NCERT Solutions for Class 9 Math Chapter 2 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among class 9 students for Math Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 9 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class 9 Math are prepared by experts and are 100% accurate.

#### Page No 32:

#### Question 1:

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) (ii) (iii)

(iv) (v)

#### Answer:

(i)

Yes, this expression is a polynomial in one variable* x*.

(ii)

Yes, this expression is a polynomial in one variable *y*.

(iii)

No. It can be observed that the exponent of variable *t* in term is , which is not a whole number. Therefore, this expression is not a polynomial.

(iv)

No. It can be observed that the exponent of variable *y* in termis −1, which is not a whole number. Therefore, this expression is not a polynomial.

(v)

No. It can be observed that this expression is a polynomial in 3 variables *x*, *y*, and *t*. Therefore, it is not a polynomial in one variable.

##### Video Solution for Polynomials (Page: 32 , Q.No.: 1)

NCERT Solution for Class 9 math - Polynomials 32 , Question 1

#### Page No 32:

#### Question 2:

Write the coefficients of in each of the following:

(i) (ii)

(iii) (iv)

#### Answer:

(i)

In the above expression, the coefficient of is 1.

(ii)

In the above expression, the coefficient of is −1.

(iii)

In the above expression, the coefficient of is.

(iv)

In the above expression, the coefficient of is 0.

##### Video Solution for Polynomials (Page: 32 , Q.No.: 2)

NCERT Solution for Class 9 math - Polynomials 32 , Question 2

#### Page No 32:

#### Question 3:

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

#### Answer:

Degree of a polynomial is the highest power of the variable in the polynomial.

Binomial has two terms in it. Therefore, binomial of degree 35 can be written as .

Monomial has only one term in it. Therefore, monomial of degree 100 can be written as *x*^{100}.

##### Video Solution for Polynomials (Page: 32 , Q.No.: 3)

NCERT Solution for Class 9 math - Polynomials 32 , Question 3

#### Page No 32:

#### Question 4:

Write the degree of each of the following polynomials:

(i) (ii)

(iii) (iv) 3

#### Answer:

Degree of a polynomial is the highest power of the variable in the polynomial.

(i)

This is a polynomial in variable *x* and the highest power of variable *x* is 3. Therefore, the degree of this polynomial is 3.

(ii)

This is a polynomial in variable *y* and the highest power of variable *y* is 2. Therefore, the degree of this polynomial is 2.

(iii)

This is a polynomial in variable *t* and the highest power of variable *t* is 1. Therefore, the degree of this polynomial is 1.

(iv) 3

This is a constant polynomial. Degree of a constant polynomial is always 0.

##### Video Solution for Polynomials (Page: 32 , Q.No.: 4)

NCERT Solution for Class 9 math - Polynomials 32 , Question 4

#### Page No 32:

#### Question 5:

Classify the following as linear, quadratic and cubic polynomial:

(i) (ii) (iii) (iv) (v)

(vi) (vii)

#### Answer:

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively.

(i) is a quadratic polynomial as its degree is 2.

(ii) is a cubic polynomial as its degree is 3.

(iii) is a quadratic polynomial as its degree is 2.

(iv) 1 + *x* is a linear polynomial as its degree is 1.

(v) is a linear polynomial as its degree is 1.

(vi) is a quadratic polynomial as its degree is 2.

(vii) is a cubic polynomial as its degree is 3.

##### Video Solution for Polynomials (Page: 32 , Q.No.: 5)

NCERT Solution for Class 9 math - Polynomials 32 , Question 5

#### Page No 34:

#### Question 1:

Find the value of the polynomial** ** at

(i) *x* = 0 (ii) *x* = −1 (iii) *x* = 2

#### Answer:

(i)

(ii)

(iii)

##### Video Solution for Polynomials (Page: 34 , Q.No.: 1)

NCERT Solution for Class 9 math - Polynomials 34 , Question 1

#### Page No 34:

#### Question 2:

Find *p*(0), *p*(1) and *p*(2) for each of the following polynomials:

(i) *p*(*y*) = *y*^{2} − *y* + 1 (ii) *p*(*t*) = 2 + *t *+ 2*t*^{2} − *t*^{3 }

(iii) *p*(*x*) = *x*^{3} (iv) *p*(*x*) = (*x* − 1) (*x* + 1)

#### Answer:

(i) *p*(*y*) = *y*^{2} − *y* + 1

*p*(0) = (0)^{2} − (0) + 1 = 1

*p*(1) = (1)^{2} − (1) + 1 = 1

*p*(2) = (2)^{2} − (2) + 1 = 3

(ii) *p*(*t*) = 2 + *t *+ 2*t*^{2} − *t*^{3}

*p*(0) = 2 + 0 + 2 (0)^{2} − (0)^{3 }= 2

*p*(1) = 2 + (1) + 2(1)^{2} − (1)^{3}

= 2 + 1 + 2 − 1 = 4

*p*(2) = 2 + 2 + 2(2)^{2} − (2)^{3}

= 2 + 2 + 8 − 8 = 4

(iii) *p*(*x*) = *x*^{3}

*p*(0) = (0)^{3} = 0

*p*(1) = (1)^{3} = 1

*p*(2) = (2)^{3} = 8

(iv) *p*(*x*) = (*x* − 1) (*x* + 1)

*p*(0) = (0 − 1) (0 + 1) = (− 1) (1) = − 1

*p*(1) = (1 − 1) (1 + 1) = 0 (2) = 0

*p*(2) = (2 − 1 ) (2 + 1) = 1(3) = 3

##### Video Solution for Polynomials (Page: 34 , Q.No.: 2)

NCERT Solution for Class 9 math - Polynomials 34 , Question 2

#### Page No 35:

#### Question 3:

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) (ii)

(iii) *p*(*x*) = *x*^{2} − 1, *x* = 1, − 1 (iv) *p*(*x*) = (*x* + 1) (*x* − 2), *x* = − 1, 2

(v) *p*(*x*) = *x*^{2 }, *x* = 0 (vi) *p*(*x*) =* lx *+ *m*

(vii) (viii)

#### Answer:

(i) If is a zero of given polynomial *p*(*x*) = 3*x* + 1, then should be 0.

Therefore, is a zero of the given polynomial.

(ii) If is a zero of polynomial *p*(*x*) = 5*x* − π , thenshould be 0.

Therefore, is not a zero of the given polynomial.

(iii) If *x* = 1 and *x* = −1 are zeroes of polynomial *p*(*x*) = *x*^{2} − 1, then *p*(1) and *p*(−1) should be 0.

Here, *p*(1) = (1)^{2} − 1 = 0, and

*p*(− 1) = (− 1)^{2} − 1 = 0

Hence, *x* = 1 and −1 are zeroes of the given polynomial.

(iv) If *x* = −1 and *x* = 2 are zeroes of polynomial *p*(*x*) = (*x* +1) (*x* − 2), then *p*(−1) and *p*(2)should be 0.

Here, *p*(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and

*p*(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0

Therefore, *x* = −1 and *x *= 2 are zeroes of the given polynomial.

(v) If *x* = 0 is a zero of polynomial *p*(*x*) = *x*^{2}, then *p*(0) should be zero.

Here, *p*(0) = (0)^{2 }= 0

Hence, *x* = 0 is a zero of the given polynomial.

(vi) If is a zero of polynomial *p*(*x*) = *lx* + *m*, then should be 0.

Here,

Therefore, is a zero of the given polynomial.

(vii) If and are zeroes of polynomial *p*(*x*) = 3*x*^{2} − 1, then

Hence, is a zero of the given polynomial. However, is not a zero of the given polynomial.

(viii) If is a zero of polynomial *p*(*x*) = 2*x* + 1, then should be 0.

Therefore, is not a zero of the given polynomial.

##### Video Solution for Polynomials (Page: 35 , Q.No.: 3)

NCERT Solution for Class 9 math - Polynomials 35 , Question 3

#### Page No 35:

#### Question 4:

Find the zero of the polynomial in each of the following cases:

(i) *p*(*x*) = *x* + 5 (ii) *p*(*x*) = *x* − 5 (iii) *p*(*x*) = 2*x* + 5

(iv) *p*(*x*) = 3*x* − 2 (v) *p*(*x*) = 3*x* (vi) *p*(*x*) = *ax*, *a *≠ 0

(vii) *p*(*x*) = *cx* + *d*, *c *≠ 0, *c*, are real numbers.

#### Answer:

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

(i) *p*(*x*) = *x* + 5

*p*(*x*) = 0

*x* + 5 = 0

*x* = − 5

Therefore, for *x* = −5, the value of the polynomial is 0 and hence, *x* = −5 is a zero of the given polynomial.

(ii) *p*(*x*) = *x* − 5

*p*(*x*) = 0

*x* − 5 = 0

*x* = 5

Therefore, for *x* = 5, the value of the polynomial is0 and hence, *x* = 5 is a zero of the given polynomial.

(iii) *p*(*x*) = 2*x* + 5

*p*(*x*) = 0

2*x* + 5 = 0

2*x* = − 5

Therefore, for, the value of the polynomial is 0 and hence, is a zero of the given polynomial.

(iv) *p*(*x*) = 3*x* − 2

*p*(*x*) = 0

3*x* − 2 = 0

Therefore, for, the value of the polynomial is 0 and hence, is a zero of the given polynomial.

(v) *p*(*x*) = 3*x*

*p*(*x*) = 0

3*x* = 0

*x* = 0

Therefore, for *x* = 0, the value of the polynomial is 0 and hence, *x* = 0 is a zero of the given polynomial.

(vi) *p*(*x*) = *ax*

*p*(*x*) = 0

*ax* = 0

*x* = 0

Therefore, for *x *= 0, the value of the polynomial is 0 and hence, *x* = 0 is a zero of the given polynomial.

(vii) *p*(*x*) = *cx* + *d*

*p*(*x*) = 0

*cx+ d* = 0

Therefore, for, the value of the polynomial is 0 and hence, * *is a zero of the given polynomial.

##### Video Solution for Polynomials (Page: 35 , Q.No.: 4)

NCERT Solution for Class 9 math - Polynomials 35 , Question 4

#### Page No 40:

#### Question 1:

Find the remainder when *x*^{3} + 3*x*^{2} + 3*x* + 1 is divided by

(i) *x* + 1 (ii) (iii) *x*

(iv) *x* + π (v) 5 + 2*x*

#### Answer:

(i) *x* + 1

By long division,

Therefore, the remainder is 0.

(ii)

By long division,

Therefore, the remainder is.

(iii) *x*

By long division,

Therefore, the remainder is 1.

(iv) *x* + π

By long division,

Therefore, the remainder is

(v) 5 + 2*x*

By long division,

Therefore, the remainder is

##### Video Solution for Polynomials (Page: 40 , Q.No.: 1)

NCERT Solution for Class 9 math - Polynomials 40 , Question 1

#### Page No 40:

#### Question 2:

Find the remainder when *x*^{3} − *ax*^{2} + 6*x* − *a *is divided by *x − a*.

#### Answer:

By long division,

Therefore, when *x*^{3} − *ax*^{2} + 6*x* − *a *is divided by *x − a*, the remainder obtained is 5*a*.

##### Video Solution for Polynomials (Page: 40 , Q.No.: 2)

NCERT Solution for Class 9 math - Polynomials 40 , Question 2

#### Page No 40:

#### Question 3:

Check whether 7 + 3*x* is a factor of 3*x*^{3} + 7*x*.

#### Answer:

Let us divide (3*x*^{3} + 7*x*) by (7 + 3*x*). If the remainder obtained is 0, then 7 + 3*x *will be a factor of 3*x*^{3} + 7*x*.

By long division,

As the remainder is not zero, therefore, 7 + 3*x* is not a factor of 3*x*^{3} + 7*x*.

##### Video Solution for Polynomials (Page: 40 , Q.No.: 3)

NCERT Solution for Class 9 math - Polynomials 40 , Question 3

#### Page No 43:

#### Question 1:

Determine which of the following polynomials has (*x* + 1) a factor:

(i) *x*^{3} + *x*^{2} + *x* + 1 (ii) *x*^{4} + *x*^{3} + *x*^{2} + *x* + 1

(iii) *x*^{4} + 3*x*^{3} + 3*x*^{2} + *x* + 1 (iv)

#### Answer:

(i) If (*x* + 1) is a factor of *p*(*x*) = *x*^{3} + *x*^{2} + *x* + 1, then *p* (−1) must be zero, otherwise (*x* + 1) is not a factor of *p*(*x*).

*p*(*x*) = *x*^{3} + *x*^{2} + *x* + 1

*p*(−1) = (−1)^{3} + (−1)^{2} + (−1) + 1

= − 1 + 1 − 1 + 1 = 0

Hence, *x* + 1 is a factor of this polynomial.

(ii) If (*x* + 1) is a factor of *p*(*x*) = *x*^{4} + *x*^{3} + *x*^{2} + *x* + 1, then *p* (−1) must be zero, otherwise (*x* + 1) is not a factor of *p*(*x*).

*p*(*x*) = *x*^{4} + *x*^{3} + *x*^{2} + *x* + 1

*p*(−1) = (−1)^{4} + (−1)^{3} + (−1)^{2} + (−1) + 1

= 1 − 1 + 1 −1 + 1 = 1

As *p*(− 1) ≠ 0,

Therefore, *x* + 1 is not a factor of this polynomial.

(iii) If (*x* + 1) is a factor of polynomial *p*(*x*) = *x*^{4} + 3*x*^{3} + 3*x*^{2} + *x* + 1, then *p*(−1) must be 0, otherwise (*x* + 1) is not a factor of this polynomial.

*p*(−1) = (−1)^{4} + 3(−1)^{3} + 3(−1)^{2} + (−1) + 1

= 1 − 3 + 3 − 1 + 1 = 1

As *p*(−1) ≠ 0,

Therefore, *x* + 1 is not a factor of this polynomial.

(iv) If(*x* + 1) is a factor of polynomial *p*(*x*) = , then *p*(−1) must be 0, otherwise (*x* + 1) is not a factor of this polynomial.

As *p*(−1) ≠ 0,

Therefore, (*x* + 1) is not a factor of this polynomial.

##### Video Solution for Polynomials (Page: 43 , Q.No.: 1)

NCERT Solution for Class 9 math - Polynomials 43 , Question 1

#### Page No 43:

#### Question 2:

Use the Factor Theorem to determine whether *g*(*x*) is a factor of *p*(*x*) in each of the following cases:

(i) *p*(*x*) = 2*x*^{3} + *x*^{2} − 2*x* − 1, *g*(*x*) = *x* + 1

(ii) *p*(*x*) = *x*^{3} + 3*x*^{2} + 3*x* + 1, *g*(*x*) = *x* + 2

(iii) *p*(*x*) = *x*^{3} − 4 *x*^{2} + *x* + 6, *g*(*x*) = *x* − 3

#### Answer:

(i) If *g*(*x*) = *x* + 1 is a factor of the given polynomial *p*(*x*), then *p*(−1) must be zero.

*p*(*x*) = 2*x*^{3} + *x*^{2} − 2*x* − 1

*p*(−1) = 2(−1)^{3} + (−1)^{2} − 2(−1) − 1

= 2(−1) + 1 + 2 − 1 = 0

Hence, *g*(*x*) = *x* + 1 is a factor of the given polynomial.

(ii) If *g*(*x*) = *x* + 2 is a factor of the given polynomial *p*(*x*), then *p*(−2) must

be 0.

*p*(*x*) = *x*^{3} +3*x*^{2} + 3*x* + 1

*p*(−2) = (−2)^{3} + 3(−2)^{2} + 3(−2) + 1

= − 8 + 12 − 6 + 1

= −1

As *p*(−2) ≠ 0,

Hence, *g*(*x*) = *x* + 2 is not a factor of the given polynomial.

(iii) If *g*(*x*) = *x* − 3 is a factor of the given polynomial *p*(*x*), then *p*(3) must

be 0.

*p*(*x*) = *x*^{3} − 4 *x*^{2} + *x* + 6

*p*(3) = (3)^{3} − 4(3)^{2} + 3 + 6

= 27 − 36 + 9 = 0

Hence, *g*(*x*) = *x* − 3 is a factor of the given polynomial.

##### Video Solution for Polynomials (Page: 43 , Q.No.: 2)

NCERT Solution for Class 9 math - Polynomials 43 , Question 2

#### Page No 44:

#### Question 3:

Find the value of *k*, if *x* − 1 is a factor of *p*(*x*) in each of the following cases:

(i) *p*(*x*) = *x*^{2} + *x* + *k *(ii)* *

(iii) * *(iv) *p*(*x*) = *kx*^{2} − 3*x* + *k*

#### Answer:

If *x* − 1 is a factor of polynomial *p*(*x*), then *p*(1) must be 0.

(i) *p*(*x*) = *x*^{2} + *x* + *k*

*p*(1) = 0

⇒ (1)^{2} + 1 + *k* = 0

⇒ 2 + *k* = 0

⇒ *k* = −2

Therefore, the value of* k *is −2*.*

(ii)* *

*p*(1) = 0

(iii)* *

*p*(1) = 0

(iv) *p*(*x*) = *kx*^{2} − 3*x* + *k*

⇒ *p*(1) = 0

⇒ *k*(1)^{2} − 3(1) + *k* = 0

⇒ *k* − 3 + *k* = 0

⇒ 2*k *− 3 = 0

Therefore, the value of* k *is.

##### Video Solution for Polynomials (Page: 44 , Q.No.: 3)

NCERT Solution for Class 9 math - Polynomials 44 , Question 3

#### Page No 44:

#### Question 4:

Factorise:

(i) 12*x*^{2} − 7*x* + 1 (ii) 2*x*^{2} + 7*x* + 3

(iii) 6*x*^{2} + 5*x* − 6 (iv) 3*x*^{2} − *x* − 4

#### Answer:

(i) 12*x*^{2} − 7*x* + 1

We can find two numbers such that *pq* = 12 × 1 = 12 and *p *+ *q* = −7. They are *p* = −4 and *q *= −3.

Here, 12*x*^{2} − 7*x* + 1 = 12*x*^{2} − 4*x *− 3*x* + 1

= 4*x *(3*x *− 1) − 1 (3*x *− 1)

= (3*x *− 1) (4*x *− 1)

(ii) 2*x*^{2} + 7*x* + 3

We can find two numbers such that *pq* = 2 × 3 = 6 and *p *+ *q* = 7.

They are *p* = 6 and *q *= 1.

Here, 2*x*^{2} + 7*x* + 3 = 2*x*^{2} + 6*x* + *x* + 3

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

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

(iii) 6*x*^{2} + 5*x* − 6

We can find two numbers such that *pq* = −36 and *p *+ *q* = 5.

They are *p* = 9 and *q *= −4.

Here,

6*x*^{2} + 5*x* − 6 = 6*x*^{2} + 9*x* − 4*x* − 6

= 3*x *(2*x *+ 3) − 2 (2*x *+ 3)

= (2*x* + 3) (3*x *− 2)

(iv) 3*x*^{2} − *x* − 4

We can find two numbers such that *pq* = 3 × (− 4) = −12

and *p *+ *q* = −1.

They are *p* = −4 and *q *= 3.

Here,

3*x*^{2} − *x* − 4 = 3*x*^{2} − 4*x* + 3*x* − 4

= *x *(3*x *− 4) + 1 (3*x *− 4)

= (3*x* − 4) (*x *+ 1)

##### Video Solution for Polynomials (Page: 44 , Q.No.: 4)

NCERT Solution for Class 9 math - Polynomials 44 , Question 4

#### Page No 44:

#### Question 5:

Factorise:

(i) *x*^{3} − 2*x*^{2} − *x* + 2 (ii) *x*^{3} + 3*x*^{2} −9*x *− 5

(iii) *x*^{3} + 13*x*^{2} + 32*x* + 20 (iv) 2*y*^{3} + *y*^{2} − 2*y* − 1

#### Answer:

(i) Let *p*(*x*) = *x*^{3} − 2*x*^{2} − *x* + 2

All the factors of 2 have to be considered. These are ± 1, ± 2.

By trial method,

*p*(−1) = (−1)^{3} − 2(−1)^{2} − (−1) + 2

= −1 − 2 + 1 + 2 = 0

Therefore, (*x* +1 ) is factor of polynomial *p*(*x*).

Let us find the quotient on dividing *x*^{3} − 2*x*^{2} − *x* + 2 by *x* + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ *x*^{3} − 2*x*^{2} − *x* + 2 = (*x* + 1) (*x*^{2} − 3*x* + 2) + 0

= (*x* + 1) [*x*^{2} − 2*x* − *x* + 2]

= (*x* + 1) [*x* (*x* − 2) − 1 (*x* − 2)]

= (*x* + 1) (*x* − 1) (*x* − 2)

= (*x* − 2) (*x* − 1) (*x* + 1)

(ii) Let *p*(*x*) = *x*^{3} − 3*x*^{2} − 9*x *− 5

All the factors of 5 have to be considered. These are ±1, ± 5.

By trial method,

*p*(−1) = (−1)^{3} − 3(−1)^{2} − 9(−1) − 5

= − 1 − 3 + 9 − 5 = 0

Therefore, *x* + 1 is a factor of this polynomial.

Let us find the quotient on dividing *x*^{3} + 3*x*^{2} − 9*x *− 5 by *x* + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ *x*^{3} − 3*x*^{2} − 9*x *− 5 = (*x *+ 1) (*x*^{2} − 4*x* − 5) + 0

= (*x *+ 1) (*x*^{2} − 5*x* + *x* − 5)

= *(x* + 1) [(*x *(*x* − 5) +1 (*x* − 5)]

= (*x* + 1) (*x* − 5) (*x* + 1)

= (*x* − 5) (*x* + 1) (*x* + 1)

(iii) Let *p*(*x*) = *x*^{3} + 13*x*^{2} + 32*x* + 20

All the factors of 20 have to be considered. Some of them are ±1,

± 2, ± 4, ± 5 ……

By trial method,

*p*(−1) = (−1)^{3} + 13(−1)^{2} + 32(−1) + 20

= − 1 +13 − 32 + 20

= 33 − 33 = 0

As *p*(−1) is zero, therefore, *x *+ 1 is a factor of this polynomial *p*(*x*).

Let us find the quotient on dividing *x*^{3} + 13*x*^{2} + 32*x* + 20 by (*x* + 1).

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

*x*^{3} + 13*x*^{2} + 32*x* + 20 = (*x *+ 1) (*x*^{2} + 12*x* + 20) + 0

= (*x *+ 1) (*x*^{2} + 10*x* + 2*x *+ 20)

= (*x* + 1) [*x* (*x *+ 10) + 2 (*x *+ 10)]

= (*x* + 1) (*x *+ 10) (*x *+ 2)

= (*x* + 1) (*x* + 2) (*x* + 10)

(iv) Let *p*(*y*) = 2*y*^{3} + *y*^{2} − 2*y* − 1

By trial method,

*p*(1) = 2 ( 1)^{3} + (1)^{2} − 2( 1) − 1

= 2 + 1 − 2 − 1= 0

Therefore, *y* − 1 is a factor of this polynomial.

Let us find the quotient on dividing 2*y*^{3} + *y*^{2} − 2*y* − 1 by *y* − 1.

*p*(*y*) = 2*y*^{3} + *y*^{2} − 2*y* − 1

= (*y *− 1) (2*y*^{2} +3y + 1)

= (*y *− 1) (2*y*^{2} +2y + *y* +1)

= (*y *− 1) [2*y *(*y *+ 1) + 1 (*y *+ 1)]

= (*y *− 1) (*y *+ 1) (2*y *+ 1)

##### Video Solution for Polynomials (Page: 44 , Q.No.: 5)

NCERT Solution for Class 9 math - Polynomials 44 , Question 5

#### Page No 48:

#### Question 1:

Use suitable identities to find the following products:

(i) (ii)

(iii) (iv)

(v)

#### Answer:

(i) By using the identity ,

(ii) By using the identity ,

(iii)

By using the identity ,

(iv) By using the identity ,

(v) By using the identity ,

##### Video Solution for Polynomials (Page: 48 , Q.No.: 1)

NCERT Solution for Class 9 math - Polynomials 48 , Question 1

#### Page No 48:

#### Question 2:

Evaluate the following products without multiplying directly:

(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96

#### Answer:

(i) 103 × 107 = (100 + 3) (100 + 7)

= (100)^{2} + (3 + 7) 100 + (3) (7)

[By using the identity, where

*x* = 100, *a* = 3, and *b* = 7]

= 10000 + 1000 + 21

= 11021

(ii) 95 × 96 = (100 − 5) (100 − 4)

= (100)^{2} + (− 5 − 4) 100 + (− 5) (− 4)

[By using the identity, where

*x* = 100, *a* = −5, and *b* = −4]

= 10000 − 900 + 20

= 9120

(iii) 104 × 96 = (100 + 4) (100 − 4)

= (100)^{2} − (4)^{2}

= 10000 − 16

= 9984

##### Video Solution for Polynomials (Page: 48 , Q.No.: 2)

NCERT Solution for Class 9 math - Polynomials 48 , Question 2

#### Page No 48:

#### Question 3:

Factorise the following using appropriate identities:

(i) 9*x*^{2} + 6*xy* + *y*^{2}

(ii)

(iii)

#### Answer:

(i)

(ii)

(iii)

##### Video Solution for Polynomials (Page: 48 , Q.No.: 3)

NCERT Solution for Class 9 math - Polynomials 48 , Question 3

#### Page No 49:

#### Question 4:

Expand each of the following, using suitable identities:

(i) (ii)

(iii) (iv)

(v) (vi)

#### Answer:

It is known that,

(i)

(ii)

(iii)

(iv)

(v)

(vi)

[[VIDEO:16294>]]

#### Page No 49:

#### Question 5:

Factorise:

(i)

(ii)

#### Answer:

It is known that,

(i)

(ii)

##### Video Solution for Polynomials (Page: 49 , Q.No.: 5)

NCERT Solution for Class 9 math - Polynomials 49 , Question 5

#### Page No 49:

#### Question 6:

Write the following cubes in expanded form:

(i) (ii)

(iii) (iv)

#### Answer:

It is known that,

(i)

(ii)

(iii)

(vi)

##### Video Solution for Polynomials (Page: 49 , Q.No.: 6)

NCERT Solution for Class 9 math - Polynomials 49 , Question 6

#### Page No 49:

#### Question 7:

Evaluate the following using suitable identities:

(i) (99)^{3} (ii) (102)^{3} (iii) (998)^{3}

#### Answer:

It is known that,

(i) (99)^{3 }= (100 − 1)^{3}

= (100)^{3} − (1)^{3} − 3(100) (1) (100 − 1)

= 1000000 − 1 − 300(99)

= 1000000 − 1 − 29700

= 970299

(ii) (102)^{3} = (100 + 2)^{3}

= (100)^{3} + (2)^{3} + 3(100) (2) (100 + 2)

= 1000000 + 8 + 600 (102)

= 1000000 + 8 + 61200

= 1061208

(iii) (998)^{3}= (1000 − 2)^{3}

= (1000)^{3} − (2)^{3} − 3(1000) (2) (1000 − 2)

= 1000000000 − 8 − 6000(998)

= 1000000000 − 8 − 5988000

= 1000000000 − 5988008

= 994011992

##### Video Solution for Polynomials (Page: 49 , Q.No.: 7)

NCERT Solution for Class 9 math - Polynomials 49 , Question 7

#### Page No 49:

#### Question 8:

Factorise each of the following:

(i) (ii)

(iii) (iv)

(v)

#### Answer:

It is known that,

(i)

(ii)

(iii)

(iv)

(v)

##### Video Solution for Polynomials (Page: 49 , Q.No.: 8)

NCERT Solution for Class 9 math - Polynomials 49 , Question 8

#### Page No 49:

#### Question 9:

Verify:

(i)

(ii)

#### Answer:

(i) It is known that,

(ii) It is known that,

##### Video Solution for Polynomials (Page: 49 , Q.No.: 9)

NCERT Solution for Class 9 math - Polynomials 49 , Question 9

#### Page No 49:

#### Question 10:

Factorise each of the following:

(i)

(ii)

[**Hint: **See question 9.]

#### Answer:

(i)

(ii)

##### Video Solution for Polynomials (Page: 49 , Q.No.: 10)

NCERT Solution for Class 9 math - Polynomials 49 , Question 10

#### Page No 49:

#### Question 11:

Factorise:

#### Answer:

It is known that,

##### Video Solution for Polynomials (Page: 49 , Q.No.: 11)

NCERT Solution for Class 9 math - Polynomials 49 , Question 11

#### Page No 49:

#### Question 12:

Verify that

#### Answer:

It is known that,

##### Video Solution for Polynomials (Page: 49 , Q.No.: 12)

NCERT Solution for Class 9 math - Polynomials 49 , Question 12

#### Page No 49:

#### Question 13:

If *x + y + z* = 0, show that ${x}^{3}+{y}^{3}+{z}^{3}=3xyz$.

#### Answer:

It is known that,

Put *x + y + z* = 0,

##### Video Solution for Polynomials (Page: 49 , Q.No.: 13)

NCERT Solution for Class 9 math - Polynomials 49 , Question 13

#### Page No 49:

#### Question 14:

Without actually calculating the cubes, find the value of each of the following:

(i)

(ii)

#### Answer:

(i)

Let *x* = −12, *y* = 7, and *z* = 5

It can be observed that,

*x* + *y* + *z* = − 12 + 7 + 5 = 0

It is known that if *x* + *y* + *z* = 0, then

∴

= −1260

(ii)

Let *x* = 28, *y* = −15, and *z* = −13

It can be observed that,

*x* + *y* + *z* = 28 + (−15) + (−13) = 28 − 28 = 0

It is known that if *x* + *y* + *z* = 0, then

##### Video Solution for Polynomials (Page: 49 , Q.No.: 14)

NCERT Solution for Class 9 math - Polynomials 49 , Question 14

#### Page No 49:

#### Question 15:

Give possible expressions for the length and breadth of each of thefollowing rectangles, in which their areas are given:

#### Answer:

Area = Length × Breadth

The expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth.

(i)

Therefore, possible length = 5*a* − 3

And, possible breadth = 5*a* − 4

(ii)

Therefore, possible length = 5*y* + 4

And, possible breadth = 7*y* − 3

##### Video Solution for Polynomials (Page: 49 , Q.No.: 15)

NCERT Solution for Class 9 math - Polynomials 49 , Question 15

#### Page No 50:

#### Question 16:

What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

#### Answer:

Volume of cuboid = Length × Breadth × Height

The expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height.

(i)

One of the possible solutions is as follows.

Length
= 3, Breadth = *x*, Height = *x* − 4

(ii)

One of the possible solutions is as follows.

Length
= 4*k*, Breadth = 3*y* + 5, Height = *y* − 1

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