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Polynomial Chapter 2 Exercise 2.4 and 2.5|NCERT Solutions for Class 9 Maths





In this page we have NCERT Solutions for Class 9 Maths Chapter 2 Polynomial for Exercise 2.4 and 2.5 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Chapter 2 Exercise 2.4

Question 1
Determine which of the following polynomials has (x+ 1) a factor:
(i)$x^3+x^2+x+ 1$
(ii)$x^4+x^3+x^2+x+ 1$
(iii)$x^4+ 3x^3+ 3x^2+x+ 1$
(iv)$x^3-x^2- (2+ \sqrt {2}) x + \sqrt {2}$
Solution
We know from remainder theorem, if (x-a) is a factor of polynomial p(x) then p(a)=0
(i) If (x+ 1) is a factor of $p(x) =x^3+x^2+x+ 1$
Then p (-1) must be zero.
Here,$p(x) =x^3+x^2+x+ 1$
$p (-1) = (-1)^3+ (-1)^2+ (-1) + 1$
$= -1 + 1 - 1 + 1 = 0$
Hence,x+ 1 is a factor of this polynomial

(ii) If (x+ 1) is a factor of $p(x) =x^4+x^3+x^2+x+ 1$
Then p (-1) must be zero.
$p (-1) = (-1)^4+ (-1)^3+ (-1)^2+ (-1) + 1$
= 1
As, $p (-1) \ne 0$
Hence,x+ 1 is not a factor of this polynomial

(iii)If (x+ 1) is a factor of polynomial $p(x) =x^4+ 3x^3+ 3x^2+x+ 1$
Then p (- 1) must be 0.
$p (-1) = (-1)^4+ 3(-1)^3+ 3(-1)^2+ (-1) + 1$
= 1
As, $p (-1) \ne 0$
Hence,x+ 1 is not a factor of this polynomial.

(iv) If (x+ 1) is a factor of polynomial
$p(x) =x^3-x^2- (2+ \sqrt {2}) x + \sqrt {2}$
Then p (- 1) must be 0.
$p (-1) = (-1)^3- (-1)^2- (2+ \sqrt {2})(-1)+\sqrt {2}$
$= -1 - 1+ 2 + \sqrt {2} + \sqrt {2} $
$=2\sqrt {2}$
As, $p(-1) \ne 0$
Hence,x+ 1 is not a factor of this polynomial.
Question 2.
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i)$p(x) = 2x^3+x^2- 2x- 1$,$g(x) =x+ 1$
(ii)$p(x) =x^3+ 3x^2+ 3x+ 1$,$g(x) =x+ 2$
(iii) $p(x) =x^3- 4x^2+x+ 6$,$g(x) =x- 3$
Solution
(i) If $g(x) =x+ 1$ is a factor of given polynomial p(x)
Then, $p(- 1)$ must be zero.
$p(x) = 2x^3+x^2- 2x- 1$
$p(- 1) = 2(-1)^3+ (-1)^2- 2(-1) - 1$
$= 2(- 1) + 1 + 2 - 1 = 0$
Hence,g(x) =x+ 1 is a factor of given polynomial.

(ii) If $g(x) =x+ 2$ is a factor of given polynomial p(x)
Then, $p(-2)$ must be 0.
$p(x) =x^3+3x^2+ 3x+ 1$
$p(-2) = (-2)^3+ 3(- 2)^2+ 3(- 2) + 1$
$= -8 + 12 - 6 + 1$
$=-1$
As, $p(-2) \ne 0$
Therefore, $g(x) =x+ 2$ is not a factor of given polynomial.

(iii) If $g(x) =x- 3$ is a factor of given polynomial $p(x)$
Then, p(3) must be 0.
$p(x) =x^3- 4x^2+x+ 6$
$p(3) = (3)^3- 4(3)^2+ 3 + 6$
$= 27 - 36 + 9 = 0$
Hence,$g(x) =x- 3$ is a factor of given polynomial.

Question 3.
Find the value of k, if x- 1 is a factor of p(x) in each of the following cases:
(i)$p(x) =x^2+x+k$
(ii)$p(x) = 2x^2+kx+ \sqrt{2}$
(iii)$p(x) =kx^2- \sqrt{2} x+ 1$
(iv)$p(x) =kx^2-3x+k$
Solution
(i) If x- 1 is a factor of polynomial $p(x) =x^2+x+k$, then
$p(1) = 0$
$(1)^2+ 1 +k= 0$
$2 +k= 0$
$k= - 2$
So, value of k is -2.

(ii) If x- 1 is a factor of polynomial $p(x) = 2x^2+kx+ \sqrt{2}$, then
$p(1) = 0$
$2(1)^2+k(1) + \sqrt{2}= 0$
$k= -2 - \sqrt{2}= -(2 + \sqrt{2})$
So, value of k is $- (2 + \sqrt{2})$.

(iii) If x- 1 is a factor of polynomial $p(x) =kx^2- \sqrt{2}x+ 1$, then
$p (1) = 0$
$k(1)^2- \sqrt{2}(1) + 1= 0$
$k= \sqrt{2}-1$
So, value of k is $\sqrt{2}-1$

(iv) If x- 1 is a factor of polynomial $p(x) =kx^2-3x+k$, then
$p(1) = 0$
$k(1)^2+ 3(1) +k= 0$
$k- 3+k= 0$
$k= \frac {3}{2}$
So, value of k is $\frac {3}{2}$

Question 4
Factorise:
(i) $12x^2+ 7x+ 1$
(ii) $2x^2+ 7x+ 3$
(iii)$6x^2+ 5x- 6$
(iv)$3x^2-x- 4$
Solution
Here we would be using splitting the middle term to factorize the polynomial
To factorise $ax^2 +bx +c$, we should write b as the sum of two numbers whose product is ac
(i) $12x^2+ 7x+ 1$
Here a=12, c=1 and b=7 So $b=7=3+4$ ,$ac=12 \times 1=12= 3 \times 4$
$12x^2+ 7x+ 1$
$=12x^2+4x+ 3x+ 1$
$= 4x(3x+ 1) + 1 (3x+ 1)$
$= (3x+1) (4x+ 1)$

(ii) $2x^2+ 7x+ 3$
Here a=2, c=3 and b=7 So $b=7=6+1$ ,$ac= 2 \times 3=6 = 6 \times 1$
$= 2x^2+ 6x+x+ 3$
$= 2x(x+ 3) + 1 (x+ 3)$
$=(x+ 3) (2x+ 1)$

(iii) $6x^2+ 5x- 6$
Here a=6, c=-6 and b=5 So, $b=5=9 + (-4)$ ,$ac= 6 \times (-6) =-36= 9 \times (-4) $
$=6x^2+ 9x- 4x- 6$
$= 3x(2x+ 3) - 2 (2x+ 3)$
$= (2x+ 3) (3x- 2)$

(iv) $3x^2-x- 4$
Here a=3, c=-4 and b=-1 So, $b=-1=-4+3$, $ac= 3 \times (-4) =-12=(-4) \times 3$
$=3x^2- 4x+ 3x- 4$
$=x(3x- 4) + 1 (3x- 4)$
$= (3x- 4) (x+ 1)$

Question 5.
Factorise:
(i) $x^3- 2x^2-x+ 2$
(ii) $x^3- 3x^2- 9x- 5$
(iii) $x^3+ 13x^2+ 32x+ 20$
(iv) $2y^3+y^2- 2y- 1$
Solution
These are cubic polynomials and can be factorized using combination of long division method , remainder theorem and split middle term method
(i) Let $p(x) =x^3- 2x^2-x+ 2$
Factors of 2 are ±1 and ± 2
By trial method, we find that
$p(-1)= (-1)^3- 2(-1)^2- (-1)+ 2=0$
So,(x+1) is factor of p(x)
. Now using the Long division method , we can find the quotient as
NCERT Solutions for Class 9 Maths Chapter 2 Polynomial Exercise 2.4 Question 5 (i)
Now, $Dividend = Divisor \times Quotient+ Remainder$
$x^3- 2x^2-x+ 2=(x+1) (x^2- 3x + 2)$
Factorizing the second part by split middle term method
$= (x+1) (x^2- x - 2x + 2)$
$= (x+1) [x(x-1) -2(x-1)]$
$= (x+1) (x-1) (x+2)$

(ii) Let $p(x) =x^3- 3x^2- 9x- 5$
Factors of 5 are ±1 and ±5
By trial method, we find that
$p(5)= 5^3- 3 \times 5^2- 9 \times 5- 5=0$
So,(x-5) is factor of p(x)
Now using the Long division method , we can find the quotient as
NCERT Solutions for Class 9 Maths Chapter 2 Polynomial Exercise 2.4 Question 5 (ii)
Now, $Dividend = Divisor \times Quotient+ Remainder$
$x^3- 3x^2- 9x- 5=(x-5) (x^2+ 2x + 1)$
Factorizing the second part by split middle term method
$= (x-5) (x^2+ x + x + 1)$
$=(x-5){x(x+1) +1(x+1)}$
$=(x-5)(x+1)(x+1)$

(iii) Let $p(x) =x^3+ 13x^2+ 32x+ 20$
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
$p(-1)= (-1)^3+ 13(-1)^2+ 32(-1)+ 20=0$
So,(x+1) is factor ofp(x)
Now using the Long division method , we can find the quotient as
NCERT Solutions for Class 9 Maths Chapter 2 Polynomial Exercise 2.4 Question 5 (iii)
Now, Dividend = Divisor \times Quotient+ Remainder
$x^3+ 13x^2+ 32x+ 20=(x+1) (x^2+ 12x + 20)$
Factorizing the second part by split middle term method
$= (x+1) (x^2+ 2x+ 10x + 20)$
$=(x-5)[x(x+2) +10(x+2)]$
$=(x-5)(x+2)(x+10)$

(iv) Let $p(y) = 2y^3+y^2- 2y- 1$
Factors of ab = 2 \times (-1) = -2 are ±1 and ±2
By trial method, we find that
$p(1)= 2(1)^3+(1)^2- 2(1)- 1=0$
So,(y-1) is factor of p(y)
Now using the Long division method , we can find the quotient as
NCERT Solutions for Class 9 Maths Chapter 2 Polynomial Exercise 2.4 Question 5 (iv)
Now, $Dividend = Divisor \times Quotient+ Remainder$
$2y^3+y^2- 2y- 1=(y-1) (2y^2+ 3y + 1)$
Factorizing the second part by split middle term method
$=(y-1)(2y^2+ 2y+ y + 1)$
$=(y-1)[2y(y+1) +1(y+1)]$
$=(y-1)(2y+1)(y+1)$



Chapter 2 Exercise 2.5

Question 1
Use suitable identities to find the following products:
(i) $(x+ 4) (x+ 10)$
(ii) $(x+ 8) (x- 10)$
(iii) $(3x+ 4) (3x- 5)$
(iv) $(y^2+ \frac {3}{2}) (y^2- \frac {3}{2})$
(v) $(3 - 2x) (3 + 2x)$
Solution
(i)Using identity, $(x+ a) (x+ b) =x^2+ (a + b)x+ ab$
Here a = 4 and b = 10
Therefore,
$(x+ 4) (x+ 10) =x^2+ (4 + 10)x+ (4 \times 10)$
$=x^2+ 14x+ 40$

(ii) $(x+ 8) (x -10)$
Using identity, $(x+ a) (x+ b) =x^2+ (a + b)x+ ab$
Here, a = 8 and b = -10
Therefore,
$(x+ 8) (x -10) =x^2+ {8 +(- 10)}x+ {8 \times (- 10)}$
$ =x^2- 2x- 80$

(iii) $(3x+ 4) (3x - 5)$
Using identity, $(x+ a) (x+ b) =x^2+ (a + b)x+ ab$
Here,x= 3x,a = 4 and b = -5
Therefore,
$(3x+ 4) (3x- 5) = (3x)^2+ {4 + (-5)}3x+ {4 \times (-5)}$
$= 9x^2+ 3x (4 - 5) - 20$
$= 9x^2- 3x- 20$

(iv) $(y^2+ \frac {3}{2}) (y^2- \frac {3}{2})$
Using identity, $(x+y) (x-y) =x^2-y^2$
Here,$x=y^2$ and $y= \frac {3}{2}$
Therefore,
$(y^2+ \frac {3}{2}) (y^2- \frac {3}{2}) = (y^2)^2- (\frac {3}{2})^2$
$=y^4- \frac {9}{4}$

(v) $(3 - 2x) (3 + 2x)$
Using identity, $(x+y) (x-y) =x^2-y^2$
Here,x= 3 andy= 2x
Therefore,
$(3 - 2x) (3 + 2x) = 3^2- (2x)^2$
$=9- 4x^2$

Question 2
Evaluate the following products without multiplying directly:
(i) $103 \times 107$
(ii) $95 \times 96$
(iii) $104 \times 96$
Solution
(i) $103 \times 107$ can be written as
$= (100+ 3) (100+ 7)$
Using identity, $(x+ a) (x+ b) =x^2+ (a + b)x+ ab$
Here,x= 100, a = 3 and b = 7
Therefore,
$103 \times 107 = (100+ 3) (100+ 7) = (100)^2+ (3 + 7)10 + (3 \times 7)$
$= 10000+ 100 + 21$
$= 10121$
(ii) $95 \times 96 = (90 + 5) (90 + 4)$
Using identity, $(x+ a) (x+ b) =x^2+ (a + b)x+ ab$
Here,x= 90, a = 5 and b = 4
Therefore,

$95 \times 96 = (90 + 5) (90 + 4) = 90^2+ 90(5 + 6) + (5 \times 6)$
$= 8100 + (11 \times 90) + 30$
$= 8100 + 990 + 30 = 9120$

(iii) $104 \times 96 = (100 + 4) (100 - 4)$
Using identity, $(x+y) (x-y) =x^2-y^2$
Here,x= 100 andy= 4
Therefore,
$104 \times 96 = (100 + 4) (100 - 4) = (100)^2-(4)^2= 10000 - 16 = 9984$

Question 3
Factorise the following using appropriate identities:
(i) $9x^2+ 6xy+y^2$
(ii) $4y^2- 4y+ 1$
(iii) $x^2- \frac {y^2}{100}$
Solution
(i)$9x^2+ 6xy + y^2 = (3x)^2+ (2 \times 3x \times y) + y^2$
Using identity, $(a + b)^2= a^2+ 2ab + b^2$
Here, a = 3x and b =y
$9x^2+ 6xy + y^= (3x)^2+ (2 \times 3x \times y) + y^2= (3x + y)^2= (3x + y) (3x + y)$

(ii)$4y^2- 4y + 1 = (2y)^2- (2 \times 2y \times 1) + 1^2$
Using identity, $(a - b)^2= a^2- 2ab + b^2$
Here, a =2y and b =1
$4y^2- 4y + 1 = (2y)^2- (2 \times 2y \times 1) + 1^2= (2y - 1)^2= (2y - 1) (2y - 1)$

(iii)$x^2- \frac {y^2}{100} = x^2- (\frac {y}{10})^2$
Using identity, $a^2- b^2= (a + b) (a - b)$
Here, a =x and b =(y/10)
$x^2- \frac {y^2}{100} = x^2- (\frac {y}{10})^2= (x- \frac {y}{10}) (x+ \frac {y}{10})$

Question 4
Expand each of the following, using suitable identities:
(i) $(x+ 2y+ 4z)^2$
(ii) $(2x-y+z)^2$
(iii) $(-2x+ 3y+ 2z)^2$
(iv) $(3a- 7b-c)^2$
(v) $(-2x+ 5y- 3z)^2$
(vi) $[\frac {1}{4}a - \frac {1}{2}b + 1]^2$
Answer
We will be using the identity
$(a + b +c)^2= a^2+ b^2+ c^2+ 2ab + 2bc + 2ca$

(i) $(x+ 2y+ 4z)^2$
Here, a =x, b = 2yand c = 4z
$(x+ 2y+ 4z)^2=x^2+(2y)^2+ (4z)^2+ (2 \times x \times 2y) + (2 \times 2y \times 4z) + (2 \times 4z \times x)$
$=x^2+ 4y^2+ 16z^2+ 4xy + 16yz + 8xz$

(ii)$(2x-y+z)^2$
Here, a =2x, b = -y and c =z
$(2x-y+z)^2=(2x)^2+ (-y)^2+ z^2+ (2 \times 2x \times -y)+ (2 \times -y \times z)+ (2 \times z \times 2x)$
$ =4x^2+ y^2+ z^2- 4xy- 2yz+ 4xz$

(iii) $(-2x+ 3y+ 2z)^2$
Here, a =-2x, b = 3y and c =2z
$(-2x + 3y + 2z)^2=(-2x)^2+ (3y)^2+ (2z)^2+ (2 \times -2x \times 3y)+ (2 \times 3y \times 2z)+ (2 \times 2z \times -2x)$
$ =4x^2+ 9y^2+ 4z^2- 12xy+ 12yz- 8xz$

(iv) $(3a- 7b-c)^2$
Here, a =3a, b = -7b and c =-c
$(3a - 7b - c)^2=(3a)^2+ (-7b)^2+ (-c)^2+ (2 \times 3a \times -7b)+ (2 \times -7b \times -c)+ (2 \times -c \times 3a)$
$ =9a^2+ 49b^2+ c^2- 42ab+ 14bc- 6ac$

(v) $(-2x+ 5y- 3z)^2$
Here, a =-2x, b =5y and c =-3z
$(-2x + 5y - 3z)^2=(-2x)^2+ (5y)^2+ (-3z)^2+ (2 \times -2x \times 5y)+ (2 \times 5y \times -3z)+ (2 \times -3z \times -2x)$
$ =4x^2+ 25y^2+ 9z^2- 20xy- 30yz+ 12xz$

(vi) $[\frac {1}{4}a - \frac {1}{2}b + 1]^2$
Here, $a = \frac {1}{4} a$, $b = \frac {-1}{2}b$ and c = 1
$[\frac {1}{4}a - \frac {1}{2}b + 1]^2=(\frac {1}{4}a)^2+ (\frac {-1}{2}b)^2+1^2+ (2 \times \frac {1}{4}a \times \frac {-1}{2}b)+ (2 \times \frac {-1}{2}b \times 1)+ (2 \times 1 \times \frac {1}{4}a)$
$=\frac {1}{16} a^2 + \frac {1}{4} b^2 +1 - \frac {1}{4} ab - b + \frac {1}{2}a$

Question 5
Factorise:
(i) $4x^2+ 9y^2+ 16z^2+ 12xy- 24yz- 16xz$
(ii) $2x^2+y^2+ 8z^2- 2\sqrt {2} xy+ 4 \sqrt {2} yz- 8xz$

Solution

(i) $4x^2+ 9y^2+ 16z^2+ 12xy- 24yz- 16xz$
Using identity, $(a + b +c)^2= a^2+ b^2+ c^2+ 2ab + 2bc + 2ca$
$4x^2+ 9y^2+ 16z^2+ 12xy - 24yz - 16xz$
$= (2x)^2+ (3y)^2+ (-4z)^2+ (2 \times 2x \times 3y) + (2 \times 3y \times -4z) + (2 \times -4z \times 2x)$
$=(2x + 3y - 4z)^2$
$=(2x + 3y - 4z) (2x + 3y - 4z)$

(ii) $2x^2+y^2+ 8z^2- 2\sqrt {2} xy+ 4 \sqrt {2} yz- 8xz$
Using identity, $(a + b +c)^2= a^2+ b^2+ c^2+ 2ab + 2bc + 2ca$
$2x^2+y^2+ 8z^2- 2\sqrt {2} xy+ 4 \sqrt {2} yz- 8xz$
$= (-\sqrt{2} x)^2+ (y)^2+ (2\sqrt {2}z)^2+ (2 \times -\sqrt {2} x \times y) + (2 \times y \times 2 \sqrt {2}z) + (2 \times 2\sqrt {2} z \times -\sqrt {2}x)$
$=(-\sqrt {2}x+ y+2\sqrt {2}z)^2$
$=(-\sqrt {2}x+ y+2\sqrt {2}z)(-\sqrt {2}x+ y+2\sqrt {2}z)$

Question 6
Write the following cubes in expanded form:
(i) $(2x + 1)^3$
(ii) $(2a - 3b)^3$
(iii) $[\frac {3}{2} x+ 1]^3$
(iv) $[x- \frac {2}{3}y]^3$
Solution
We will be using the identity
$(a + b)^3= a^3+ b^3+ 3ab (a + b)$
$(a - b)^3= a^3- b^3- 3ab (a - b)$

(i) $(2x+ 1)^3$
$=(2x)^3+ 1^3+(3 \times 2x \times 1) (2x + 1)$
$= 8x^3+ 1 +6x (2x + 1)$
$=8x^3+12x^2+6x+ 1$

(ii) $(2a - 3b)^3$
$= (2a)^3- (3b)^3- (3 \times 2a \times 3b) (2a - 3b)$
$= 8a^3- 27b^3- 18ab (2a - 3b)$
$=8a^3- 27b^3- 36a^2b+ 54ab^2$

(iii) $[\frac {3}{2} x+ 1]^3$
$= (\frac {3}{2} x)^3+ 1^3+(3 \times \frac {3}{2} x \times 1) (\frac {3}{2} x + 1)
$= \frac {27}{8}x^3+ 1 +\frac {9}{2} x (\frac {3}{2} x + 1)$
$=\frac {27}{8} x^3+ 1 +\frac {27}{4} x^2+ \frac {9}{2} x$
$=\frac {27}{8}x^3+\frac {27}{4}x^2+ \frac {9}{2} x+ 1$

(iv) $[x- \frac {2}{3}y]^3= (x)^3- (\frac {2}{3}y)^3- (3 \times x \times \frac {2}{3}y) (x - \frac {2}{3}y)$
$= x^3- \frac {8}{27}y^3- 2xy (x - \frac {2}{3}y)$
$=x^3- \frac {8}{27}y^3- 2x^2y+ \frac {4}{3}xy^2$

Question 7
Find the value of the following using suitable identities:
(i) $(99)^3$
(ii) $(102)^3$
(iii) $(998)^3$
Answer
We will be using the identity
$(a + b)^3= a^3+ b^3+ 3ab (a + b)$
$(a - b)^3= a^3- b^3- 3ab (a - b)$

(i) $(99)^3= (100 - 1)^3$
$= (100)^3- 1^3- (3 \times 100 \times 1) (100 - 1)$
$= 1000000 - 1 - 300(100 - 1)$
$=1000000 - 1 - 30000 + 300$
$= 970299$

(ii) $(102)^3= (100 + 2)^3$
$= (100)^3+ 2^3+ (3 \times 100 \times 2) (100 + 2)$
$= 1000000+ 8+ 600(100 + 2)$
$=1000000+ 8+ 60000 + 1200$
$= 1061208$

(iii) $(998)^3$
$= (1000)^3- 2^3- (3 \times 1000 \times 2) (1000 - 2)$
$= 100000000 - 8 - 6000(1000 - 2)$
$=100000000 - 8- 600000 + 12000$
$= 994011992$

Question 8
Factorise each of the following:
(i) $8a^3+ b^3+ 12a^2b + 6ab^2$
(ii) $8a^3- b^3- 12a^2b + 6ab^2$
(iii) $27 - 125a^3- 135a + 225a^2 $
(iv) $64a^3- 27b^3- 144a^2b + 108ab^2$
(v) $27p^3- \frac {1}{216} - \frac {9}{2} p^2 + \frac {1}{4} p$

Solution
(i)$8a^3+ b^3+ 12a^2b + 6ab^2$
Using identity, $(a + b)^3= a^3+ b^3+ 3a^2b+ 3ab^2$
$8a^3+ b^3+ 12a^2b + 6ab^2$
$=(2a)^3+ b^3+3(2a)^2b+3(2a) (b)^2$
$= (2a + b)^3$
$= (2a + b) (2a + b) (2a + b)$

(ii) $8a^3- b^3- 12a^2b + 6ab^2$
Using identity, $(a - b)^3= a^3- b^3- 3a^2b + 3ab^2$
$8a^3- b^3- 12a^2b + 6ab^2=(2a)^3- b^3-3(2a)^2b+3(2a) (b)^2$
$= (2a - b)^3$
$= (2a - b) (2a - b) (2a - b)$

(iii) $27 - 125a^3- 135a + 225a^2$
Using identity, $(a - b)^3= a^3- b^3- 3a^2b + 3ab^2$
$27 - 125a^3- 135a + 225a^2=3^3- (5a)^3-3(3)^2(5a)+3(3) (5a)^2$
$= (3 - 5a)^3$
$=(3 - 5a) (3 - 5a) (3 - 5a)$

(iv) $64a^3- 27b^3- 144a^2b + 108ab^2$
Using identity, $(a - b)^3= a^3- b^3- 3a^2b + 3ab^2$
$64a^3- 27b^3- 144a^2b + 108ab^2=(4a)^3- (3b)^3-3(4a)^2(3b)+3(4a) (3b)^2$
$= (4a - 3b)^3$
$= (4a - 3b) (4a - 3b) (4a - 3b)$

(v)$27p^3- \frac {1}{216} - \frac {9}{2} p^2 + \frac {1}{4} p$

Using identity, $(a - b)^3= a^3- b^3- 3a^2b + 3ab^2$
$27p^3- \frac {1}{216} - \frac {9}{2} p^2 + \frac {1}{4} p$
$=(3p)^3- (\frac {1}{6})^3-3(3p)^2(\frac {1}{6})+3(3p) (\frac {1}{6})^2$
$= (3p - \frac {1}{6})^3$
$= (3p - \frac {1}{6}) (3p - \frac {1}{6}) (3p - \frac {1}{6})$

Question 9
Verify the below identities
(i)$x^3+y^3= (x+y) (x^2-xy+ y^2)$
(ii)$x^3-y^3= (x-y) (x^2+xy+ y^2)$
Solution

(i)$x^3+y^3= (x+y) (x^2-xy+ y^2)$
We know that,
$(x + y)^3= x^3+ y^3+3xy (x + y)$
Taking $3xy (x + y)$ on another side, we get
$x^3+ y^3= (x + y)^3-3xy (x + y)$
Taking (x + y) common factor
$= (x + y) [(x + y)^2-3xy]$
$= (x + y)[(x^2+ y^2+ 2xy) -3xy]$
$= (x + y)(x^2+ y^2- xy)$

(ii)$x^3-y^3= (x-y) (x^2+xy+ y^2)$
We know that,
$(x - y)^3= x^3- y^3-3xy(x - y)$
$x^3- y^3= (x - y)^3+3xy(x - y)$
Taking $(x-y)$ common factor
$= (x - y)[(x - y)^2+3xy]$
$= (x - y)[(x^2+ y^2- 2xy)+3xy]$
$= (x + y)(x^2+ y^2+ xy)$
Question 10
Factorise each of the following:
(i) $27y^3+ 125z^3$
(ii) $64m^3- 343n^3$
Solution
(i) $27y^3+ 125z^3$
Using identity proved in the above question, $x^3+y^3= (x+y) (x^2-xy+ y^2)$
$27y^3+ 125z^3=(3y)^3+ (5z)^3$
$=(3y+5z) [(3y)^2-(3y)(5z)+(5z)^2]$
$=(3y+5z) (9y^2- 15yz+25z^2)$

(ii) $64m^3- 343n^3$
Using identity proved in the above question, $x^3-y^3= (x-y) (x^2+xy+ y^2)$
$64m^3- 343n^3=(4m)^3- (7n)^3$
$=(4m+7n) [(4m)^2+(4m) (7n)+(7n)^2]$
$= (4m+7n) (16m^2+ 28mn +49n^2)$

Question 11
Factorise: 27x^3+y^3+z^3- 9xyz

Solution
$27x^3+ y^3+ z^3- 9xyz = (3x)^3+y^3+z^3- 3 \times 3xyz$
Using identity
$x^3+y^3+z^3- 3xyz = (x + y + z) (x^2+y^2+z^2-xy - yz - xz)$
Therefore,
$27x^3+ y^3+ z^3- 9xyz$
$=(3x + y + z) [(3x)^2+y^2+z^2-3xy - yz - 3xz]$
$=(3x + y + z) (9x^2+y^2+z^2-3xy - yz - 3xz)$

Question 12
Verify that: $x^3+ y^3+ z^3-3xyz =\frac {1}{2}(x + y + z) [(x-y)^2+(y - z)^2+(z - x)^2]$
Solution
We know that,
$x^3+y^3+z^3- 3xyz = (x + y + z) (x^2+y^2+z^2-xy - yz - xz)$
Multiplying and dividing by 2 on Right hand side
$= \frac {1}{2} \times (x + y + z) \times 2 \times (x^2+y^2+z^2-xy - yz - xz)$
$= \frac {1}{2} (x + y + z) (2x^2+ 2y^2 + 2z^2- 2xy - 2yz - 2xz)$
$=\frac {1}{2} (x + y + z) [(x^2+y^2-2xy) + (y^2+z^2- 2yz) + (x^2+z^2- 2xz)]$
$= \frac {1}{2} (x + y + z) [(x - y)^2+(y - z)^2+(z - x)^2]$

Question 13
If $x + y + z =0$, show that $x^3+y^3+z^3= 3xyz$
Solution
We know that,
$x^3+y^3+z^3- 3xyz = (x + y + z) (x^2+y^2+z^2-xy - yz - xz)$
Now put $(x + y + z) =0$,
$x^3+y^3+z^3- 3xyz = (0) (x^2+y^2+z^2-xy - yz - xz)$
$x^3+y^3+z^3- 3xyz = 0$
$x^3+y^3+z^3= 3xyz$

Question 14
Without calculating the cubes, find the value of each of the following:
(i) $(-12)^3+ (7)^3+ (5)^3$
(ii) $(28)^3+ (-15)^3+ (-13)^3$
Answer
From previous question, we know that
If $x + y + z =0$, then $x^3+y^3+z^3= 3xyz$

(i) $(-12)^3+ (7)^3+ (5)^3$
Let x=-12,y= 7 andz= 5
We observed that,$x + y + z= -12 + 7 + 5 = 0$

Therefore,
$(-12)^3+ (7)^3+ (5)^3= 3(-12)(7)(5) = -1260$

(ii)$(28)^3+ (-15)^3+ (-13)^3$
Let x=28,y= -15 andz= -13
We observed that,$x + y + z= 28 - 15 - 13 = 0$

Therefore.
$(28)^3+ (-15)^3+ (-13)^3= 3(28)(-15)(-13) = 16380$

Question 15
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

Solution
(i) Area: $25a^2- 35a + 12$

Now $Area= Length \times Breadth$
So, factorizing Area
$25a^2- 35a + 12$
By Split middle method $=25a^2- 15a -20a + 12$
$= 5a (5a - 3) - 4(5a - 3)$
$=(5a - 4) (5a - 3)$
Possible expression for length= $5a - 4$
Possible expression for breadth=$5a - 3$

(ii)Area: $35y^2+ 13y- 12$
$35 y^2+ 13y - 12$
By Split middle method $=35y^2- 15y + 28y - 12$
$= 5y (7y - 3) + 4(7y - 3)$
$=(5y+ 4) (7y - 3)$
Possible expression for length=$(5y+ 4)$
Possible expression for breadth=$(7y - 3)$


Question 16
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
NCERT Solutions for Class 9 Maths Chapter 2 Polynomial Exercise 2.5
Answer
(i) Volume: $3x^2- 12x$
$\text {Volume of Cuboids}= Length \times Breadth \times height$
Now, $3x^2- 12x$
$= 3x (x - 4)$
Possible expression for length=$3$
Possible expression for breadth=$x$
Possible expression for height=$(x - 4)$

(ii) Volume: $12ky^2+8ky -20k$
$\text {Volume of Cuboids}= Length \times Breadth \times height$
Now,
$12ky^2+ 8ky - 20k$
$= 4k (3y^2+ 2y - 5)$
$= 4k (3y^2+5y - 3y - 5)$
$=4k [y (3y +5) - 1(3y + 5)]$
$=4k (3y +5) (y - 1)$
Possible expression for length= $4k$
Possible expression for breadth=$(3y +5)$
Possible expression for height=$(y - 1)$


Download NCERT Polynomials Exercise 2.5 and 2.6 as pdf
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Reference Books for class 9 Math

Given below are the links of some of the reference books for class 9 Math.

  1. Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
  2. Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
  3. Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
  4. Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.



Class 9 Maths Class 9 Science





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