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# Short Answers - Polynomials Class 9 Notes | EduRev

## Mathematics (Maths) Class 9

Created by: Full Circle

## Class 9 : Short Answers - Polynomials Class 9 Notes | EduRev

The document Short Answers - Polynomials Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. Write the numerical co-efficient and degree of each term of: Solution:

 Term Numerical  co-efficient Degree x/2 1/2 1 -3x2 -3 2 5/2 3 -5x4 -5 4

Question 2. Find the remainder when x3 – ax2 + 4x – a is divided by (x – a).
Solution:
p(x) = x– ax2 + 4x – a (x – a) = 0
⇒ x = a
∴ p(a) = (a)3 – a(a)2 + 4(a) – a = a3 – a+ 4a – a = 4a – a = 3a
∴ The required remainder = 3a

Question 3. When the polynomial kx3 + 9x2 + 4x – 8 is divided x + 3, then a remainder 7 is obtained.
Find the value of k.
Solution:
Here, p(x) = kx3 + 9x+ 4x – 8
Since, Divisor = x + 3
∴ x + 3 = 0
⇒ x = –3
∴ p(–3) = 7
⇒ k(–3)3 + 9(–3)+ 4(–3) – 8 = 7
⇒ –27k + 81 – 12 – 8 = 7
⇒ –27k = 7 – 81 + 12 + 8
⇒ –27k = 27 – 81
⇒ –27k = –54
⇒ k= -(54/27) = 2
Thus, the required value of k = 2.

Question 4. (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)?
Solution:
(a) Here p(x) = x+ 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

Question 5. Factorise x2 – x – 12.
Solution:
We have x2 – x – 12 = x– 4x + 3x – 12
= x(x – 4) + 3(x – 4)
= (x – 4)(x + 3)
Thus, x– x – 12 = (x – 4)(x + 3)

Question 6. If x + (1/2)x  = 5, then find the value of x2 Solution
: We have x + 1/2)x  = 5

Squaring both sides, we get ⇒ ⇒ ⇒ Thus, the requiewd value of Question 1. Check whether (x – 1) is a factor of the polynomial x3 – 27x+ 8x + 18.
Solution:
Here, p(x) = x– 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.

Question 2. Find the remainder when f(x) = Solution: Here f(x) = Divisor = x+(2/3)

since, x+(2/3) = 0 ⇒ x = -2/3

∴ Remainder = f (-2/3)

i.e.

Remainder  Thus,  the required remainder =  (10/27)

Question 3. Find the value of k, if (x – k) is a factor of x6 – kx5 + x4 – kx3 + 3x – k + 4.
Solution:
Here p(x) = x6 – kx5 + x– 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
or k– k6 + k4 – k4 + 3k – k + 4 = 0
or 2k + 4 = 0
or 2k = –4
or k = (-4/2)
= –2
Thus, the required value of k is –2.

Question 4. Factorise: 9a2 – 9b+ 6a + 1
Solution:
9a2 – 9b2 + 6a + 1
= [9a2 + 6a + 1] – 9b
= [(3a)2 + 2(3a)(1) + (1)2] – (3b)
= (3a + 1)2 – (3b)2
= [(3a + 1) + 3b][(3a + 1) – 3b]    [using x– y2 = (x – y)(x + y)
= (3a + 1 + 3b)(3a + 1 – 3b)

Question 5. Find the value of x3 + y3 – 12xy + 64, when x + y = –4.
Solution:
x3 + y– 12xy + 64
= (x)3 + (y)3 + (4)– 3(x)(y)(4)
= [x2 + y2 + 42 – xy – y. 4 – 4 .  x](x + y + 4)
= [x2 + y2 + 16 – xy – 4y – 4x][x + y + 4]                    ...(1)
Since, x + y = –4 ∴ x + y + 4 = 0                                 ...(2)

From (1) and (2), we have x3 + y– 12xy + 64
= [x+ y2 + 16 – xy – 4y – 4x] = 0
Thus, x3 + y3 – 12xy + 64 = 0.

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