NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials

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NCERT Exemplar Class 9 Maths Solutions Polynomials

Exercise 2.1: Multiple Choice Questions (MCQs)

Question 1:
Which one of the following is a polynomial?

Solution:

Question 2:
√2 is a polynomial of degree
(a) 2             (b) 0           (c) 1          (d)½
Solution:
(b) √2 = -√2x°. Hence, √2 is a polynomial of degree 0, because exponent of x is 0.

Question 3:
Degree of the polynomial 4x⁴ + Ox³ + Ox⁵ + 5x+ 7 is
(a) 4             (b) 5         (c) 3            (d) 7
Solution:
(a) Degree of 4x⁴ + Ox³ + Ox⁵ + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4,
Hence, the degree of a polynomial is 4.

Question 4:
Degree of the zero polynomial is
(a) 0                                                   (b) 1
(c) any natural number                 (d) not defined
Solution:
(d) The degree of zero polynomial is not defined, because in zero polynomial, the coefficient of any variable is zero i.e., Ox² or Ox⁵,etc.
Hence, we cannot exactly determine the degree of variable.

Question 5:
If p (x) = x2 – 2√2x + 1, then p (2√2) is equal to
(a) 0                 (b) 1                   (c) 4√2              (d) 8 √2 +1
Solution:
(b) Given, p(x) = x² – 2√2x + 1 …(i)
On putting x = 2√2 in Eq. (i), we get
P(2√2) = (2√2)²– (2√2)(2√2) + 1 =8-8+1=1

Question 6:
The value of the polynomial 5x – 4x² + 3, when x = – 1 is
(a)-6          (b) 6           (c) 2                    (d) -2
Solution:
(a) Let p (x) = 5x – 4x² + 3 …(i)
On putting x = -1 in Eq. (i), we get
p(-1) = 5(-1) -4(-1)² + 3= -5-4+3 = -6

Question 7:
If p (x) = x + 3, then p(x)+ p (- x) is equal to
(a) 3               (b) 2x            (c) 0                   (d) 6
Solution:
(d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3
Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6

Question 8:
Zero of the zero polynomial is
(a) 0        (b) 1           (c) any real number               (d) not defined
Solution:
(c) Zero of the zero polynomial is any real number.
e.g., Let us consider zero polynomial be 0(x-k), where k is a real number For determining the zero, putx-k = 0=>x = k Hence, zero of the zero polynomial be any real number.

Question 9:
Zero of the polynomial p(x)=2x+5 is
(a) -2/5       (b) -5/2         (c)2/5            (d)5/2
Solution:
(b) Given, p(x) = 2x+5
For zero of the polynomial, put p(x) = 0   ∴ 2x + 5 = 0
=> -5/2
Hence, zero of the polynomial p(x) is -5/2.

Question 10:
One of the zeroes of the polynomial 2x² + 7x – 4 is
(a) 2              (b)½        (c)-1              (d)-2
Thinking Process
(i) Firstly, determine the factor by using splitting method.
(ii) Further, put the factors equals to zero, then determine the values of x.
Solution:
(b) Let p (x) = 2x² + 7x-4
= 2x² + 8x-x-4 [by splitting middle term]
= 2x(x+ 4)-1(x+ 4)
= (2x-1)(x+ 4)
For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0
=> 2x-1 = 0 and x+4 = 0
=> x = ½ and x = -4
Hence, one of the zeroes of the polynomial p(x) is ½.

Question 11:
If x⁵¹ + 51 is divided by x + 1, then the remainder is
(a) 0        (b) 1             (c) 49             (d) 50
Solution:
(d) Let p(x) = x⁵¹ + 51 . …(i)
When we divide p(x) by x+1, we get the remainder p(-1)
On putting x= -1 in Eq. (i), we get p(-1) = (-1)⁵¹ + 51
= -1 + 51 = 50
Hence, the remainder is 50.

Question 12:
If x + 1 is a factor of the polynomial 2x² + kx, then the value of k is
(a) -3      (b) 4      (c) 2            (d)-2
Solution:
(c) Let p(x) = 2x² + kx
Since, (x + 1) is a factor of p(x), then
p(-1)=0
2(-1)2 + k(-1) = 0
=> 2-k = 0 => k= 2
Hence, the value of k is 2.

Question 13:
x + 1 is a factor of the polynomial
(a) x³ + x² – x +1         (b) x³ + x² + x+1
(c) x⁴ + x³ + x² +1      (d) x⁴ + 3x³ + 3x² + x +1
Solution:
(b) Let assume (x + 1) is a factor of x³ + x² + x+1.
So, x = -1 is zero of x³ + x² + x+1
(-1)³ + (-1)² + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true.

Question 14:
One of the factors of (25x² – 1) + (1 + 5x)² is
(a) 5 + x               (b) 5 – x         (c) 5x -1           (d) 10x
Solution:
(d) Now, (25x² -1) + (1 + 5x)²
= 25x² -1 + 1 + 25x² + 10x [using identity, (a + b)² = a² + b² + 2ab]
= 50x² + 10x = 10x (5x+ 1)
Hence, one of the factor of given polynomial is 10x.

Question 15:
The value of 249² – 248² is
(a) 1²        (b) 477           (c) 487           (d) 497
Solution:
(d) Now, 249² – 248² = (249 + 248) (249 – 248) [using identity, a² – b² = (a – b)(a + b)]
= 497 x 1 = 497.

Question 16:
The factorization of 4x² + 8x+ 3 is
(a) (x +1) (x + 3)             (b) (2x+1) (2x + 3)
(c) (2x + 2) (2x + 5)       (d) (2x -1) (2x – 3)
Solution:
(b) Now, 4x² + 8x + 3= 4x² + 6x + 2x + 3 [by splitting middle term]
= 2x(2x+ 3) + 1 (2x+ 3)
= (2x + 3) (2x + 1)

Question 17:
Which of the following is a factor of (x+ y)³ – (x³ + y³)?
(a) x² + y² + 2 xy      (b) x² + y² – xy         (c) xy²            (d) 3xy
Solution:
(d) Now, (x+ y)3 – (x³ + y³) = (x + y) – (x + y)(x²– xy + y²)
[using identity, a³ + b³ = (a + b)(a² -ab+ b²)] = (x+ y)[(x+ y)² -(x² -xy+ y²)]
= (x+ y)(x²+ y²+ 2xy- x²+ xy- y²)
[using identity, (a + b)² = a² + b² + 2 ab)]
= (x + y) (3xy)
Hence, one of the factor of given polynomial is 3xy.

Question 18:
The coefficient of x in the expansion of (x + 3)³ is
(a) 1     (b) 9     (c) 18     (d) 27
Solution:
(d) Now, (x + 3)³ = x³ + 3³ + 3x (3)(x + 3)
[using identity, (a + b)³ = a³ + b³ + 3ab (a + b)]
= x³ + 27 + 9x (x + 3)
= x³ +27 + 9x²+27x Hence, the coefficient of x in (x + 3)³ is 27.

Question 19:

Solution:

Question 20:

Solution:

Question 21:
If a + b + c =0, then a³+b³ + c³ is equal to
(a) 0                   (b) abc
(c) 3abc             (d) 2abc
Solution:
(d) Now, a³+b³ + c³= (a+ b + c) (a² + b² + c² – ab – be – ca) + 3abc
[using identity, a³+b³ + c³ – 3 abc = (a + b + c)(a² + b² + c² – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given]
a³+b³ + c³ = 3abc

Exercise 2.2: Short Answer Type Questions

Question 1:
Which of the following expressions are polynomials? Justify your answer,

Solution:

Question 2:
Write whether the following statements are true or false. Justify your answer. ’
(i) A Binomial can have atmost two terms.
(ii) Every polynomial is a Binomial.
(iii) A binomial ipay have degree 5.
(iv) Zero of a polynomial is always 0.
(v) A polynomial cannot have more than one zero.
(vi) The degree of the sum of two polynomials each of degree 5 is always 5.
Solution:
(i) False, because a binomial has exactly two terms.
(ii) False, because every polynomial is not a binomial .
e.g., (a) 3x² + 4x + 5 [polynomial but hot a binomial]
(b) 3x² + 5 [polynomial and also a binomial]
(Hi) True, because a binomial is a polynomial whose degree is a whole number greater than equal to one. So, it may have degree 5.
(iv) False, because zero of a polynomial can be any real number e.g., p(x) = x – 2, then 2 is a zero of polynomial p(x).
(v) False, because a polynomial can have any number of zeroes. It depends upon the degree of the polynomial
e.g., p(x) = x² -2, as degree pf p(x) is 2 ,so it has two degree, so it has two zeroes i.e., √2 and —√2.
(vi) False, because the sum of any two polynomials of same degree is not always same degree.
e.g., Let f(x) = x⁴ + 2 and g(x) = -x⁴ + 4x³ + 2x

∴ Sum of two polynomials,
f{x) + g(x) = x⁴ + 2 + (-x⁴ + 4x³ + 2x)
= 4x³ + 2x + 2 which is not a polynomial of degree 4.

Exercise 2.3: Short Answer Type Questions

Question 1:
Classify the following polynomials as polynomials in one variable, two variables etc.
(i) x² + x +1           (ii) y³ – 5y
(iii) xy+ yz +zx     (iv) x² – Zxy + y² +1
Solution:
(i) Polynomial x²+ x+ 1 is a one variable polynomial, because it contains only one variable i.e., x.
(ii) Polynomial y³ – 5y is a one variable polynomial, because it contains only one variable i.e., y.
(iii) Polynomial xy+ yz+ zx is a three variables polynomial, because it contains three variables x, y and z.
(iv) Polynomial  x² – Zxy + y² +1 is a two variables pplynomial, because it contains two variables x and y.

Question 2:
Determine the degree of each of the following polynomials.
(i) 2x-1                       (ii) -10
(iii) x³-9x+ 3x⁵        (iv) y³(1-y⁴)
Solution:
(i) Degree of polynomial 2x-1 is one, Decause the maximum exponent of x is one.
(ii) Degree of polynomial -10 or -10x° is zero, because the exponent of x is zero.
(iii) Degree of polynomial x3 – 9x + 3xs is five, because the maximum exponent of x is five.
(iv) Degree of polynomial y³(1-y⁴) or y³ – y⁷ is seven, because the maximum exponent of y is seven.

Question 3:

Solution:

Question 4:
Write the coefficient of x² in each of the following

Solution:

Question 5:
Classify the following as a constant, linear, quadratic and cubic polynomials

Thinking Process
(i) Firstly check the maximum exponent of the variable..
(ii) If the maximum exponent of a variable is 0, then it is a constant polynomial.
(iii) If the maximum exponent of a variable is 1, then it is a linear polynomial.
(iv) If the maximum exponent of a variable is 2, then it is a quadratic polynomial.
(v) If the maximum exponent of a variable is 3, then it is a cubic polynomial.
Solution:

(i) Polynomial 2 – x² + x³ is a cubic polynomial, because maximum exponent of x is 3.
(ii) Polynomial 3x³ is a cublic polynomial, because maximum exponent of x is 3.
(iii) Polynomial 5t -√7 is a linear polynomial, because maximum exponent of t is 1.
(iv) Polynomial 4- 5y² is a quadratic polynomial, because maximum exponent of y is 2.
(v) Polynomial 3 is a constant polynomial, because the exponent of variable is 0. ’
(vi) Polynomial 2 + x  is a linear polynomial, because maximum exponent of x is 1.
(vii) Polynomial y³ – y is a cubic polynomial, because maximum exponent of y is 3.
(viii) Polynomial 1 + x+ x² is a quadratic polynomial, because maximum exponent of xis 2.
(ix) Polynomial t² is a quadratic polynomial, because maximum exponent of t is 2.
(x) Polynomial √2x-1 is a linear polynomial, because maximum exponent of xis 1.

Question 6:
Give an example of a polynomial, which is
(i) monomial of degree 1.
(ii) -binomial of degree 20.
(iii) trinomial of degree 2.
Solution:
(i) The example of monomial of degree 1 is 5y or 10x.
(ii) The example of binomial of degree 20 is 6x²⁰ + x¹¹ or x²⁰ +1
(iii) The example of trinomial of degree 2 is x² – 5x+ 4 or 2x² -x-1

Question 7:
Find the value of the polynomial 3x³ – 4x² + 7x – 5, when x = 3 and also when x = -3.
Solution:
Let p(x) =3x³ – 4x² + 7x – 5
At x= 3, p(3) = 3(3)³ – 4(3)² + 7(3) – 5
= 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61
At x = -3, p(-3)= 3(—3)³ – 4(-3)² + 7(-3)- 5
= 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143
Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

Question 8:
If p (x) = x² – 4x + 3, then evaluate p(2) – p (-1) + p ( ½).
Solution:

Question 9:
Find p(0), p( 1) and p(-2) for the following polynomials
(i) p(x) = 10x – 4x² – 3 (ii) p(y) = (y + 2)(y – 2)
Solution:
(i) Given, polynomial is
p(x) = 10x – 4x² – 3
On putting x = 0,1 and – 2, respectively in Eq. (i), we get p(0) = 10(0)-4(0)² -3 = 0-0-3= -3
p(1) = 10 (1) — 4 (1 )² -3
= 10-4-3= 10-7= 3
and p(-2) =10 (-2)- 4 (-2)² – 3
= -20-4×4-3 =-20-16-3=-39
Hence, the values of p(0), p(1) and p(-2) are respectively, -3,3 and – 39.
(ii) Given, polynomial isp(y) = (y+2)(y-2)
On putting y =0,1 and -2, respectively in Eq. (i), we get p(0) =(0+2)(0-2)= -4
p(1) = (1 + 2)(1-2)
= 3 x (-1) = -3
and p(-2) = (-2 + 2)(-2 -2)
=0 (-4) = 0
Hence, the values of p(0),p(1) and p(-2) are respectively,-4,-3 and 0.

Question 10:
Verify whether the following are true or false.
(i) -3 is a zero of at – 3
(ii) -1/3 is a zero of 3x+1
(iii) -4/5 is a zero of 4 – 5y
(iv) 0 and 2 are the zeroes of t² – 2t
(v) -3 is a zero of y² +y-6
Solution:

Question 11:
Find the zeroes of the polynomial in each of the following,
(i) p(x)=x – 4           (ii) g(x)= 3 – 6x
(iii) q(x) = 2x – 7    (iv) h(y) = 2y
Solution:
(i) Given, polynomial is
p(x) = x- 4
For zero of polynomial, put p(x) = x-4 = 0
=> x= 4
Hence, zero of polynomial is 4.
(ii) Given, polynomial is
g(x) = 3-6x
For zero of polynomial, put g(x) = 0
3-6x= 0 => 6x =3 => x=1/2.
Hence, zero of polynomial is X
(iii) Given, polynomial is q(x) = 2x -7 For zero of polynomial, put q(x) =  2x-7 = 0
2x= 7 => x =7/2
Hence, zero of polynomial is
(iv) Given polynomial h(y) = 2 y For zero of polynomial, put h(y) = 0
2y= 0
Hence, the zero of polynomial is 0,

Question 12:
Find the zeroes of the polynomial p(x)= (x – 2)² – (x+ 2)².
Solution:
Given, polynomial is p(x) = (x – 2)² – (x+ 2)²
For zeroes of polynomial, put p(x) = 0
(x – 2)² – (x+ 2)²=0
(x-2 + x+2)(x-2-x-2) = 0 [using identity, a²-b² =(a-b)(a + b)] =» (2x)(-4) = 0

Question 13:
By actual division, find the quotient and the remainder when the first
polynomial is divided by the second polynomial x⁴ + 1 and x-1.
Solution:

Question 14:
By remainder theorem, find the remainder when p(x) is divided by g(x)
(i) p(x) = x³-2x²-4x-1, g(x)=x + 1
(ii) p(x) = x³ -3x² + 4x + 50, g(x)= x – 3
(iii) p(x) = x³ – 12x² + 14x -3, g(x)= 2x – 1 – 1
(iv) ‘p(x) = x³-6x²+2x-4, g(x) = 1 -(3/2) x
Solution:

Question 15:
Check whether p(x) is a multiple of g(x) or not
(i) p(x) = x³ – 5x² + 4x – 3,  g(x) = x – 2.
(ii) p(x) = 2x³ – 11x² – 4x+ 5,  g(x) = 2x + l

Thinking Process
(i) Firstly, find the zero of g(x) and then put the value of in p(x) and simplify it.
(ii) lf p(a) = Q then p(x) is a multiple of g(x) and f(p(a) # Q then p(x) is not a multiple of g(x) where ‘a’ is a zero of g (x)).
Solution:

Question 16:
Show that,
(i) x + 3 is a factor of 69 + 11c – x² + x³
(ii) 2x – 3 is a factor of x + 2x³ -9x² +12
Solution:

Question 17:
Determine which of the following polynomial has x – 2 a factor (i) 3x² + 6x – 24 (ii) 4x²+ x – 2
Solution:

Question 18:
Show that p-1 is a factor of p¹⁰ -1 and also of p¹¹ -1.
Solution:
Let g (p) = p¹⁰ -1
and h(p) = p¹¹ -1.
On putting p=1 in Eq. (i), we get
g(1)=1¹⁰-1= 1-1=0 Hence, p-1 is a factor of g(p).
Again, putting p = 1 in Eq. (ii), we get
h (1) = (1)¹¹ —1 = 1 —1 = 0 Hence, p -1 is a factor of h(p).

Question 19:
For what value of m is x³ -2mx² +16 divisible by x + 2?
Solution:
Let p(x) = x³ -2mx² +16
Since, p(x) is divisible by (x+2), then remainder = 0
P(-2) = 0
=> (-2)³ -2m(-2)² + 16=0
=> -8-8m+16=0
=> 8=8 m
m = 1
Hence, the value of m is 1 .

Question 20:
If x + 2a is a factor of a⁵ -4a²x³ +2x + 2a +3, then find the value of a.
Solution:
Let p(x) =a⁵ -4a²x³ +2x + 2a +3
Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (-2a)⁵ – 4a² (-2a)³ + 2(-2a) + 2a + 3 = 0 => -32a⁵ + 32a⁵ -4a + 2a+ 3 = 0
=> -2a + 3=0
2a =3
a = 3/2.
Hence, the value of a is 3/2.

Question 21:
Find the value of m, so that 2x -1 be a factor of
8x⁴ +4x³ -16x² +10x+07.
Solution:

Question 22:
If x +1 is a factor of ax³ +x² -2x+4a-9, then find the value of a.
Solution:

Question 23:
Factorise
(i) x² + 9x +18        (ii) 6x² +7x -3
(iii) 2x² -7x.-15       (iv) 84-2r-2r²
Solution:

Question 24:
Factorise
(i) 2x³ -3x² -17x + 30         (ii) x³ -6x² +11 x-6
(iii) x³ + x² – 4x – 4             (iv) 3x³ – x² – 3x +1
Thinking Process
(i) Firstly, find the prime factors of constant term in given polynomial.
(ii) Substitute these factors in place of x in given polynomial and find the minimum factor which satisfies given polynomial and write it in the form of a linear polynomial.
(iii) Now, adjust the given polynomial in such a way that it becomes the product of two factors, one of them is a linear polynomial and other is a quadratic polynomial.
(iv) Further, determine the factor of quadratic polynomial by splitting the middle term.
Solution:

Question 25:
Using suitable identity, evaluate the following
(i) 103³           (ii) 101 x 102       (iii) 999²
Thinking Process
Firstlyadjust the given number into two number such that one is a multiple of 10 and use the proper identity.
Solution:

Question 26:
Factorise the following

Solution:

Question 27:
Factorise the following
(i) 9x² -12x+ 3          (ii) 9x² -12xy + 4
Solution:

Question 28:
Expand the following
(i) (4a-b + 2c)²           (ii) (3a – 5b – c)²
(iii) (-x + 2y-3z)²
Solution:

Question 29:
Factorise the following
(i) 9x² +4y² + 16z² +12xy-16yz -24xz
(ii) 25x² + 16y² + 4z² – 40xy + 16yz – 20xz
iii) 16x² + 4y² + 9z² – 16xy – 12yz + 24xz
Solution:

Question 30:
If a + b + c = 9 and ab + bc + ca = 26, find a² + b² +c².
Solution:

Question 31:
Expand the following

Solution:

Question 32:
Factorise the following

Solution:

Question 33:
Find the following products

Solution:

Question 34:
Factorise
(i) 1 + 64x³          (ii) a³ -2√2b³
Thinking Process
(i) Firstly, adjust the given number either in the farm 0f a³ + b³ or a³ -b³
(ii) Further, use any of the identities i.e., a³ + b³ =(a + b)(a²+b²-ab) and a³ -b³ =(a-b)(a² + b² + ab), then simplify it, to get the factor.
Solution:

Question 35:
Fi nd (2x – y + 3z) (4x² + y² + 9z² + 2xy + 3yz – 6xz).
Solution:

Question 36:
Factorise
(i) a³ -8b³ -64c³ -2Aabc
(ii) 2√2a³ +8b³ -27c³ +18√2abc
Solution:

Question 37:
Without actually calculating the cubes, find the value of 36xy-36xy = 0

Thinking Process
In this method firstly check the values of a + b+ c, then . if a + b+c = Q, now use the identity a³ + b³ + c³ = 3abc.
Solution:

Question 38:
Without finding the cubes, factorise (x- 2y)³ + (2y – 3z)³ + (3z – x)³.
Solution:
We know that,
a³ + b³ + c³ – 3 abc = (a + b + c)(a² + b² + c² -ab-bc-ca)
Also, if a + b + c = 0, then a³ + b³ + c³ = 3abc
Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0
Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

Question 39:
Find the value of
(i) x³ +y³ -12xy + 64,when x+y = -4.
(ii) x³ -8y³ -36xy-216,when x = 2y + 6.
Solution:

Question 40:
Give possible expression for the length and breadth of the rectangle whose area is given by 4a² +4a – 3.
Solution:
Given, area of rectangle = 4a² + 6a-2a-3
= 4a² + 4a – 3 [by splitting middle term]
= 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3

Exercise 2.4: Long Answer Type Questions

Question 1:
If the polynomials az³ +4z² + 3z-4 and z³-4z + o leave the same remainder when divided by z – 3, find the value of a.
Solution:
Let p₁(z) = az³ +4z² + 3z-4 and p₂(z) = z³-4z + o
When we divide p₁(z) by z – 3, then we get the remainder p,(3).
Now, p₁(3) = a(3)3 + 4(3)2 + 3(3) – 4
= 27a+ 36+ 9-4= 27a+ 41 When we divide p₂(z) by z-3 then we get the remainder p₂(3).
Now, p₂(3) = (3)³-4(3)+a
= 27-12 + a = 15+a According to’ the question, both the remainders are same.
p₁(3)= p₂(3)
27a+41 = 15+a
27a-a = 15-41 .
26a = 26 a = -1

Question 2:
The polynomial p{x) = x⁴ -2x³ + 3x² -ax+3a-7 when divided by x+1 leaves the remainder 19. Find the values of a. Also, find the remainder when p(x) is divided by x+ 2.
Solution:

Question 3:
If both x – 2 and x -(1/2)  are factors of px²+ 5x+r, then show that p = r.
Solution:

Question 4:
Without actual division, prove that 2x⁴ – 5x³ + 2x² – x+ 2 is divisible by x²-3x+2
Thinking Process
(i) firstly, determine the factors of quadratic polynomial by splitting middle term.
(ii) The two different values of zeroes put in biquadratic polynomial.
(iii) In both the case if remainder is zero, then biquadratic polynomial is divisible by
quadratic polynomial.
Solution:
Let p(x) = 2x⁴ – 5x³ + 2x² – x+ 2 firstly, factorise x²-3x+2.
Now, x²-3x+2 = x²-2x-x+2 [by splitting middle term]
= x(x-2)-1 (x-2)= (x-1)(x-2)
Hence, 0 of x²-3x+2 are land 2.
We have to prove that, 2x⁴ – 5x³ + 2x² – x+ 2 is divisible by x²-3x+2 i.e., to prove that p (1) =0 and p(2) =0
Now, p(1) = 2(1)⁴ – 5(1)³ + 2(1)² -1 + 2 =2-5+2-1+2=6-6=0
and p(2) = 2(2)⁴ – 5(2)³ + 2(2)² – 2 + 2 = 2x¹⁶-5x⁸+2x⁴+ 0 = 32 – 40 + 8 = 40 – 40 =0
Hence, p(x) is divisible by x²-3x+2.

Question 5:
Simplify (2x- 5y)³ – (2x+ 5y)³.
Solution:
(2x -5y)³ – (2x + 5y)³ = [(2x)³ – (5y)³ – 3(2x)(5y)(2x – 5y)]
-[(2x)³ + (5y)³ + 3(2x)(5y)(2x+5y)]
[using identity, (a – b)³ = a³ -b³ – 3ab  and (a + b)³ =a³ +b³ + 3ab ] = (2x)³ – (5y)³ – 30xy(2x – 5y) – (2x)³ – (5y)³ – 30xy (2x + 5y)
= -2 (5y)³ – 30xy(2x – 5y + 2x + 5y)
= -2 x 125y³ – 30xy(4x) = -250y³ -120x²y

Question 6:
Multiply x² + 4y² + z² + 2xy + xz – 2yz by (-z + x-2y).
Solution:

Question 7:

Solution:

Question 8:
If a+b+c= 5 and ab+bc+ca =10, then prove that a³ +b³ +c³  – 3abc = -25.
Solution:

Question 9:
Prove that (a +b +c)³ -a³ -b³ – c³ =3(a +b)(b +c)(c +a).
Solution:

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