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Important Questions Class 9 Mathematics Chapter 2 Polynomials.
Mathematics Chapter 2 of Class 9 is about Polynomials. Polynomial consists of two terms, namely Poly (meaning “many”) and Nominal (meaning “terms.”). A polynomial is explained as an expression which is composed of variables, constants and exponents that are combined using mathematical operations like addition, subtraction, multiplication and division (No division operation by a variable). Based on the number of terms present in the expression, it is classified as monomial, binomial, and trinomial.
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Our Mathematics experts believe that students must practice questions regularly to perform better in exams. For this purpose, they have prepared the Important Questions Class 9 Mathematics Chapter 2 to help students get access to questions from all the topics of the Polynomials. The questions are followed by their stepbystep answers, which will further help students to revise the chapter. The questions are curated from various sources such as the NCERT textbook and exemplar book, CBSE past years’ question papers, and other reference materials.
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Important Questions Class 9 Mathematics Chapter 2 – With Solutions
Our inhouse Mathematics faculty experts have collated a complete list of Important Questions Class 9 Mathematics Chapter 2 by referring to various sources. The subject experts have meticulously prepared illustration for individual questions that will enable students to comprehend the notions used in each question. Furthermore, the questions are selected in a way that would cover all the topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak topics. And enhance their preparation by further concentrating on weaker sections of the chapter.
Given below are a few of the questions and answers from our question bank of Important Questions Class 9 Mathematics Chapter 2:
Question 1: Calculate the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12.
Answer 1: Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y)² = 12²
=> 9x² + 12xy + 4y² = 144
=>9x² + 4y² = 144 – 12xy
From the questions, xy = 6
So,
9x² + 4y² = 144 – 72
Thus, the value of 9x² + 4y² = 72
Question 2:Evaluate the following using suitable identity
(102) ³
Answer 2: We can write 102 as 100+2
Using identity,(x+y) ³ = x ³ +y ³ +3xy(x+y)
(100+2) ³ =(100) ³ +2 ³ +(3×100×2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
Question 3:Without any actual division, prove that the following 2x⁴
– 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.
[Hint: Factorise x² – 3x + 2]
Answer 3: x²3x+2
x²2x1x+2
x(x2)1(x2)
(x2)(x1)
Therefore,(x2)(x1)are the factors.
Considering (x2),
x2=0
x=2
Then, p(x) becomes,
p(x)=2
p(x)=2x⁴5x³+2x²x+2
p(2)=2(2)⁴5(2)³+2(2)²2+2
=3240+8
= 40+40=0
Therefore, (x2) is a factor.
Considering (x1),
x1=0
x=1
Then, p(x) becomes,
p(x)=1
p(x)=2x⁴5x³+2x²x+2
p(1)=2(1)⁴5(1)³+2(1)²1+2
=25+21+2
=66
=0
Therefore, (x1) is a factor.
Question 4: Using the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case
(i) p(x) = 2x³+x²–2x–1, g(x) = x+1
Answer 4:p(x) = 2x³+x²–2x–1, g(x) = x+1
g(x) = 0
⇒ x+1 = 0
⇒ x = −1
∴Zero of g(x) is 1.
Now,
p(−1) = 2(−1)³+(−1)²–2(−1)–1
= −2+1+2−1
= 0
∴By the given factor theorem, g(x) is a factor of p(x).
Question 5: Obtain an example of a monomial and a binomial having degrees of 82 and 99, respectively.
Answer 5: An example of a monomial having a degree of 82 = x⁸²
An example of a binomial having a required degree of 99 = x⁹⁹ + 7
Question 6: If the two x – 2 and x – ½ are the given factors of px²
+ 5x + r, show that p = r.
Answer 6: Given, f(x) = px²+5x+r and factors are x2, x – ½
g1(x) = 0,
x – 2 = 0
x = 2
Substituting x = 2 in place of the equation, we get
f(x) = px²+5x+r
f(2) = p(2)²+5(2)+r=0
= 4p + 10 + r = 0 … eq.(i)
x – ½ = 0
x = ½
Substituting x = ½ in place of the equation, we get,
f(x) = px²+5x+r
f( ½ ) = p( ½ )² + 5( ½ ) + r =0
= p/4 + 5/2 + r = 0
= p + 10 + 4r = 0 … eq(ii)
On solving eq(i) and eq(ii),
We get,
4p + r = – 10 and p + 4r = – 10
the RHS of both equations are the same,
We get,
4p + r = p + 4r
3p=3r
p = r.
Hence Proved.
Question 7: Identify constant, linear, quadratic, cubic and quartic polynomials from the following.
(i) – 7 + x
(ii) 6y
(iii) – ? ³
(iv) 1 – y – ? ³
(v) x – ? ³ + ?⁴
(vi) 1 + x + ?²
(vii) 6?²
(viii) 13
(ix) –p
Answer 7: (i) – 7 + x
The degree of – 7 + x is 1.
Hence, it is a linear polynomial.
(ii) 6y
The degree of 6y is 1.
Therefore, it is a linear polynomial.
(iii) – ? ³
We know that the degree of – ? ³ is 3.
Therefore, it is a cubic polynomial.
(iv) 1 – y – ? ³
We know that the degree of 1 – y – ? ³ is 3.
Therefore, it is a cubic polynomial.
(v) x – ? ³ + ?⁴
We know that the degree of x – ? ³ + ?⁴ is 4.
Therefore, it is a quartic polynomial.
(vi) 1 + x + ?²
We know that the degree of 1 + x + ?² is 2.
Therefore, it is a quadratic polynomial.
(vii) 6?²
We know that the degree of 6?² is 2.
Therefore, it is a quadratic polynomial.
(viii) 13
We know that 13 is a constant.
Therefore, it is a constant polynomial.
(ix) – p
We know that the degree of –p is 1.
Therefore, it is a linear polynomial.
Question 8: Observe the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.
Answer 8: Let the polynomial be f(x) = 5x – 4x² + 3
Now, for x = 2,
f(2) = 5(2) – 4(2)² + 3
=> f(2) = 10 – 16 + 3 = –3
Or, the value of the polynomial 5x – 4x² + 3 at x = 2 is 3.
Similarly, for x = –1,
f(–1) = 5(–1) – 4(–1)² + 3
=> f(–1) = –5 –4 + 3 = 6
The value of the polynomial 5x – 4x² + 3 at x = 1 is 6.
Question 9:Expanding each of the following, using all the suitable identities:
(i) (x+2y+4z)²
(ii) (2x−y+z)²
(iii) (−2x+3y+2z)²
(iv) (3a –7b–c)²
(v) (–2x+5y–3z)²
Answer 9: (i) (x+2y+4z)²
Using identity, (x+y+z)² = x²+²+z²+2xy+2yz+2zx
Here, x = x
y = 2y
z = 4z
(x+2y+4z)² = x²+(2y)²+(4z)²+(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x²+4y²+16z²+4xy+16yz+8xz
(ii) (2x−y+z)²
Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx
Here, x = 2x
y = −y
z = z
(2x−y+z)² = (2x)²+(−y)²+z²+(2×2x×−y)+(2×−y×z)+(2×z×2x)
= 4x²+y²+z²–4xy–2yz+4xz
(iii) (−2x+3y+2z)²
Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx
Here, x = −2x
y = 3y
z = 2z
(−2x+3y+2z)² = (−2x)²+(3y)²+(2z)²+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)
= 4x²+9y²+4z²–12xy+12yz–8xz
(iv) (3a –7b–c)²
Using identity (x+y+z)²= x²+y²+z²+2xy+2yz+2zx
Here, x = 3a
y = – 7b
z = – c
(3a –7b– c)² = (3a)²+(– 7b)²+(– c)²+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)
= 9a² + 49b² + c²– 42ab+14bc–6ca
(v) (–2x+5y–3z)²
Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx
Here, x = –2x
y = 5y
z = – 3z
(–2x+5y–3z)² = (–2x)² + (5y)² + (–3z)² + (2 × –2x × 5y) + (2 × 5y× – 3z)+(2×–3z ×–2x)
= 4x²+25y² +9z²– 20xy–30yz+12zx
Question 10: If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + leave the same remainder when divided by z – 3, find the value of a.
Answer 10: Zero of the polynomial,
g1(z) = 0
z3 = 0
z = 3
Hence, zero of g(z) = – 2a
Let p(z) = az³+4z²+3z4
Now, substituting the given value of z = 3 in p(z), we get,
p(3) = a (3)³ + 4 (3)² + 3 (3) – 4
⇒p(3) = 27a+36+94
⇒p(3) = 27a+41
Let h(z) = z³4z+a
Now, by substituting the value of z = 3 in h(z), we get,
h(3) = (3)³4(3)+a
⇒h(3) = 2712+a
⇒h(3) = 15+a
As per the question,
The two polynomials, p(z) and h(z), leave the same remainder when divided by z3
So, h(3)=p(3)
⇒15+a = 27a+41
⇒1541 = 27a – a
⇒26 = 26a
⇒a = 1
Question 11: Compute the perimeter of a rectangle whose area is 25x² – 35x + 12.
Answer 11: Area of rectangle = 25x² – 35x + 12
We know the area of a rectangle = length × breadth
So, by factoring 25x² – 35x + 12, the length and breadth can be obtained.
25x² – 35x + 12 = 25x² – 15x – 20x + 12
=> 25x² – 35x + 12 = 5x(5x – 3) – 4(5x – 3)
=> 25x² – 35x + 12 = (5x – 3)(5x – 4)
Thus, the length and breadth of a rectangle are (5x – 3)(5x – 4).
So, the perimeter = 2(length + breadth)
Therefore, the perimeter of the given rectangle = 2[(5x – 3)+(5x – 4)]
= 2(5x – 3 + 5x – 4)
= 2(10x – 7)
= 20x – 14
Hence, the perimeter of the rectangle = 20x – 14
Question 12: 2x²+y²+²–2√2xy+4√2yz–8xz
Answer 12: Using identity, (x +y+z)² = x²+y²+z²+2xy+2yz+2zx
We can say that, x²+²+²+2xy+2yz+2zx = (x+y+z)²
2x²+y²+8z²–2√2xy+4√2yz–8xz
= (√2x)²+(y)²+(2√2z)²+(2×√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
= (−√2x+y+2√2z)²
= (−√2x+y+2√2z)(−√2x+y+2√2z)
Question 13: If ? + 2? is a factor of ? ⁵ – 4?²?³ + 2? + 2? + 3, find a.
Answer 13: According to the question,
Let p(x) = x ⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a
g(x) = 0
⟹ x + 2a = 0
⟹ x = – 2a
Hence, zero of g(x) = – 2a
As per the factor theorem,
If g(x) is a factor of p(x), then p( – 2a) = 0
So, substituting the value of x in p(x), we get,
p ( – 2a) = ( – 2a) ⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0
⟹ – 32a ⁵ + 32a ⁵ – 2a + 3 = 0
⟹ – 2a = – 3
⟹ a = 3/2
Question 14: Find the value of x³+ y ³ + z ³ – 3xyz if x² + y² + z² = 83 and x + y + z = 1
Answer 14: Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
So,
(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)
From the question, x² + y² + z²= 83 and x + y + z = 15
So,
152 = 83 + 2(xy + yz + xz)
=> 225 – 83 = 2(xy + yz + xz)
Or, xy + yz + xz = 142/2 = 71
Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),
x ³ + y ³ + z ³ – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))
Now,
x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71
So, x ³ + y ³ + z ³ – 3xyz = 15(83 – 71)
=> x ³ + y ³ + z ³ – 3xyz = 15 × 12
Or, x ³ + y ³ + z ³ – 3xyz = 180
Question 15:Verify that:
(i) x³+y³ = (x+y)(x²–xy+y²)
(ii) x³–y³ = (x–y)(x²+xy+y²)
Answer 15:(i) x³+y³ = (x+y)(x²–xy+y²)
We know that (x+y)³= x³+y³+3xy(x+y)
⇒ x³+y³ = (x+y)³–3xy(x+y)
⇒ x³+y³ = (x+y)[(x+y)²–3xy]
Taking (x+y) common ⇒ x³+y³ = (x+y)[(x²+y²+2xy)–3xy]
⇒ x³+y³ = (x+y)(x²+y²–xy)
(ii) x³–y³ = (x–y)(x²+xy+y²)
We know that (x–y)³ = x³–y³–3xy(x–y)
⇒ x³−y³ = (x–y)³+3xy(x–y)
⇒ x³−y³ = (x–y)[(x–y)²+3xy]
Taking (x+y) common ⇒ x³−y³ = (x–y)[(x²+y²–2xy)+3xy]
⇒ x³+y³ = (x–y)(x²+y²+xy)
Question 16: For what value of m is ?³ – 2??² + 16 divisible by x + 2?
Answer 16: According to the question,
Let p(x) = x³ – 2mx² + 16, and g(x) = x + 2
g(x) = 0
⟹ x + 2 = 0
⟹ x = – 2
Hence, zero of g(x) = – 2
As per the factor theorem,
if p(x) is divisible by g(x), then the remainder of p(−2) should be zero.
Thus, substituting the value of x in p(x), we obtain,
p( – 2) = 0
⟹ ( – 2)³ – 2m( – 2)² + 16 = 0
⟹ 0 – 8 – 8m + 16 = 0
⟹ 8m = 8
⟹ m = 1
Question 17:If a + b + c = 15 and a² + b² + c² = 83, find the value of a³ + b³ + c³ – 3abc.
Answer 17: We know that,
a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) ….(i)
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca ….(ii)
Given, a + b + c = 15 and a² + b² + c² = 83
From (ii), we have
152 = 83 + 2(ab + bc + ca)
⇒ 225 – 83 = 2(ab + bc + ca)
⇒ 142/2 = ab + bc + ca
⇒ ab + bc + ca = 71
Now, (i) can be written as
a³ + b³ + c³ – 3abc = (a + b + c)[(a² + b² + c² ) – (ab + bc + ca)]
a³ + b³+ c³ – 3abc = 15 × [83 – 71] = 15 × 12 = 180.
Question 18: Factorise: 27x³+y³+z³–9xyz
Answer 18: The expression27x³+y³+z³–9xyz can be written as (3x)³+y³+z³–3(3x)(y)(z)
27x³+y³+z³–9xyz = (3x)³+y³+z³–3(3x)(y)(z)
We know that x³+y³+³–3xyz = (x+y+z)(x²+y²+z²–xy –yz–zx)
27x³+y³+z³–9xyz = (3x)³+y³+z³–3(3x)(y)(z)
= (3x+y+z)[(3x)²+y²+z²–3xy–yz–3xz]
= (3x+y+z)(9x²+y²+²–3xy–yz–3xz)
Question 19: If (x – 1/x) = 4, then evaluate (x² + 1/x²) and (x⁴ + 1/x⁴).
Answer 19: Given, (x – 1/x) = 4
Squaring both sides, we get,
(x – 1/x)² = 16
⇒ x² – 2.x.1/x + 1/x² = 16
⇒ x² – 2 + 1/x² = 16
⇒ x² + 1/x² = 16 + 2 = 18
∴ (x² + 1/x²) = 18 ….(i)
Again, squaring both sides of (i), we get
(x² + 1/x²)² = 324
⇒ x⁴ + 2.x².1/x² + 1/x⁴= 324
⇒ x⁴ + 2 + 1/x⁴ = 324
⇒ x⁴ + 1/x⁴ = 324 – 2 = 322
∴ (x⁴ + 1/x⁴) = 322.
Question 20: Factorise
64m³–343n³
Answer 20: The expression 64m³–343n³ can be written as (4m)³–(7n)³
64m³–343n³ =(4m)³–(7n)³
We know that x³–y³ = (x–y)(x²+xy+y²)
64m³–343n³ = (4m)³–(7n)³
= (4m7n)[(4m)²+(4m)(7n)+(7n)²]
= (4m7n)(16m²+28mn+49n²)
Question 21: Find out the values of a and b so that (2x³ + ax² + x + b) has (x + 2) and (2x – 1) as factors.
Answer 21: Let p(x) = 2x³ + ax² + x + b. Then, p( –2) = and p(½) = 0.
p(2) = 2(2)³ + a(2)² + 2 + b = 0
⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)
p(½) = 2(½)³ + a(½)² + (½) + b = 0
⇒ a + 4b = –3 ….(ii)
On solving (i) and (ii), we get a = 5 and b = –2.
Hence, a = 5 and b = –2.
Question 22: Explain that p – 1 is a factor of p¹⁰ – 1 and p¹¹ – 1.
Answer 22: According to the question,
Let h(p) = ?¹⁰ − 1,and g(p) = ? – 1
zero of g(p) ⇒ g(p) = 0
p – 1 = 0
p = 1
Therefore, zero of g(x) = 1
We know that,
According to the factor theorem, if g(p) is a factor of h(p), then h(1) should be zero
So,
h(1) = (1) ¹⁰ − 1 = 1 − 1 = 0
⟹ g (p) is a factor of h(p).
Here, we have h(p) = ?¹¹ − 1, g (p) = ? – 1
Putting g (p) = 0 ⟹ ? − 1 = 0 ⟹ ? = 1
As per the factor theorem, if g (p) is a factor of h(p),
Then h(1) = 0
⟹ (1) ¹¹ – 1 = 0
Hence, g(p) = ? – 1 is the factor of h(p) = ? ¹⁰ – 1
Question 23: Examine whether (7 + 3x) is a factor of (3×3 + 7x).
Answer 23: Let p(x) = 3×3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.
By the remainder theorem, p(x) is divided by g(x), and then the remainder is p(–7/3).
Now, p(–7/3) = 3(–7/3)3 + 7(–7/3) = –490/9 ≠ 0.
∴ g(x) is not a factor of p(x).
Question 24:Prove that:
x³+y³+z³–3xyz = (1/2) (x+y+z)[(x–y)²+(y–z)²+(z–x)²]
Answer 24: We know that,
x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²–xy–yz–xz)
⇒ x³+y³+z³–3xyz = (1/2)(x+y+z)[2(x²+y²+z²–xy–yz–xz)]
= (1/2)(x+y+z)(2×2+2y²+²–2xy–2yz–2xz)
= (1/2)(x+y+z)[(x²+y²−2xy)+(y²+z²–2yz)+(x²+z²–2xz)]
= (1/2)(x+y+z)[(x–y)²+(y–z)²+(z–x)²]
Question 25: Find out which of the following polynomials has x – 2 a factor:
(i) 3?² + 6?−24.
(ii) 4?² + ?−2.
Answer 25: (i) According to the question,
Let p(x) =3?² + 6?−24 and g(x) = x – 2
g(x) = x – 2
zero of g(x) ⇒ g(x) = 0
x – 2 = 0
x = 2
Hence, zero of g(x) = 2
Thus, substituting the value of x in p(x), we get,
p(2) = 3(2)² + 6 (2) – 24
= 12 + 12 – 24
= 0
the remainder = zero,
We can derive that,
g(x) = x – 2 is factor of p(x) = 3?² + 6?−24
(ii) According to the question,
Let p(x) = 4?² + ?−2 and g(x) = x – 2
g(x) = x – 2
zero of g(x) ⇒ g(x) = 0
x – 2 = 0
x = 2
Hence, zero of g(x) = 2
Thus, substituting the value of x in p(x), we get,
p(2) = 4(2)² + 2−2
= 16 ≠ 0
Since the remainder ≠ zero,
We can say that,
g(x) = x – 2 is not a factor of p(x) = 4?² + ?−2
Question 26: Factorise x² + 1/x² + 2 – 2x – 2/x.
Answer 26: x² + 1/x² + 2 – 2x – 2/x = (x² + 1/x² + 2) – 2(x + 1/x)
= (x + 1/x)² – 2(x + 1/x)
= (x + 1/x)(x + 1/x – 2).
Question 27: Factorise
8a³+b³+12a²b+6ab²
Answer 27: The expression, 8a³+b³+12a²b+6ab² can be written as (2a)³+b³+3(2a)²b+3(2a)(b)²
8a³+b³+12a²b+6ab² = (2a)³+b³+3(2a)²b+3(2a)(b)²
= (2a+b)³
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)³ = x³+y³+3xy(x+y) is used.
Question 28: By Remainder Theorem, find out the remainder when p(x) is divided by g(x), where
(i) p(?) = ?³ – 2?² – 4? – 1, g(?) = ? + 1
(ii) p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3
(iii) p(?) = 4?³ – 12?² + 14? – 3, g(?) = 2? – 1
(iv) p(?) = ?³ – 6?² + 2? – 4, g(?) = 1 – 3/2 ?
Answer 28: (i) Given p(x) = ?³ – 2?² – 4? – 1 and g(x) = x + 1
Here zero of g(x) = – 1
By applying the remainder theorem
P(x) divided by g(x) = p( – 1)
P ( – 1) = ( – 1)³ – 2 ( – 1)² – 4 ( – 1) – 1 = 0
Therefore, the remainder = 0
(ii) given p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3
Here zero of g(x) = 3
By applying the remainder theorem p(x) divided by g(x) = p(3)
p(3) = 3³ – 3 × (3)² + 4 × 3 + 50 = 62
Therefore, the remainder = 62
(iii) p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1
Here zero of g(x) = ½
By applying the remainder theorem p(x) divided by g(x) = p (½)
P( ½ ) = 4( ½ )³ – 12( ½ )² + 14 ( ½ ) – 3
= 4/8 – 12/4 + 14/2 – 3
= ½ + 1
= 3/2
Hence, the remainder = 3/2
(iv) p(?) = ?³ – 6?² + 2? – 4, g(?) = 1 – 3/2 ?
so, zero of g(x) = 2/3
By applying the remainder theorem p(x) divided by g(x) = p(2/3)
p(2/3) = (2/3)³ – 6(2/3)² + 2(2/3) – 4
= – 136/27
Therefore, the remainder = – 136/27
Question 29:Factorise x² – 1 – 2a – a².
Answer 29: x² – 1 – 2a – a² = x² – (1 + 2a + a²)
= x² – (1 + a)²
= [x – (1 – a)][x + 1 + a]
= (x – 1 – a)(x + 1 + a)
∴ x² – 1 – 2a – a² = (x – 1 – a)(x + 1 + a).
Question 30:Evaluate the following using suitable identity
(998)³
Answer 30: We can write 99 as 1000–2
Using identity,(x–y)³ = x³ –y³ –3xy(x–y)
(998)³ =(1000–2)³
=(1000)³ –2³ –(3×1000×2)(1000–2)
= 1000000000–8–6000(1000– 2)
= 1000000000–8 6000000+12000
= 994011992
Question 31: Find the zeroes of the polynomial:
p(?)= (? –2)² −(? + 2)²
Answer 31: p(x) = (? –2)² −(? + 2)²
We know that,
Zero of the polynomial p(x) = 0
Hence, we get,
⇒ (x–2)² −(x + 2)² = 0
Expanding using the identity, a² – b² = (a – b) (a + b)
⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0
⇒ 2x ( – 4) = 0
⇒ – 8 x= 0
Therefore, the zero of the polynomial = 0
Benefits Of Solving Important Questions Class 9 Mathematics Chapter 2
Consistently solving questions is a vital element of mastering Mathematics. By solving Mathematics Class 9 Chapter 2 important questions, students can get a further understanding of the polynomials chapter.
A few other advantages of solving Important Questions Class 9 Mathematics Chapter 2 are:
 Class 9 Mathematics Chapter 2 important questions provide details about the types of questions that may be expected in the exams, which increases their confidence in achieving a high grade. .
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Q.1 By actual division, find the quotient and the remainder when x^{5} + 1 is divided by x 1
Marks:3
Ans
$\begin{array}{l}\text{}{\mathrm{x}}^{4}+{\mathrm{x}}^{3}+{\mathrm{x}}^{2}+\mathrm{x}+1\\ \mathrm{x}1\overline{){\mathrm{x}}^{5}+1\text{}}\\ \text{}{\mathrm{x}}^{5}{\mathrm{x}}^{4}\\ \text{}\text{+}\\ \text{}\overline{\begin{array}{l}{\text{x}}^{4}+1\\ {\text{x}}^{4}{\mathrm{x}}^{3}\\ \text{+}\\ \overline{\begin{array}{l}{\text{x}}^{3}\text{+ 1}\\ {\text{x}}^{3}{\mathrm{x}}^{2}\\ \text{+}\end{array}}\\ \text{}\overline{\begin{array}{l}{\text{x}}^{2}+1\\ {\text{x}}^{2}\mathrm{x}\\ \text{+}\\ \overline{\begin{array}{l}\text{x + 1}\\ \text{x 1}\\ \text{+}\\ \overline{\text{2}}\end{array}}\end{array}}\end{array}}\\ \mathrm{Quotient}:\text{}{\mathrm{x}}^{4}+{\mathrm{x}}^{3}+{\mathrm{x}}^{2}+\mathrm{x}+1\\ \mathrm{Remainder}:\text{2}\end{array}$
Q.2 Find the value of k if x 5 is a factor of kx^{2} + 3x + 7.
Marks:2
Ans
$\begin{array}{l}\mathrm{Zero}\text{of x 5 is 5}\mathrm{as}\text{}\mathrm{x}\text{}\text{5}=\text{}0\text{}\mathrm{gives}\text{}\mathrm{x}\text{}=\text{5}.\\ {\text{p(x) = kx}}^{2}+3\mathrm{x}+7\\ \mathrm{p}\left(5\right)=0\\ 25\mathrm{k}+15+7=0\\ 25\mathrm{k}+22=0\\ \mathrm{k}=\frac{22}{25}\end{array}$
Q.3 If x + y + z = 6 and xy + yz + zx = 11, then find the value of x^{2} + y^{2} + z^{2}_{.}
Marks:3
Ans
$\begin{array}{l}\mathrm{Given}:\text{}\mathrm{x}+\mathrm{y}+\mathrm{z}=6\text{and}\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=11\\ \mathrm{Squaring}\text{}\mathrm{both}\text{}\mathrm{sides},\text{}\mathrm{we}\text{}\mathrm{get}\\ \text{}{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}^{2}={\left(6\right)}^{2}\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\mathrm{xy}+2\mathrm{yz}+2\mathrm{zx}=36\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)\text{}=\text{}36\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\left(11\right)\text{}=\text{}36\text{}\left[\text{Since}\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\text{}=\text{}11\right]\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+22=36\\ \text{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}\text{}=\text{}3622\text{}=\text{}14\end{array}$
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FAQs (Frequently Asked Questions)
1. What are the four types of polynomials?
The 4 types of polynomials are zero polynomial, linear polynomial, quadratic polynomial, and cubic polynomial.
2. Where can I get important questions for Class 9 Mathematics Chapter 2 online?
On the Extramarks website, you can find all of the important questions for Class 9 Mathematics Chapter 2, along with their answers. On the website, you can also find important questions and NCERT solutions for all classes from 1 to 12.
3. What are the important chapters in Class 9 Mathematics?
The NCERT Mathematics book has 15 chapters. Each chapter is equally important when it comes to learning the fundamentals and taking the test. Additionally, because CBSE does not specify the distribution of marks for each chapter, students are advised to fully study all chapters. Each and every chapter must be completely understood to acquire a good grade in exams.
All the fifteen chapters of CBSE Class 9 Mathematics syllabus are given below:
 Chapter – Number Systems
 Chapter – Polynomials
 Chapter – Coordinate Geometry
 Chapter – Linear Equations In Two Variables
 Chapter – Introduction To Euclid’s Geometry
 Chapter – Lines And Angles
 Chapter – Triangles
 Chapter – Quadrilaterals
 Chapter – Areas Of Parallelograms And Triangles
 Chapter – Circles
 Chapter – Constructions
 Chapter – Heron’s Formula
 Chapter – Surface Areas And Volumes
 Chapter – Statistics
 Chapter – Probability