Please see Polynomials Exam Questions Class 9 Mathematics below. These important questions with solutions have been prepared based on the latest examination guidelines and syllabus issued by CBSE, NCERT, and KVS. We have provided Class 9 Mathematics Questions and answers for all chapters in your NCERT Book for Class 9 Mathematics. These solved problems for Polynomials in Class 9 Mathematics will help you to score more marks in upcoming examinations.
Exam Questions Polynomials Class 9 Mathematics
Very Short Answer Type Questions:
Question. Write the zeroes of the polynomial p(x) = x(x – 2)(x – 3).
Ans. For zeroes, put p(x) = 0
∴ x(x – 2)(x – 3) = 0,
Therefore, x = 0, 2, 3
Question. Write an example of a constant polynomial.
Ans. Constant polynomial is 7.
Question. Find the value of m, if x + 4 is a factor of the polynomial x2 + 3x + m.
Ans. Given, x + 4 is a factor of x2 + 3x + m = p(x)
∴ p(– 4) = 0
or, 16 – 12 + m = 0
or, m = – 4
Question. Write the expression which represents a polynomial.
Ans. 3 x2 – x – 1.
Question. Write the coefficient of x2 in the expansion of (x – 2)3.
Ans. (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. If x + 1/x = 4, then calculate the value of x2 + 1/x2
Ans. x2+ 1/x2= (x + 1/x)2 − 2(x)(1/x)
= (4)2 − 2 = 16 − 2 = 14
Question. Write any polynomial in one variable.
Ans. √2 x2 + 3x or √3 y2 + 3y.
Question. In the expression x2 + π /2 x – 7, what is the coefficient of x ?
Ans. Coefficient of x in expression x2 + π /2 x – 7 is π /2
Question. Factorize : 20x2 – 9x + 1
Ans. 20x2 – 9x + 1 = 20x2 – 5x – 4x + 1
= 5x(4x – 1) – 1(4x – 1)
= (4x – 1)(5x – 1)
Short Answer Type Questions:
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.
Ans. 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. If x2+ 1/x2 = 98. Find the value of x3+ 1/x3
Ans. 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. Factorize : x3 – 3x2 – 9x – 5.
Ans. 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. Factorize : 6x2 + 7x – 3
Ans. 6x2 + 7x – 3 = 6x2 + 9x – 2x – 3 (By splitting middle term)
= 3x(2x + 3) – 1 (2x + 3)
= (2x + 3) (3x – 1)
Question. Using a suitable identity, find (98)3.
Ans. (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
Question. Find the value of k, so that polynomial x3 + 3x2 – kx – 3 has one factor as x + 3.
Ans. 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. For what value of k, is the polynomial p(x) = 2x3 – kx2 + 3x+ 10 exactly divisible by (x + 2) ?
Ans. 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. Find the value of ‘k’ if (x – 1) is a factor of p(x) = 2x2 + kx + √2 .
Ans. 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. Evaluate 1113, using a suitable identity.
Ans. 1113 = (100 + 11)3
= (100)3 + 3(100)2 (11) + 3(100) (11)2 + (11)3
= 1367631
Question. Factorize : (x + y)3 – (x3+ y3)
Ans. (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)
Long Answer Type Questions:
Question. Factorize : x3 – 12x2 + 47x – 60
Ans. 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)
Question. If x + 1/x = 5, evaluate x2 + 1/x2.
Ans. 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 = √3 , evaluate x3 + 1/x3.
Ans. 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. Prove that x3+ y3 + z3 – 3xyz = 1/2(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Ans. 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. If x − 1/x =2 , find x4 + 1/x4
Ans. 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. Prove that (a2– b2 )3 + (b2 – c2 )3 + (c2 – a2)3 = 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a).
Ans. 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)