NCERT solution class 9 chapter 2 Polynomials exercise 2.4 mathematics

EXERCISE 2.4

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Q 1, Ex 2.4 - Polynomials - Chapter 2 - Maths Class 9th - NCERT

Q 1 (i and ii), Ex 2.4, Page No 43, Polynomials, Class 9th Maths Solutions

Q 1 (iii and iv), Ex 2.4, Page No 43, Polynomials, Class 9th Maths

Page No 40:

Question 1:

Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) $x-\frac{1}{2}$
(iii) x
(iv) x + π
(v) 5 + 2x

Answer:

(i) x + 1
By long division,

Therefore, the remainder is 0.

(ii) $x-\frac{1}{2}$
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 $-\pi^{3}+3 \pi^{2}-3 \pi+1$

(v) 5 + 2x
By long division,

Therefore, the remainder is $-\frac{27}{8}$

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Question 2:

Find the remainder when x³ − ax² + 6x − a is divided by x − a.

Answer:

By long division,

Therefore, when x³ − ax² + 6x − a is divided by x − a, the remainder obtained is 5a.

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Q 3, Ex. 2.4, Page No 44, Polynomials, Maths Class 9th

Question 3:

Check whether 7 + 3x is a factor of 3x³ + 7x.

Answer:

Let us divide (3x³ + 7x) by (7 + 3x). If the remainder obtained is 0, then 7 + 3x will be a factor of
3x³ + 7x.

By long division,

As the remainder is not zero, therefore, 7 + 3x is not a factor of 3x³ + 7x.

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Q 4(i) and (ii), Ex 2.4, Page No 44, Polynomials, Class 9th Maths

Q 4 (iii & iv), Ex 2.4, Page No 44, Polynomials, Class 9th NCERT Mathematics

Page No 49:

Question 4:

Factorise:
(i) 12x² − 7x + 1 (ii) 2x² + 7x + 3
(iii) 6x² + 5x − 6 (iv) 3x² − x − 4

Answer 4:

(i) 12x² − 7x + 1
We can find two numbers such that pq = 12 × 1 = 12 and p + q = −7. They are p = −4 and q = −3.
Here, 12x² − 7x + 1 = 12x² − 4x − 3x + 1
= 4x (3x − 1) − 1 (3x − 1)
= (3x − 1) (4x − 1)

(ii) 2x² + 7x + 3
We can find two numbers such that pq = 2 × 3 = 6 and p + q = 7.
They are p = 6 and q = 1.
Here, 2x² + 7x + 3 = 2x² + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x+ 1)

(iii) 6x² + 5x − 6
We can find two numbers such that pq = −36 and p + q = 5.
They are p = 9 and q = −4.
Here,
6x² + 5x − 6 = 6x² + 9x − 4x − 6
= 3x (2x + 3) − 2 (2x + 3)
= (2x + 3) (3x − 2)
(iv) 3x² − 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,
3x² − x − 4 = 3x² − 4x + 3x − 4
= x (3x − 4) + 1 (3x − 4)
= (3x − 4) (x + 1)

Q 5(i), Ex 2.4, Page No 44, Polynomials, Class 9th Maths

Q 5(ii), Ex 2.4, Page No 44, Polynomials, NCERT Solutions Class 9th

Q 5(iii), Ex 2.4, Page No 44, Polynomials, NCERT Solutions Class 9th

Question 5:

Factorise:
(i) x³ − 2x² − x + 2 (ii) x³ + 3x² −9x − 5
(iii) x³ + 13x² + 32x + 20 (iv) 2y³ + y² − 2y − 1

Answer 5:

(i) Let p(x) = x³ − 2x² − 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 x³ − 2x² − x + 2 by x + 1.
By long division,

It is known that,
Dividend = Divisor × Quotient + Remainder
∴ x³ − 2x² − x + 2 = (x + 1) (x² − 3x + 2) + 0
= (x + 1) [x² − 2x − 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) = x³ − 3x² − 9x − 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 x³ + 3x² − 9x − 5 by x + 1.
By long division,

It is known that,
Dividend = Divisor × Quotient + Remainder
∴ x³ − 3x² − 9x − 5 = (x + 1) (x² − 4x − 5) + 0
= (x + 1) (x² − 5x + 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) = x³ + 13x² + 32x + 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 x³ + 13x² + 32x + 20 by (x + 1).
By long division,

It is known that,
Dividend = Divisor × Quotient + Remainder
x³ + 13x² + 32x + 20 = (x + 1) (x² + 12x + 20) + 0
= (x + 1) (x² + 10x + 2x + 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) = 2y³ + y² − 2y − 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 2y³ + y² − 2y − 1 by y ­− 1.

p(y) = 2y³ + y² − 2y − 1
= (y − 1) (2y² +3y + 1)
= (y − 1) (2y² +2y + y +1)
= (y − 1) [2y (y + 1) + 1 (y + 1)]
= (y − 1) (y + 1) (2y + 1)