 6, *g*(*x*) = 2 – *x*^{2}

Quotient = -*x*^{2} -2 and remainder -5*x* 10

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

**Answer**

(i) *t*^{2} – 3, 2*t*^{4} 3*t*^{3} – 2*t*^{2} – 9*t* – 12

*t*^{2} – 3 exactly divides 2*t*^{4} 3*t*^{3} – 2*t*^{2} – 9*t* – 12 leaving no remainder. Hence, it is a factor of 2*t*^{4} 3*t*^{3} – 2*t*^{2} – 9*t* – 12.

(ii) *x*^{2} 3*x* 1, 3*x*^{4} 5*x*^{3} – 7*x*^{2} 2*x* 2

*x*^{2} 3*x* 1 exactly divides 3*x*^{4} 5*x*^{3} – 7*x*^{2} 2*x* 2 leaving no remainder. Hence, it is factor of 3*x*^{4} 5*x*^{3} – 7*x*^{2} 2*x* 2.

(iii) *x*^{3} – 3*x* 1, *x*^{5} – 4*x*^{3} *x*^{2} 3*x* 1

*x*^{2} 3*x* 1 exactly divides 3*x*^{4} 5*x*^{3} – 7*x*^{2} 2*x* 2 leaving no remainder. Hence, it is factor of 3*x*^{4} 5*x*^{3} – 7*x*^{2} 2*x* 2.

(iii) *x*^{3} – 3*x* 1, *x*^{5} – 4*x*^{3} *x*^{2} 3*x* 1

*x*^{3} – 3*x* 1 didn't divides exactly *x*^{5} – 4*x*^{3} *x*^{2} 3*x* 1 and leaves 2 as remainder. Hence, it not a factor of *x*^{5} – 4*x*^{3} *x*^{2} 3*x* 1.

**3. Obtain all other zeroes of 3 x^{4} 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are √(5/3) and - √(5/3).**

Since the two zeroes are √(5/3) and - √(5/3).

We factorize *x*^{2} + 2*x +* 1

= (*x * + 1)^{2}

Therefore, its zero is given by *x* + 1 = 0

*x* = -1

As it has the term (*x +* 1)^{2} , therefore, there will be 2 zeroes at *x* = - 1.

Hence, the zeroes of the given polynomial are √(5/3) and - √(5/3), - 1 and - 1.**4. On dividing x^{3} - 3x^{2} x 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x 4, respectively. Find g(x).**

Here in the given question,

Dividend = *x*^{3} - 3*x*^{2} *x* 2

Quotient = *x* - 2

Remainder = -2*x* 4

Divisor = *g*(*x*)

We know that,

Dividend = Quotient × Divisor Remainder

⇒ *x*^{3} - 3*x*^{2} *x* 2 = (*x* - 2) × *g*(*x*) (-2*x* 4)⇒ *x*^{3} - 3*x*^{2} *x* 2 - (-2*x* 4) = (*x* - 2) × *g*(*x*)

⇒ *x*^{3} - 3*x*^{2} 3*x* - 2 = (*x* - 2) × *g*(*x*)

⇒ *g*(*x*) = (*x*^{3} - 3*x*^{2} 3*x* - 2)/(*x* - 2)

∴ *g*(*x*) = (*x*^{2} - *x* + 1)**5.Give examples of polynomial 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

(i) Let us assume the division of 6

Here,

Degree of

Checking for division algorithm,

Or, 6

Hence, division algorithm is satisfied.

(ii) Let us assume the division of *x*^{3} *x* by *x*^{2},

Here, *p*(*x*) = *x*^{3} *x**g*(*x*) = *x*^{2}*q*(*x*) = *x* and *r*(*x*) = *x*

Clearly, the degree of *q*(*x*) and *r*(*x*) is the same i.e., 1.

Checking for division algorithm,*p*(*x*) = *g*(*x*) × *q*(*x*) *r*(*x*)*x*^{3} *x* = (*x*^{2} ) × *x* *x**x*^{3} *x* = *x*^{3} *x*

Thus, the division algorithm is satisfied.

(iii) Let us assume the division of *x*^{3} 1 by *x*^{2}.

Here, *p*(*x*) = *x*^{3} 1

g(x) = x^{2}*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*)*x*^{3} 1 = (*x*^{2} ) × *x * 1*x*^{3} 1 = *x*^{3} 1

Thus, the division algorithm is satisfied.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!