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**Q.1. Determine which of the following polynomials has (x + 1) a factor:(i) x ^{3}+x^{2}+x+1**

Let p(x) = x

The zero of x+1 is -1. [x+1 = 0 means x = -1]

p(−1) = (−1)

= −1+1−1+1

= 0

∴ By factor theorem, x+1 is a factor of x

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

Solution:

Let p(x)= x

The zero of x+1 is -1. . [x+1= 0 means x = -1]

p(−1) = (−1)4+(−1)

= 1−1+1−1+1

= 1 ≠ 0

∴ By factor theorem, x+1 is not a factor of x^{4}+ x^{3}+ x^{2}+x+1

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

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

The zero of x+1 is -1.

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

=1−3+3−1+1

=1 ≠ 0

∴ By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1

**(iv) x ^{3} – x^{2}– (2+√2)x +√2**

Let p(x) = x

The zero of x+1 is -1.

p(−1) = (-1)

= 2√2 ≠ 0

∴ By factor theorem, x+1 is not a factor of x

**Q.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) = 2x3+x2–2x–1, g(x) = x+1****Solution:**

p(x) = 2x^{3}+x^{2}–2x–1, g(x) = x+1

g(x) = 0

⇒ x+1 = 0

⇒ x = −1

∴ Zero of g(x) is -1.

Now,

p(−1) = 2(−1)^{3}+(−1)^{2}–2(−1)–1

= −2+1+2−1

= 0

∴ By factor theorem, g(x) is a factor of p(x).

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

p(x) = x

g(x) = 0

⇒ x+2 = 0

⇒ x = −2

∴ Zero of g(x) is -2.

Now,

p(−2) = (−2)

= −8+12−6+1

= −1 ≠ 0

∴ By factor theorem, g(x) is not a factor of p(x).

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

**Solution:**

p(x) = x^{3}–4x^{2}+x+6, g(x) = x -3

g(x) = 0

⇒ x−3 = 0

⇒ x = 3

∴ Zero of g(x) is 3.

Now,

p(3) = (3)^{3}−4(3)^{2}+(3)+6

= 27−36+3+6

= 0

∴ By factor theorem, g(x) is a factor of p(x).

**Q.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**

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

⇒ (1)^{2}+(1)+k = 0

⇒ 1+1+k = 0

⇒ 2+k = 0

⇒ k = −2

**(ii) p(x) = 2x ^{2}+kx+√2**

If x-1 is a factor of p(x), then p(1)=0

⇒ 2(1)

⇒ 2+k+√2 = 0

⇒ k = −(2+√2)

**(iii) p(x) = kx ^{2}–√2x+1**

If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

⇒ k(1)^{2}-√2(1)+1=0

⇒ k = √2-1

**(iv) p(x)=kx ^{2}–3x+k**

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

⇒ k(1)^{2}–3(1)+k = 0

⇒ k−3+k = 0

⇒ 2k−3 = 0

⇒ k= 3/2

**Q.4. Factorize:****(i) 12x ^{2}–7x+1**

Using the splitting the middle term method,

We have to find a number whose sum = -7 and product =1×12 = 12

We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]

12x

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

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

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

**Solution:**

Using the splitting the middle term method,

We have to find a number whose sum = 7 and product = 2×3 = 6

We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]

2x^{2}+7x+3 = 2x^{2}+6x+1x+3

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

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

**(iii) 6x ^{2}+5x-6 **

**Solution:**

Using the splitting the middle term method,

We have to find a number whose sum = 5 and product = 6×-6 = -36

We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]

6x^{2}+5x-6 = 6x^{2}+9x–4x–6

= 3x(2x+3)–2(2x+3)

= (2x+3)(3x–2)

**(iv) 3x ^{2}–x–4 **

Using the splitting the middle term method,

We have to find a number whose sum = -1 and product = 3×-4 = -12

We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]

3x

= 3x^{2}–4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

**5. Factorize:****(i) x ^{3}–2x^{2}–x+2**

**Solution:**

Let p(x) = x^{3}–2x^{2}–x+2

Factors of 2 are ±1 and ± 2

Now,

p(x) = x^{3}–2x^{2}–x+2

p(−1) = (−1)^{3}–2(−1)^{2}–(−1)+2

= −1−2+1+2

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x^{2}–3x+2) = (x+1)(x^{2}–x–2x+2)

= (x+1)(x(x−1)−2(x−1))

= (x+1)(x−1)(x-2)

**(ii) x ^{3}–3x^{2}–9x–5**

Let p(x) = x

Factors of 5 are ±1 and ±5

By trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

Now,

p(x) = x^{3}–3x^{2}–9x–5

p(5) = (5)^{3}–3(5)^{2}–9(5)–5

= 125−75−45−5

= 0

Therefore, (x-5) is the factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

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

= (x−5)(x(x+1)+1(x+1))

= (x−5)(x+1)(x+1)

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

Let p(x) = x

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

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

p(-1) = (−1)^{3}+13(−1)^{2}+32(−1)+20

= −1+13−32+20

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient +Remainder

(x+1)(x^{2}+12x+20) = (x+1)(x^{2}+2x+10x+20)

= (x−5)x(x+2)+10(x+2)

= (x−5)(x+2)(x+10)

**(iv) 2y ^{3}+y^{2}–2y–1**

Let p(y) = 2y

Factors = 2×(−1)= -2 are ±1 and ±2

By trial method, we find that

p(1) = 0

So, (y-1) is factor of p(y)

Now,

p(y) = 2y^{3}+y^{2}–2y–1

p(1) = 2(1)^{3}+(1)^{2}–2(1)–1

= 2+1−2

= 0

Therefore, (y-1) is the factor of p(y)

Now, Dividend = Divisor × Quotient + Remainder

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

= (y−1)(2y(y+1)+1(y+1))

= (y−1)(2y+1)(y+1)

**Check out the NCERT Solutions of all the exercises of Polynomials:**

Exercise 2.1. NCERT Solutions: Polynomials

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