CBSE Class 9 Mathematics Polynomials Worksheet Set 09

Very Short Answer Type Questions:

Question. In the expression x² + π /2 x – 7, what is the coefficient of x ?
Answer. Coefficient of x in expression x² + π /2 x – 7 is π /2

Question. Write the coefficient of x² in the expansion of (x – 2)³.
Answer. (x – 2)³ = (x)³ – (2)³ – 3 × x × 2(x – 2)
= x³ – 8 – 6x² + 12x
Coefficient of x² in the expansion of (x – 2)³ = – 6

Question. Factorize : 20x² – 9x + 1
Answer. 20x² – 9x + 1 = 20x² – 5x – 4x + 1
= 5x(4x – 1) – 1(4x – 1)
= (4x – 1)(5x – 1)

Question. Write the zeroes of the polynomial p(x) = x(x – 2)(x – 3).
Answer. For zeroes, put p(x) = 0
∴ x(x – 2)(x – 3) = 0,
Therefore, x = 0, 2, 3

Question. If x + 1/x = 4, then calculate the value of x² + 1/x²
Answer. x²+ 1/x²= (x + 1/x)² − 2(x)(1/x)
= (4)² − 2 = 16 − 2 = 14

Question. Write the expression which represents a polynomial.
Answer. 3 x² – x – 1.

Question. Write an example of a constant polynomial.
Answer. Constant polynomial is 7.

Question. Write any polynomial in one variable.
Answer. √2 x² + 3x or √3 y² + 3y.

Question. Find the value of m, if x + 4 is a factor of the polynomial x² + 3x + m.
Answer. Given, x + 4 is a factor of x² + 3x + m = p(x)
∴ p(– 4) = 0
or, 16 – 12 + m = 0
or, m = – 4

Short Answer Type Questions:

Question. Factorize : 6x² + 7x – 3
Answer. 6x² + 7x – 3 = 6x² + 9x – 2x – 3 (By splitting middle term)
= 3x(2x + 3) – 1 (2x + 3)
= (2x + 3) (3x – 1)

Question. Factorize : x³ – 3x² – 9x – 5.
Answer. Let p(x) = x³– 3x² – 9x – 5,
Since, p (– 1) = – 1 – 3 + 9 – 5 = 0
Therefore, (x + 1) is a factor of x³– 3x² – 9x – 5. 
∴ (x³– 3x² – 9x – 5) = (x + 1)(x² – 4x – 5) 
Now, x²– 4x – 5 = x²– 5x + x – 5
= x(x – 5) + 1(x – 5)
= (x + 1)(x – 5) 1
∴ p(x) = (x + 1)(x + 1)(x – 5) 

Question. For what value of k, is the polynomial p(x) = 2x³ – kx² + 3x+ 10 exactly divisible by (x + 2) ?
Answer. Since, (x + 2) is a factor of p(x).
Thus, p(–2) = 0
or, 2(–2)³ – k(–2)² + 3(–2) +10 = 0
or, – 16 – 4k – 6 + 10 = 0
k = – 3

Question. If y = 2 and y = 0 are the zeroes of the polynomial f(y) = 2y³ – 5y² + ay + b, find the value of a and b.
Answer. Given, f(y) = 2y³ – 5y² + ay + b
∴ f(2) = 2(2)3 – 5(2)² + a(2) + b = 0
or, 16 – 20 + 2a + b = 0 
or, 2a + b = 4 …(i)
and f(0) = b = 0
From (i), 2a + 0 = 4
or, a = 2
∴ a = 2, b = 0

Question. Find the value of k, so that polynomial x³ + 3x² – kx – 3 has one factor as x + 3.
Answer. Let f(x) = x³+ 3x² – kx – 3
Since, (x + 3) is a factor of f(x).
Then, f(– 3) = 0
or, (– 3)³ + 3(– 3)² – k(–3) – 3 = 0
or, – 27 + 27 + 3k – 3 = 0
or, 3k – 3 = 0
or, k = 1

Question. Evaluate 111³, using a suitable identity.
Answer. 111³ = (100 + 11)³
= (100)³ + 3(100)² (11) + 3(100) (11)² + (11)³
= 1367631

Question. Factorize : (x + y)³ – (x³+ y³)
Answer. (x + y)³ – (x³+ y³)
= (x + y)³ – (x + y) (x² + y² – xy)
= (x + y) {(x + y)² – (x² + y² – xy)}
= (x + y) (x²+ y² + 2xy – x²– y²+ xy)
= (x + y) (3xy)
= 3xy(x + y)

Question. If x²+ 1/x² = 98. Find the value of x³+ 1/x³
Answer. x²+ 1/x² = 98
(x + 1/x)² = 98 + 2 = 100
x + 1/x = 10
Now, x³+ 1/x³= (x + 1/x)(x²− x.1/x + 1/x² )
= (10)(x²+ 1/x² − 1)
= (10)(98 − 1)
= 10 x 97 = 970

Question. Find the value of ‘k’ if (x – 1) is a factor of p(x) = 2x² + kx + √2 .
Answer. Given, p(x) = 2x² + kx + √
Since, (x – 1) is a factor of p(x), then p(1) = 0.
∴ 2 (1)² + k(1) + √2 = 0
or, 2 + k + √2 = 0
or, k = –2 – √2

Question. Using a suitable identity, find (98)³.
Answer. (98)³ = (100 – 2)³
= (100)³ – (2)³ – 3 × 100 × 2(100 – 2)
= 1000000 – 8 – 600 × (100 – 2)
= 1000000 – 8 – 60000 + 1200
= 1000000 – 58808 = 941192

Long Answer Type Questions:  

Question. Prove that x³+ y³ + z³ – 3xyz = 1/2(x + y + z) [(x – y)² + (y – z)² + (z – x)²]
Answer. RHS = 1/2(x + y + z) [(x – y)²+ (y – z)² + (z – x)²]
= 1/2(x + y + z) [ x² + y² − 2xy + y²+ z² − 2yz + z² + x²− 2zx]
= 1/2(x + y + z)[2x² + 2y² + 2z² – 2xy – 2yz – 2zx]
= 1/2(x + y + z). 2[x² + y² + z² – xy – yz – zx]
= x³+ y³ + z³ – 3xyz
[Using identity x³+ y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)]
= LHS

Question. Prove that (a²– b² )³ + (b² – c² )³ + (c² – a²)³ = 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a).
Answer. Let x = a²– b² , y = b² – c² , z = c² – a²
Now, x + y + z = a²– b² + b² – c² + c² – a²= 0
∴ x + y + z = 0
or, x³ + y³ + z3 = 3xyz
i.e., (a²– b² )³ + (b² – c² )³ + (c² – a²)³
= 3(a²– b² )(b² – c² )(c² – a²)
= 3(a + b)(a – b)(b + c)(b – c)(c + a)(c – a)
= 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a)

Question. If x + 1/x = 5, evaluate x² + 1/x².
Answer. x + 1/x = 5
On squaring both sides, we get 
(x + 1/x)²= 5²
or, x² +( 1/x)² + 2 × x + 1/x = 25 [(a + b)² = a² + b² + 2ab]
or, x² + 1/x² + 2 = 25
or, x² + 1/x² = 25 − 2 
or x² + 1/x² = 23

Question. If x − 1/x =2 , find x⁴ + 1/x⁴ 
Answer. x² + 1/x² = (x − 1/x)² + 2
= 4 + 2 = 6
(x² + 1/x²)² = (x⁴ + 1/x⁴) + 2
or, (6)² = x⁴ + 1/x⁴ + 2
or, 36 − 2 = x⁴ + 1/x⁴ 
or, x⁴ + 1/x⁴ = 34

Question. If x + 1/x = √3 , evaluate x³ + 1/x³.
Answer. x + 1/x = √3
Cubing both sides, we get
(x + 1/x)³= (√3)³
or, x³ + 1/x³ + 3x. 1/x(x + 1/x) = 3√3
or, x³ + 1/x³ + 3(√3) = 3√3
or, x³ + 1/x³ = 0

Question. Factorize : x³ – 12x² + 47x – 60
Answer. Factor of 60 = (±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±30, ±60,)
p(x) = x³ – 12x² + 47x – 60
p(3) = (3)³ – 12(3)² + 47(3) – 60
= 27 – 108 + 141 – 60
= 168 – 168 = 0
∴ x = 3 is a zero of p(x) or (x – 3) is a factor of p(x).
x³ – 12x² + 47x – 60
= x²(x – 3) – 9x(x – 3) + 20(x – 3)
= (x – 3)(x² – 9x + 20)
= (x – 3)(x² – 5x – 4x + 20)
= (x – 3)[x(x – 5) – 4(x – 5)]
= (x – 3)(x – 4)(x – 5)