Maths Assignment Class 9th
Chapter 2 Polynomials
Mathematics Assignment for Class IX Ch -2, Polynomials, strictly according to the CBSE syllabus. Math worksheet based on the topic Polynomial.
Mathematics Assignment on Polynomials
Class IX Chapter 2
1. Find the value of p(x) = 3x2 - 2x + 8 at
a) x = 0 Ans : 8b) x = -2 Ans : 24
c) x = 3 Ans : 29
d) x = -5 Ans : 73
2. Find the remainder when x3 - 3x2 + 3x - 1 is divided by
a) x - 1 Ans 0
b) x + 1 Ans -8
f) 3-x Ans 8
Q 3. a) Show that 2x - 3
is a factor of P(x) = 2x3 - 9x2 + x + 12.Ans: P(3/2) = 0 ⇒ 2x - 3 is the factor of P(x)
b) Show that x + 2 is a factor of P(x) = x4 - x2 - 12.Ans: P(-2) = 0 ⇒ x + 2 is the factor of P(x)
Hint: If remainder = 0 then given function is the factor of the given polynomial otherwise not.
4. Find the value of k if x + 1 is a factor of
a) 3x2 – kx + √3 Ans -3 - √3
b) √2x2 – kx + k + 2 Ans c) k – kx2 + Kx Ans: 0
d) x2 - 3x + kx Ans: 4
Q5) Find “a” if x+1 is a
factor of P(x) = ax3 - 9x2 +
x + 6a Ans:- 2
Solution Hint: Let x + 1 = 0 ⇒ x = - 1
Now find P(-1) by putting x = -1 in p(x)
If P(-1) = 0 then x + 1 is the factor of p(x) otherwise not
Q 6. Factorize:a) x2 + 11x + 30 : Ans (x + 5)(x + 6)
b) a2 - 16a + 63 :
Ans (a - 7)(a - 9)
c)15 - 2x - x2 :
Ans (3 - x)(5 + x)
d) 3 + 5x - 2x2 : Ans (1+ 2x)(3 - x)
e) 3m2 - 20m - 7 : Ans (3m+1)(m - 7)
f) 24m2 + m - 23 : Ans
(m + 1)(24m - 23)
g) 2x2 + 11xy - 21y :
Ans (x + 7y)(2x - 3y)
h) 7(x + y)2 + 48(x + y) - 7 : Ans (x + y + 7)(7x + 7y - 1)
I) 2m4n - 6m2n
+ 4n : Ans 2n(m + 1)(m - 1)(m2 - 2)
j) 36(a + 3b)2 - 36(2a - 3b)2 : Ans 108a(- a + 6b)
k) 2x3 - 9x2 +x + 12 : Ans (x + 1)(x - 4)(2x - 3)
l) x3 -3x2 -10x+24 : Ans (x + 3)(x - 4)(3x - 2)
m) 2x3 -7x2 - 3x + 18 : Ans (x - 2)(x - 3)(2x + 3)
n) 3x3 + x2 - 38x + 24 : Ans (x - 3)(x + 4)(3x - 2)
o)(x3 - (y - z)3) : Ans (x – y + z)(x2 + y2 + z2
- 2yz – xy - xz)
7. Evaluate by using identity:
a) 92 × 108 Ans:- 9936
b) 6.7 × 5.3 Ans:- 35.51
c)104 × 107 Ans:- 11128
8.Factorize:
a) 8x3 - 125y3 - z3 - 30xyz
b) 125x3 - 8 + 27y3 + 90xy
c) 3√3a3
- 5√5b3 - 8c3 - 6√15abc
d) (5a - 4b)3
+ (4b - 6c)3 + (6c - 5a)3
e) a3 + (a - 1)3 + (1 - 2a)3
f) (3x + 3y - 5z)3 + (2y + 3z - 5x)3 + (2z + 2x - 5y)3
g) (2x + y)3 - (x + y)3
h)
64m6-n6 : Ans:
(2mn + n)(4m - 2mn + n)(2m - n)(4m + 2mn + n)
Question 9.Evaluate by using identity:a) (-15)3 + (-13)3 + (28)3 Ans: 16380
b) (75)3 +
(-50)3 + (-25)3 Ans:
281250
Question 10: For what value of a if 2x3 + ax2 + 11x + a + 3 is divided by 2x - 1
Ans[a = -7 ]
Question 11: Find the remainder when x4 + 3x2 + 3x + 1 is divided by x + 1
[Ans : 2]
Question 12.
If (x - 2)and(x + 3) are factors of p(x) = ax3 + 3x2 - bx - 12. Find a and b.
Ans: a = 1, b = 4
Solution Hint :
x - 2 is the factor of P(x) ⇒ P(2) = 0
a(2)3 + 3(2)2 - b(2) - 12 = 0
8a -2b = 0
4a -b = 0 ....... (i)
x + 3 is the factor of P(x) ⇒ P(-3) = 0
a(-3)3 + 3(-3)2 - b(-3) - 12 = 0
-27a + 27 + 3b - 12 = 0
-9a +b = -5 ....... (ii)
Eqn.(i) -Eqn(ii) we get
4a - b = 0
-9a + b = -5
----------------
-5a = -5
⇒ a = 1
Putting a = 1 in eqn. (i) we get
4(1) -b = 0 ⇒ b = 4
Question 13.
If x - 5 is a factor of x3 + ax2 + bx - 20 and leaves a remainder -2 when divided by x - 3. Find a and b.
Ans (-9, 24)
Solution Hint
x - 5 is the factor of P(x) ⇒ P(5) = 0
(5)3 + a(5)2 + b(5) - 20 = 0
5a + b = -21 ...... (i)
Also P(3) = -2
(3)3 + a(3)2 + b(3) - 20 = 0
3a + b = -3 ........ (ii)
Eqn.(i) -Eqn.(ii) we get
5a + b = -21
3a + b = ∓ 3
-----------------
2a = -18
a = - 9
Putting a = -9 in eqn. (ii) we get
3(-9) + b = - 3 ⇒ b = 24
If x3 + ax2+ bx + 6 has (x - 2) as a factor and leaves remainder 3, when divided by (x – 3), find a and b
Ans[a = -3, b = -1]
HIGHER ORDER THINKING SKILL
HOTS
Q 14.) If a3 + b3 + c3 = 3abc and a + b + c = 0 then show that
16. Evaluate:
THANKS FOR YOUR VISIT
PLEASE COMMENT BELOW
Very very nice assigament, thank ku so much.
ReplyDelete