Polynomials: 9 Math


mathematics Class Nine

NCERT Exercise 2.3: 9th math

NCERT Exercise 2.3 Polynomials class ninth math Question (1) Find the remainder when x3 + 3x2 + 3x +1 is divided by

(i) x + 1

Answer

Given, p(x) = x3 + 3x2 + 3x +1

Is to be divided by x + 1

Let, x + 1 = 0

Thus, x = –1

Thus, zeros of x + 1 = –1

Thus, p(–1) = (–1)3 + 3(–1)2 + 3(–1) + 1

= –1 + 3×1 –3 + 1

= 0

Thus, Remainder = 0 Answer

Alternate Method

9 math polynomials solution of ncert exercise 2.3 question 1-i

Thus, Remainder = 0

(ii) `x-1/2`

Answer

Given, p(x) = x3 + 3x2 + 3x +1

Is to be divided by `x-1/2`

Let, `x-1/2=0`

Thus, x = 1/2

Thus, zeros of x – 1/2 = 1/2

Thus, `p(1/2)`

9 math polynomials solution of ncert exercise 2.3 question 1-iia

Thus, remainder = 27/8 Answer

Alternate method

9 math polynomials solution of ncert exercise 2.3 question 1-ii

Thus, remainder = 27/8 Answer

(iii) `x`

Answer

Given, p(x) = x3 + 3x2 + 3x +1

Is to be divided by `x`

Let x = 0

Thus, zero of x = 0

Now, according to remainder theorem

p(0) = 03 + 3(0)2 + 3 × 0 + 1

= 0 + 0 + 0 + 1

⇒ p(0) = 1

Thus, remainder = 1 Answer

Alternate method

9 math polynomials solution of ncert exercise 2.3 question 1-iii

(iv) `x+pi`

Answer

Given, p(x) = x3 + 3x2 + 3x +1

Is to be divided by `x+pi`

Let, `x+pi=0`

Thus, `x=-pi`

Thus, zeros of `x+pi=-pi`

Now, according to remainder theorem `p(pi)`

9 math polynomials solution of ncert exercise 2.3 question 1-iva

Thus, remainder `=-pi^2+3pi^2-3pi+1` Answer

Alternate method

9 math polynomials solution of ncert exercise 2.3 question 1-iv

Thus, remainder `=-pi^2+3pi^2-3pi+1` Answer

(v) 5 + 2 x

Answer

Given, p(x) = x3 + 3x2 + 3x +1

Is to be divided by 5 + 2 x

Let, 5 + 2 x = 0

⇒ 2 x = –5

⇒ x = –5/2

Thus, zeros of 5 + 2x = –5/2

Thus, `p(-5/2)`

9 math polynomials solution of ncert exercise 2.3 question 1-va

Thus, remainder `=-27/8` Answer

Alternate Method

9 math polynomials solution of ncert exercise 2.3 question 1-v

Thus, remainder `=-27/8` Answer

NCERT Exercise 2.3 Polynomials class ninth math Question (2) Find the remainder when x3 – ax2 + 6x – a is divided by x – a.

Answer

Given, p(x) = x3 – ax2 + 6x – a

Is to be divided by x – a

Let, x – a = 0

Thus, x = a

Thus, according to Remainder Theorem,

P(a) = a3 – a × a2 + 6 a – a

= a3 – a3 + 6a – a

= 6a – a

= 5a

Thus, Remainder = 5a Answer

Alternate method

9 math polynomials solution of ncert exercise 2.3 question 2

Thus, Remainder = 5a Answer

NCERT Exercise 2.3 Polynomials class ninth math Question (3) Check whether 7 + 3x is a factor of 3x3 + 7x

Answer

Given, 3x3 + 7x

Then whether 7 + 3x is a factor of given polynomial or not?

Let 7 + 3x = 0

Thus, 3x = –7

`=>x = -7/3`

Thus, zeros of 7 + 3x is `-7/3`.

Now, according to remainder theorem

9 math polynomials solution of ncert exercise 2.3 question 3a

Thus, remainder `=-490/9`

Since, here remainder is not equal to zero, thus, 7 + 3x is not a factor of 3x3 + 7x.

Thus, Answer is No.

Alternate Method

9 math polynomials solution of ncert exercise 2.3 question 3

Thus, remainder `=-490/9`

Since, here remainder is not equal to zero, thus, 7 + 3x is not a factor of 3x3 + 7x.

Thus, Answer is No.

9 math polynomials solution of ncert exercise 2.3

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