Short Answer Type Questions: Polynomials

Q1. Write the numerical co-efficient and degree of each term of: x2 - 3x2 + 52 x3 - 5x4

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)