Class 9 Maths Chapter 2 Question Answers - Polynomials

Q1. Write the numerical co-efficient and degree of each term of:

Sol:

Q2. Write the coefficient of x² in each of the following:

(i) 9 – 12x + x³

(ii) ∏/6 x² – 3x + 4

(iii) √3x – 7

Sol:
(i) 9 – 12x + x³
Coefficient of x² =0
(ii) ∏/6 x² – 3x + 4
Coefficient of x² = ∏/6
(iii) √3x – 7
Coefficient of x² = 0

Q3. Factorise z²  – 4z –12

Sol: To factorize this expression, we need to find two numbers α and β such that α + β = -4 and αβ = -12

z² – 6z + 2z – 12

z(z – 6) + 2(z – 6)

(z + 2)(z – 6)

Q4. (a) For what value of k, the polynomial x² + (4 – k)x + 2 is divisible by x – 2?
(b) For what value of ‘m’ is x³ – 2mx² + 16 is divisible by (x + 2)?

Sol: (a) Here p(x) = x² + 4x – kx + 2
If p(x) is exactly divisible by x – 2, then p(2) = 0
i.e. (2)² + 4(2) – k(2) + 2 = 0
⇒ 4 + 8 – 2k + 2 = 0
⇒ 14 – 2k = 0
⇒ 2k = 14
⇒ k= 14/2 = 7
Thus, the required value of k is 7.

(b) Here, p(x) = x³ – 2mx² + 16
∴ p(–2) = (–2)³ –2(–2)²m + 16
⇒ –8 –8m + 16
⇒ –8m + 8
Since, p(x) is divisible by x + 2
∴ p(–2) = 0
or –8m + 8 = 0
⇒ m = 1

Q5. Factorize x² – x – 12.

Sol: We have  x² – x – 12
⇒ x² – 4x + 3x – 12
⇒ x(x – 4) + 3(x – 4)
⇒ (x – 4)(x + 3)
Thus, x² – x – 12 = (x – 4)(x + 3)

Q6.Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:

(i) f(x) = 3x + 1, x = −1/3

(ii) f(x) = x² – 1, x = 1,−1

Sol: (i)f(x) = 3x + 1, x = −1/3

f(x) = 3x + 1
Substitute x = −1/3 in f(x)
f( −1/3) = 3(−1/3) + 1
= -1 + 1
= 0
Since, the result is 0, so x = −1/3 is the root of 3x + 1

(ii) f(x) = x² – 1, x = 1,−1

f(x) = x² – 1
Given that x = (1 , -1)
Substitute x = 1 in f(x)
f(1) = 1² – 1
= 1 – 1
= 0
Now, substitute x = (-1) in f(x)
f(-1) = (−1)² – 1
= 1 – 1
= 0
Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x² – 1

Q7. Check whether (x – 1) is a factor of the polynomial x³ – 27x² + 8x + 18.

Sol: Here, p(x) = x³ – 27x² + 8x + 18, (x – 1) will be a factor of p(x) only if (x – 1) divides p(x) leaving a remainder 0.
For  x – 1 = 0
⇒ x = 1
∴ p(1) = (1)³ – 27(1)² + 8(1) + 18
⇒ 1 – 27 + 8 + 18
⇒ 27 – 27
⇒ 0
Since, p(1) = 0
∴ (x – 1) is a factor of p(x).
Thus, (x – 1) is a factor of x³ – 27x² + 8x + 18.

Q8.Evaluate each of the following using identities:

(i) (2x – 1/x)²

(ii) (2x + y) (2x – y)

Sol:
(i)(2x – 1/x)²
[Use identity: (a – b)² = a² + b² – 2ab ]
(2x – 1/x)² = (2x)² + (1/x)² – 2 (2x)(1/x)
= 4x² + 1/x² – 4
(ii)(2x + y) (2x – y)
[Use identity: (a – b)(a + b) = a² – b² ]
(2x + y) (2x – y) = (2x )² – (y)²
= 4x² – y²

Q9. Find the value of k, if (x – k) is a factor of x⁶ – kx⁵ + x⁴ – kx³ + 3x – k + 4.

Sol: Here, p(x) = x⁶ – kx⁵ + x⁴ – kx³ + 3x – k + 4
If (x – k) is a factor of p(x), then p(k) = 0
i.e (k)⁶ – k(k⁵) + k⁴ – k(k³) + 3k – k + 4 = 0
⇒ k⁶ – k⁶ + k⁴ – k⁴ + 3k – k + 4 = 0
⇒ 2k + 4 = 0
⇒ 2k = – 4
⇒ k = (-4/2) = –2
Thus, the required value of k is –2.

Q10. Factorize: 9a² – 9b² + 6a + 1

Sol: 9a² – 9b² + 6a + 1
⇒ [9a² + 6a + 1] – 9b² 
⇒ [(3a)² + 2(3a)(1) + (1)2] – (3b)² 
⇒ (3a + 1)² – (3b)²
⇒ [(3a + 1) + 3b][(3a + 1) – 3b] {using x² – y² = (x – y)(x + y)}
⇒ (3a + 1 + 3b)(3a + 1 – 3b)