# Polynomials : Exercise 2.4 (Mathematics NCERT Class 9th)

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

Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) **

Â Â Â Â Â Â Â Â Â Â Â Â Â **(ii) **

Â Â Â Â Â Â Â Â Â Â Â Â Â **(iii) **

Â Â Â Â Â Â Â Â Â Â Â Â Â **(iv) Â **

** Sol.**

**(i)** In order to prove that x + 1 is a factor of , it is sufficient to show that

Now,

Hence, (x + 1) is a factor of Â

**(ii)** In order to prove that (x + 1) is a factor of Â , it is sufficient to show that p (â€“1) = 0 .

Now, Â

Therefore (x + 1) is not a factor of Â

**(iii)** In order to prove that (x + 1) is a factor of , it is sufficient to show that p(â€“1) = 0.

Now,

Therefore (x + 1) is not a factor of Â .

**(iv)** In order to prove that (x + 1) is a factor of Â , Â it is sufficient to show that p(-1)= 0

Therefore Â (x + 1) is not a factor of Â

**Q.2Â Â Â Â Â Use the factor theorem to determine whether g(x) is a factor of p (x) in each of the following cases :Â Â Â Â Â Â Â Â Â Â Â **

**(i)**

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â

**(ii)**

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â

**(iii)**

**Sol.****(i)** In order to prove that g(x) = x + 1 is a factor of , it is sufficient to show that p(â€“1) = 0.

Now,

Therefore g(x) is a factor of p(x).

**(ii)** In order to prove that g(x) = x + 2 is a factor of Â , it is sufficient to show that p(â€“2) = 0

Now,

Therefore g(x) is not a factor of p(x) .

**(iii)** In order to prove that g(x) = x â€“ 3 is a factor of . It is sufficient to show that p(+3) = 0

Now, Â

36 - 36 = 0

Therefore g (x) is Â a factor of p(x).

**Q.3Â Â Â Â Â Find the value of k, if x -1 is a factor of p (x) in each of the following cases :Â Â Â Â Â Â Â Â Â Â Â **

**(i)**

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â

**(ii)**

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â

**(iii)**

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â

**(iv)**

**Sol**

**.****(i)** If (x -1) is a factor of , then

Hence, k = -2

**(ii)** If (x - 1) is a factor of Â , then

**(iii)** If (x -1) is a factor of Â , Â then

**(iv)** If (x -1) is a factor of , then

**Q.4Â Â Â Â Â Factorise :**

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii) **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â

**(iii) Â Â Â Â Â Â Â Â Â Â Â Â Â**

Â Â Â Â Â Â Â Â (iv)

Â Â Â Â Â Â Â Â (iv)

**Sol.****(i)** Here p + q = coeff. of x = - 7

pq = coeff. of constant term

= 12 Ã— 1 = 12

Therefore p + q = â€“ 7 = â€“ 4 â€“ 3

and pq = 12 = (â€“ 4)(â€“3)

Therefore

** (ii)** Here p + q = coeff. of x = 7

pq = coeff. of constant term Â

= 2 Ã— 3 = 6

Therefore p + q = 7 = 1 + 6

and pq = 6 = 1 Ã— 6

Therefore

** (iii)** Here p + q = coeff. of x = 5

pq = coeff. of constant term

= 6 Ã— (â€“ 6)Â = â€“ 36

Therefore p + q = 5 = 9 + (â€“ 4)

and pq = â€“ 36 = 9 Ã— (â€“ 4)

Therefore

** (iv)** Here p + q = coeff. of x = â€“1

pq = coeff. of constant term

= 3 Ã— (â€“ 4) = â€“ 12

Therefore p + q = â€“1 = 3 + (â€“ 4)

and pq = â€“ 12 = 3 Ã— (â€“4)

Therefore

**Q.5Â Â Â Â Â Factorise :**

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii) Â Â Â Â Â Â Â Â Â Â Â **

**(iii) Â Â Â Â Â Â Â Â**

Â Â Â Â Â Â Â Â Â Â Â (iv)

Â Â Â Â Â Â Â Â Â Â Â (iv)

**Sol.Â****(i)** Let

The constant term in f(x) is + 2 and factors of + 2 areÂ

Putting x = 1 in f(x),we haveÂ

Therefore (xâ€“1) is a factor of f(x) .Â

putting x = â€“ 1 in f(x) , we haveÂ

Â

Â

Therefore x + 1 is a factor of f (x).Â

Putting x = 2 in f(x), we haveÂ

Therefore (x + 2) is a factor of f(x)

Putting x = â€“ 2 in f(x), we have

Therefore x - 2 is not a factor of f (x).

Therefore The factors of f(x) are (x â€“ 1), (x + 1) and (x â€“ 2).

Let f(x) = k (x â€“ 1) (x + 1) (x â€“ 2)

Putting x = 0 on both sides, we have

2 = k (â€“1) (1) (â€“2) k = 1

Therefore

**(ii)** Let

We shall look for all factors of â€“ 5. Therefore are

By trial, we find p (â€“1) = â€“ 1 â€“ 3 + 9 â€“ 5 = 0 . So (x + 1) is a factor of p(x) .

Now, divide p (x) by (x + 1)

Therefore

Therefore

** (iii)** Let

We shall look for all factors of + 20, these are

By trial , we findÂ

p(â€“ 2) = â€“ 8 + 52 â€“ 64 + 20 = 0Â

Therefore (x + 2) is a factor of p(x)Â

Now, divide p (x) by x + 2

Therefore Â

** (iv)** Let

By trial we find p(1) = 2 + 1 â€“ 2 â€“ 1 = 0Â

So, (y â€“ 1) is a factor of p(y)

Now, divide p(y) by y â€“ 1

Therefore