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#### CLASS 9 NCERT MATHS CHAP 2 POLYNOMIALS

### NCERT Solution Chapter 2: Polynomials

# TEXTBOOKS EXERCISE – 2.4

**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} + 3x^{3} + 3x^{2} +
x + 1

(iv) x^{3} – x^{2} – (2 +√ 2 ) x + √ 2

**Sol. **To have (x +
1) as a factor, substituting x =
–1 must give p(–1) = 0.

(i) x^{3} + x^{2} +
x + 1

= (–1)^{3} + (–1)^{2} + (–1) + 1

= – 1 + 1 – 1 + 1 = 0

Therefore, x + 1 is a factor of x^{3} + x^{2} + x + 1

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

= (–1)^{4} + (–1)^{3} + (–1)^{2} + (–1) + 1

= 1 – 1 + 1 – 1 + 1 = 1

Remainder is not 0, Therefore, (*x *+ 1) is not its factor.

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

= (–1)^{4} + 3(–1)^{3} + 3(–1)^{2} + (–1) + 1

= 1 – 3 + 3 – 1 + 1 = 1. Remainder is not 0, therefore, (*x *+ 1) is not its factor

(iv) x^{3} – x^{2} –
(2 + √2 ) x + √2

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

= – 1 – 1 + 2 +√ 2 + √2 = 2 √2

Remainder is not 0, therefore (x + 1) is not a factor.

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

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

(iii) p(x) = x3 – 4x^{2} + x + 6, g(x) = x – 3

**Sol. **(i) g(x) = x + 1. x = – 1
to be substituted in p(x) = 2x^{3} + x^{2} –
2x – 1

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

= – 2 + 1 + 2 – 1 = 0.

So, g(x) is a factor of p(x).

(ii) g(x) = x + 2, substitute x = – 2 in p(x)

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

p(–2) = (–2)^{3} + 3(–2)^{2} + 3(–2) + 1 = – 8 + 12 – 6 + 1 = –1.

So, g(x) is not a factor of p(x)

(iii) g(x) = x – 3 substitute x = 3 in (x).

p(x) = x^{3} – 4x^{2} +
x + 6

p(3) = (3)^{3} – 4(3)^{2} + 3 + 6

= 27 – 36 + 3 + 6 = 0.

Therefore, g(x) is a factor of x^{3} – 4x^{2} + x + 6.

**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) p(x) = 2x^{2} + kx + √2

(iii) p(x) = kx^{2} – √2 x + 1

(iv) p(x) = kx^{2} – 3x + k

**Sol. **(x –
1) is a factor, so we substitute x =
1 in each case and solve for k by
making p(1) equal to 0.

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

p(1) = 1 + 1 + k = 0 ⇒ k = –2

(ii) p(x) = 2x^{2} + kx + √2

p(1) = 2 × 1^{2} + k × 1 + √2 = 0

⇒ 2 + k + 2 = 0

⇒ k = –2 – √2 = – (2 + √2 )

(iii) p(x) = kx^{2} – √2 x + 1

p(1) = k – √2 + 1 = 0

⇒ k = √2 – 1

(iv) p(x) = kx^{2} – 3x + k

p(1) = k – 3 + k = 0 ⇒ 2k – 3 = 0

⇒ k = 3/2

**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

**Sol. **(i) 12*x*^{2} –
7*x *+ 1

= 12*x*^{2} – 4*x *– 3*x
*+ 1

= 4*x *(3*x *– 1) – 1 (3*x *– 1) = (4*x *– 1)
(3*x *– 1)

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

= 2*x*2
+ 6*x *+ *x *+ 3

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

(iii) 6*x*^{2} + 5*x *– 6

= 6*x*^{2} + 9*x *– 4*x
*– 6

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

(iv) 3*x*^{2} – *x *– 4

= 3*x*^{2} – 4*x *+ 3*x
*– 4

= *x *(3*x *– 4) + 1 (3*x *– 4) = (*x *+ 1) (3*x *– 4)

**5. **Factorise :

(i) x^{3} – 2x^{2} – x + 2

(ii) x^{3} – 3x^{2} – 9x – 5

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

(iv) 2y^{3} + y^{2} –
2y – 1

**Sol. **(i) p(x) = x^{3} – 2x^{2} –
x + 2

Let us guess a factor (x – a) and choose value of a arbitrarily as 1.

Now, putting this value in p(x).

p(1) = 1 – 2 – 1 + 2 = 0

So, (x – 1) is a factor of p(x)

Now, x^{3} – 2x^{2} –
x + 2

= x^{3 }– x^{2} – x^{2} +
x – 2x + 2

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

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

= (x – 1) (x + 1) (x – 2)

So, x^{3} – 2x^{2} – x + 2

(ii) p(x) = x^{3} – 3x^{2} –
9x – 5

Take a factor (x – a). a should be a factor of 5, i.e., ± 1 or ± 5.

For (x – 1), a = 1

p(1) = (1)^{3} – (–3) 1^{2} – 9 × 1 – 5

= 1 – 3 – 9 – 5 = – 16.

So, (x – 1) is not a factor of p(x).

For a = 5

p(5) = (5)^{3 }– 3(5)^{2} – 9 (5) – 5 = 125 –
75 – 45 – 5 = 0.

Therefore, (x – 5) is a factor of

x^{3} – 3x^{2} – 9x – 5.

Now, x^{3} – 3x^{2} –
9x – 5

= x^{3} –
5x^{2} +
2x^{2} –
10x + x – 5 = x^{2} (x – 5) + 2x (x – 5) + 1 (x – 5)

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

= (x – 5) (x + 1) (x + 1)

So, x^{3} – 3x^{2} – 9x – 5 = (x – 5) (x + 1) (x + 1).

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

Let a factor be (*x *– *a*). *a *should be a factor
of 20 which are ± 1, ± 2, ± 4, ± 5, ± 10.

For *x *– 1 = 0 ⇒ *x *= 1

Now, *p*(1) = 1 + 13 + 32 + 20 = 66 ≠ 0

Hence, (*x *– 1) is not a factor of *p*(*x*).

Again, for *x *+ 1 = 0 ⇒ *x *= – 1

Now, *p *(–1) = –1 + 13 – 32 + 20

= –33 + 33 = 0

Hence, (*x *+ 1) is a factors of *p*(*x*).

Now, *x*^{3} + 13*x*^{2} + 32*x *+ 20

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

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

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

(iv) p(y) = 2y^{3} + y^{2} –
2y – 1

Factors of –1 are ± 1, ± 2.

p(1) = 2 × 1^{3} + 1^{2} –
2 × 1 – 1 = 2 + 1 – 2 – 1 = 0.

Therefore, (y – 1) is a factor of p(y).

Now, 2y^{3} + y^{2} –
2y – 1

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

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

= (y – 1) {2y (y + 1) + 1 (y + 1) = (y – 1) (y + 1) (2y + 1)

Therefore, 2y^{3} + y^{2} – 2y – 1 = (y – 1) (2y + 1) (y + 1).