Math Assignment Class IX Ch-2

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 : 8

b) 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
$c)\: \: x-\frac{1}{2}\: \: \:\: \: Ans\: \frac{-1}{2}$
$d)\: \: 2x+1\: \: \:\: \: Ans\: \frac{-27}{8}$

$e)\: \: 5-2x\: \: \:\: \: Ans\: \frac{27}{8}$

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) 3x – kx + √3 Ans -3 - √3
b) √2x – kx + k + 2 Ans $-\frac{1}{2}(2+\sqrt{2})$

c) k – kx + Kx Ans 0
d) x - 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) 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

⇒ a = 1

Putting a = 1 in eqn. (i) we get

4(1) -b = 0 ⇒ b = 4

Q 7. 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

⇒ a = -9

Putting a = -9 in eqn. (ii) we get

3(-9) + b = -3 ⇒ b = 24

Q 8. 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 (16a+3b)(33b-4a)

k) x³+ 3x²- 4x - 12 : Ans (x - 2)(x + 2)(x + 3)

l) 2x³- 9x² +x + 12 : Ans (x + 1)(x - 4)(2x - 3)

m) x³-3x²-10x+24 : Ans (x + 3)(x - 4)(3x - 2)

n) 2x³-7x²- 3x + 18 : Ans (x - 2)(x - 3)(2x + 3)

o) 3x³+ x² - 38x + 24 : Ans (x - 3)(x + 4)(3x - 2)

p)(x³- (y - z)³) : Ans (x – y + z)(x²+ y²+ z²- 2yz – xy - xz)

$q)\: \: 27p^{3}-\frac{1}{216}-\frac{9}{2}p^{2}+\frac{1}{4}p\: \: \: Ans:\; \left ( 3p-\frac{1}{6} \right )^{3}$

9. Evaluate by using identity:

a) 92×108 Ans:- 9936

b) 6.7×5.3 Ans:- 35.51

c)104×107 Ans:- 11128

10.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)

11.Evaluate by using identity:

a) (-15)³ + (-13)³ + (28)³ Ans: 16380
b) (75)³ + (-50)³ + (-25)³ Ans: 281250
12.Simplify:

$\left ( 1-\frac{1}{x+1} \right )\left ( 1-\frac{1}{x+2} \right )\left ( 1-\frac{1}{x+3} \right )\: ......\: \left ( 1-\frac{1}{x+10} \right )$ $Ans\: \left ( \frac{x}{x+10} \right )$

Q13.If a = 1 - √2, find the value of $\left ( a-\frac{1}{a} \right )^{3}$ Ans: 8

Q 14.) If a³+ b³+ c³= 3abc and a + b + c = 0 then show that

$\frac{(b+c)^{2}}{3bc}+\frac{(a+c)^{2}}{3ac}+\frac{(b+a)^{2}}{3ba}=1$

Q15.) If $x^{2}+\frac{1}{x^{2}}=14 ,\: \: find\: \: x^{3}+\frac{1}{x^{3}}$ $Ans\: \: \pm 52$

16. Evaluate:

$\left ( \frac{x^{a}}{x^{b}} \right )^{a+b}\times \left ( \frac{x^{b}}{x^{c}} \right )^{b+c}\times \left ( \frac{x^{c}}{x^{a}} \right )^{c+a}\: \: \: Ans:\: 1$