Polynomials
NCERT Solutions • Class 9 Maths • Chapter 2Exercise 2.4
1. Use suitable identities to find the following products:
(i) $(x + 4)(x + 10)$
Identity: $(x+a)(x+b) = x^2 + (a+b)x + ab$ $= x^2 + (4+10)x + (4 \times 10) =$ $x^2 + 14x + 40$
Identity: $(x+a)(x+b) = x^2 + (a+b)x + ab$ $= x^2 + (4+10)x + (4 \times 10) =$ $x^2 + 14x + 40$
(ii) $(x + 8)(x – 10)$
$= x^2 + (8 + (-10))x + (8 \times -10) = x^2 – 2x – 80$
$x^2 – 2x – 80$
$= x^2 + (8 + (-10))x + (8 \times -10) = x^2 – 2x – 80$
$x^2 – 2x – 80$
(iii) $(3x + 4)(3x – 5)$
$= (3x)^2 + (4-5)(3x) + (4)(-5) = 9x^2 + (-1)(3x) – 20$
$9x^2 – 3x – 20$
$= (3x)^2 + (4-5)(3x) + (4)(-5) = 9x^2 + (-1)(3x) – 20$
$9x^2 – 3x – 20$
(iv) $(y^2 + \frac{3}{2})(y^2 – \frac{3}{2})$
Identity: $(a+b)(a-b) = a^2 – b^2$ $= (y^2)^2 – (\frac{3}{2})^2 =$ $y^4 – \frac{9}{4}$
Identity: $(a+b)(a-b) = a^2 – b^2$ $= (y^2)^2 – (\frac{3}{2})^2 =$ $y^4 – \frac{9}{4}$
(v) $(3 – 2x)(3 + 2x)$
$= (3)^2 – (2x)^2 =$ $9 – 4x^2$
$= (3)^2 – (2x)^2 =$ $9 – 4x^2$
2. Evaluate the following products without multiplying directly:
(i) $103 \times 107$
$= (100+3)(100+7) = 100^2 + (3+7)100 + (3 \times 7)$
$= 10000 + 1000 + 21 =$ $11021$
$= (100+3)(100+7) = 100^2 + (3+7)100 + (3 \times 7)$
$= 10000 + 1000 + 21 =$ $11021$
(ii) $95 \times 96$
$= (100-5)(100-4) = 100^2 + (-5-4)100 + (-5 \times -4)$
$= 10000 – 900 + 20 =$ $9120$
$= (100-5)(100-4) = 100^2 + (-5-4)100 + (-5 \times -4)$
$= 10000 – 900 + 20 =$ $9120$
(iii) $104 \times 96$
$= (100+4)(100-4) = 100^2 – 4^2$
$= 10000 – 16 =$ $9984$
$= (100+4)(100-4) = 100^2 – 4^2$
$= 10000 – 16 =$ $9984$
3. Factorise the following using appropriate identities:
(i) $9x^2 + 6xy + y^2$
$= (3x)^2 + 2(3x)(y) + (y)^2$ $\rightarrow$ Identity: $(a+b)^2 = a^2 + 2ab + b^2$
$=$ $(3x + y)^2$
$= (3x)^2 + 2(3x)(y) + (y)^2$ $\rightarrow$ Identity: $(a+b)^2 = a^2 + 2ab + b^2$
$=$ $(3x + y)^2$
(ii) $4y^2 – 4y + 1$
$= (2y)^2 – 2(2y)(1) + (1)^2$ $\rightarrow$ Identity: $(a-b)^2 = a^2 – 2ab + b^2$
$=$ $(2y – 1)^2$
$= (2y)^2 – 2(2y)(1) + (1)^2$ $\rightarrow$ Identity: $(a-b)^2 = a^2 – 2ab + b^2$
$=$ $(2y – 1)^2$
(iii) $x^2 – \frac{y^2}{100}$
$= (x)^2 – (\frac{y}{10})^2$ $\rightarrow$ Identity: $a^2 – b^2 = (a-b)(a+b)$
$=$ $(x – \frac{y}{10})(x + \frac{y}{10})$
$= (x)^2 – (\frac{y}{10})^2$ $\rightarrow$ Identity: $a^2 – b^2 = (a-b)(a+b)$
$=$ $(x – \frac{y}{10})(x + \frac{y}{10})$
4. Expand each of the following, using suitable identities:
(i) $(x + 2y + 4z)^2$
Identity: $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$ $= x^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$
$=$ $x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$
Identity: $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$ $= x^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$
$=$ $x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$
(ii) $(2x – y + z)^2$
$= (2x)^2 + (-y)^2 + z^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$
$=$ $4x^2 + y^2 + z^2 – 4xy – 2yz + 4zx$
$= (2x)^2 + (-y)^2 + z^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$
$=$ $4x^2 + y^2 + z^2 – 4xy – 2yz + 4zx$
(iii) $(-2x + 3y + 2z)^2$
$= (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)$
$=$ $4x^2 + 9y^2 + 4z^2 – 12xy + 12yz – 8zx$
$= (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)$
$=$ $4x^2 + 9y^2 + 4z^2 – 12xy + 12yz – 8zx$
(iv) $(3a – 7b – c)^2$
$= (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$
$=$ $9a^2 + 49b^2 + c^2 – 42ab + 14bc – 6ca$
$= (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$
$=$ $9a^2 + 49b^2 + c^2 – 42ab + 14bc – 6ca$
(v) $(-2x + 5y – 3z)^2$
$= (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)$
$=$ $4x^2 + 25y^2 + 9z^2 – 20xy – 30yz + 12zx$
$= (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)$
$=$ $4x^2 + 25y^2 + 9z^2 – 20xy – 30yz + 12zx$
(vi) $[\frac{1}{4}a – \frac{1}{2}b + 1]^2$
$= (\frac{a}{4})^2 + (-\frac{b}{2})^2 + 1^2 + 2(\frac{a}{4})(-\frac{b}{2}) + 2(-\frac{b}{2})(1) + 2(1)(\frac{a}{4})$
$=$ $\frac{a^2}{16} + \frac{b^2}{4} + 1 – \frac{ab}{4} – b + \frac{a}{2}$
$= (\frac{a}{4})^2 + (-\frac{b}{2})^2 + 1^2 + 2(\frac{a}{4})(-\frac{b}{2}) + 2(-\frac{b}{2})(1) + 2(1)(\frac{a}{4})$
$=$ $\frac{a^2}{16} + \frac{b^2}{4} + 1 – \frac{ab}{4} – b + \frac{a}{2}$
5. Factorise:
(i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
Note: $yz$ and $xz$ terms are negative, implying $z$ term is negative.
$= (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$
$=$ $(2x + 3y – 4z)^2$
Note: $yz$ and $xz$ terms are negative, implying $z$ term is negative.
$= (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$
$=$ $(2x + 3y – 4z)^2$
(ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
Note: $xy$ and $xz$ terms are negative, implying $x$ term is negative.
$= (-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 + 2(-\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) + 2(2\sqrt{2}z)(-\sqrt{2}x)$
$=$ $(-\sqrt{2}x + y + 2\sqrt{2}z)^2$
Note: $xy$ and $xz$ terms are negative, implying $x$ term is negative.
$= (-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 + 2(-\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) + 2(2\sqrt{2}z)(-\sqrt{2}x)$
$=$ $(-\sqrt{2}x + y + 2\sqrt{2}z)^2$
Polynomials
NCERT Solutions • Class 9 Maths • Chapter 2Exercise 2.4
6. Write the following cubes in expanded form:
(i) $(2x + 1)^3$
Identity: $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ $= (2x)^3 + 1^3 + 3(2x)(1)(2x + 1)$
$= 8x^3 + 1 + 6x(2x + 1) =$ $8x^3 + 12x^2 + 6x + 1$
Identity: $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ $= (2x)^3 + 1^3 + 3(2x)(1)(2x + 1)$
$= 8x^3 + 1 + 6x(2x + 1) =$ $8x^3 + 12x^2 + 6x + 1$
(ii) $(2a – 3b)^3$
Identity: $(a-b)^3 = a^3 – b^3 – 3ab(a-b)$ $= (2a)^3 – (3b)^3 – 3(2a)(3b)(2a – 3b)$
$= 8a^3 – 27b^3 – 18ab(2a – 3b) =$ $8a^3 – 27b^3 – 36a^2b + 54ab^2$
Identity: $(a-b)^3 = a^3 – b^3 – 3ab(a-b)$ $= (2a)^3 – (3b)^3 – 3(2a)(3b)(2a – 3b)$
$= 8a^3 – 27b^3 – 18ab(2a – 3b) =$ $8a^3 – 27b^3 – 36a^2b + 54ab^2$
(iii) $[\frac{3}{2}x + 1]^3$
$= (\frac{3}{2}x)^3 + 1^3 + 3(\frac{3}{2}x)(1)(\frac{3}{2}x + 1)$
$= \frac{27}{8}x^3 + 1 + \frac{9}{2}x(\frac{3}{2}x + 1) =$ $\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1$
$= (\frac{3}{2}x)^3 + 1^3 + 3(\frac{3}{2}x)(1)(\frac{3}{2}x + 1)$
$= \frac{27}{8}x^3 + 1 + \frac{9}{2}x(\frac{3}{2}x + 1) =$ $\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1$
(iv) $[x – \frac{2}{3}y]^3$
$= x^3 – (\frac{2}{3}y)^3 – 3(x)(\frac{2}{3}y)(x – \frac{2}{3}y)$
$= x^3 – \frac{8}{27}y^3 – 2xy(x – \frac{2}{3}y) =$ $x^3 – \frac{8}{27}y^3 – 2x^2y + \frac{4}{3}xy^2$
$= x^3 – (\frac{2}{3}y)^3 – 3(x)(\frac{2}{3}y)(x – \frac{2}{3}y)$
$= x^3 – \frac{8}{27}y^3 – 2xy(x – \frac{2}{3}y) =$ $x^3 – \frac{8}{27}y^3 – 2x^2y + \frac{4}{3}xy^2$
7. Evaluate the following using suitable identities:
(i) $(99)^3$
$= (100 – 1)^3 = 100^3 – 1^3 – 3(100)(1)(100 – 1)$
$= 1000000 – 1 – 300(99) = 1000000 – 1 – 29700 =$ $970299$
$= (100 – 1)^3 = 100^3 – 1^3 – 3(100)(1)(100 – 1)$
$= 1000000 – 1 – 300(99) = 1000000 – 1 – 29700 =$ $970299$
(ii) $(102)^3$
$= (100 + 2)^3 = 100^3 + 2^3 + 3(100)(2)(100 + 2)$
$= 1000000 + 8 + 600(102) = 1000000 + 8 + 61200 =$ $1061208$
$= (100 + 2)^3 = 100^3 + 2^3 + 3(100)(2)(100 + 2)$
$= 1000000 + 8 + 600(102) = 1000000 + 8 + 61200 =$ $1061208$
(iii) $(998)^3$
$= (1000 – 2)^3 = 1000^3 – 2^3 – 3(1000)(2)(1000 – 2)$
$= 1000000000 – 8 – 6000(998) = 1000000000 – 8 – 5988000 =$ $994011992$
$= (1000 – 2)^3 = 1000^3 – 2^3 – 3(1000)(2)(1000 – 2)$
$= 1000000000 – 8 – 6000(998) = 1000000000 – 8 – 5988000 =$ $994011992$
8. Factorise each of the following:
(i) $8a^3 + b^3 + 12a^2b + 6ab^2$
$= (2a)^3 + b^3 + 3(2a)^2(b) + 3(2a)(b)^2$
$=$ $(2a + b)^3$
$= (2a)^3 + b^3 + 3(2a)^2(b) + 3(2a)(b)^2$
$=$ $(2a + b)^3$
(ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
$= (2a)^3 – b^3 – 3(2a)^2(b) + 3(2a)(b)^2$
$=$ $(2a – b)^3$
$= (2a)^3 – b^3 – 3(2a)^2(b) + 3(2a)(b)^2$
$=$ $(2a – b)^3$
(iii) $27 – 125a^3 – 135a + 225a^2$
$= 3^3 – (5a)^3 – 3(3)^2(5a) + 3(3)(5a)^2$
$=$ $(3 – 5a)^3$
$= 3^3 – (5a)^3 – 3(3)^2(5a) + 3(3)(5a)^2$
$=$ $(3 – 5a)^3$
(iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
$= (4a)^3 – (3b)^3 – 3(4a)^2(3b) + 3(4a)(3b)^2$
$=$ $(4a – 3b)^3$
$= (4a)^3 – (3b)^3 – 3(4a)^2(3b) + 3(4a)(3b)^2$
$=$ $(4a – 3b)^3$
(v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
$= (3p)^3 – (\frac{1}{6})^3 – 3(3p)^2(\frac{1}{6}) + 3(3p)(\frac{1}{6})^2$
$=$ $(3p – \frac{1}{6})^3$
$= (3p)^3 – (\frac{1}{6})^3 – 3(3p)^2(\frac{1}{6}) + 3(3p)(\frac{1}{6})^2$
$=$ $(3p – \frac{1}{6})^3$
9. Verify:
(i) $x^3 + y^3 = (x + y)(x^2 – xy + y^2)$
RHS $= x(x^2 – xy + y^2) + y(x^2 – xy + y^2)$
$= x^3 – x^2y + xy^2 + x^2y – xy^2 + y^3 = x^3 + y^3 =$ LHS. Verified.
RHS $= x(x^2 – xy + y^2) + y(x^2 – xy + y^2)$
$= x^3 – x^2y + xy^2 + x^2y – xy^2 + y^3 = x^3 + y^3 =$ LHS. Verified.
(ii) $x^3 – y^3 = (x – y)(x^2 + xy + y^2)$
RHS $= x(x^2 + xy + y^2) – y(x^2 + xy + y^2)$
$= x^3 + x^2y + xy^2 – x^2y – xy^2 – y^3 = x^3 – y^3 =$ LHS. Verified.
RHS $= x(x^2 + xy + y^2) – y(x^2 + xy + y^2)$
$= x^3 + x^2y + xy^2 – x^2y – xy^2 – y^3 = x^3 – y^3 =$ LHS. Verified.
10. Factorise each of the following:
(i) $27y^3 + 125z^3$
$= (3y)^3 + (5z)^3$. Using $a^3+b^3 = (a+b)(a^2-ab+b^2)$
$= (3y + 5z)((3y)^2 – (3y)(5z) + (5z)^2) =$ $(3y + 5z)(9y^2 – 15yz + 25z^2)$
$= (3y)^3 + (5z)^3$. Using $a^3+b^3 = (a+b)(a^2-ab+b^2)$
$= (3y + 5z)((3y)^2 – (3y)(5z) + (5z)^2) =$ $(3y + 5z)(9y^2 – 15yz + 25z^2)$
(ii) $64m^3 – 343n^3$
$= (4m)^3 – (7n)^3$. Using $a^3-b^3 = (a-b)(a^2+ab+b^2)$
$= (4m – 7n)((4m)^2 + (4m)(7n) + (7n)^2) =$ $(4m – 7n)(16m^2 + 28mn + 49n^2)$
$= (4m)^3 – (7n)^3$. Using $a^3-b^3 = (a-b)(a^2+ab+b^2)$
$= (4m – 7n)((4m)^2 + (4m)(7n) + (7n)^2) =$ $(4m – 7n)(16m^2 + 28mn + 49n^2)$
11. Factorise: $27x^3 + y^3 + z^3 – 9xyz$
$= (3x)^3 + y^3 + z^3 – 3(3x)(y)(z)$
Using $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$= (3x + y + z)((3x)^2 + y^2 + z^2 – (3x)y – yz – z(3x))$
$=$ $(3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)$
Using $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$= (3x + y + z)((3x)^2 + y^2 + z^2 – (3x)y – yz – z(3x))$
$=$ $(3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)$
12. Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2]$
RHS $= \frac{1}{2}(x+y+z)(x^2 – 2xy + y^2 + y^2 – 2yz + z^2 + z^2 – 2zx + x^2)$
$= \frac{1}{2}(x+y+z)(2x^2 + 2y^2 + 2z^2 – 2xy – 2yz – 2zx)$
$= (x+y+z)(x^2 + y^2 + z^2 – xy – yz – zx) =$ LHS. Verified.
$= \frac{1}{2}(x+y+z)(2x^2 + 2y^2 + 2z^2 – 2xy – 2yz – 2zx)$
$= (x+y+z)(x^2 + y^2 + z^2 – xy – yz – zx) =$ LHS. Verified.
13. If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
We know: $x^3 + y^3 + z^3 – 3xyz = (x+y+z)(x^2 + y^2 + z^2 – xy – yz – zx)$
Since $x+y+z = 0$, RHS becomes $0 \times (\dots) = 0$.
So, $x^3 + y^3 + z^3 – 3xyz = 0 \Rightarrow$ $x^3 + y^3 + z^3 = 3xyz$.
Since $x+y+z = 0$, RHS becomes $0 \times (\dots) = 0$.
So, $x^3 + y^3 + z^3 – 3xyz = 0 \Rightarrow$ $x^3 + y^3 + z^3 = 3xyz$.
14. Without actually calculating the cubes, find the value of each:
(i) $(-12)^3 + (7)^3 + (5)^3$
Let $x=-12, y=7, z=5$. Since $x+y+z = -12+7+5 = 0$, we use $x^3+y^3+z^3 = 3xyz$.
$= 3(-12)(7)(5) = 3(-420) =$ $-1260$
Let $x=-12, y=7, z=5$. Since $x+y+z = -12+7+5 = 0$, we use $x^3+y^3+z^3 = 3xyz$.
$= 3(-12)(7)(5) = 3(-420) =$ $-1260$
(ii) $(28)^3 + (-15)^3 + (-13)^3$
Let $x=28, y=-15, z=-13$. Since $x+y+z = 28-15-13 = 0$.
$= 3(28)(-15)(-13) = 3(28)(195) = 84(195) =$ $16380$
Let $x=28, y=-15, z=-13$. Since $x+y+z = 28-15-13 = 0$.
$= 3(28)(-15)(-13) = 3(28)(195) = 84(195) =$ $16380$
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area: $25a^2 – 35a + 12$
Factorise by splitting middle term: $25 \times 12 = 300$. Factors summing to -35 are -20, -15.
$= 25a^2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)$
Possible Dimensions: Length: $(5a-3)$, Breadth: $(5a-4)$
Factorise by splitting middle term: $25 \times 12 = 300$. Factors summing to -35 are -20, -15.
$= 25a^2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)$
Possible Dimensions: Length: $(5a-3)$, Breadth: $(5a-4)$
(ii) Area: $35y^2 + 13y – 12$
Factorise: $35 \times -12 = -420$. Factors summing to +13 are +28, -15.
$= 35y^2 + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3)$
Possible Dimensions: Length: $(7y-3)$, Breadth: $(5y+4)$
Factorise: $35 \times -12 = -420$. Factors summing to +13 are +28, -15.
$= 35y^2 + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3)$
Possible Dimensions: Length: $(7y-3)$, Breadth: $(5y+4)$
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: $3x^2 – 12x$
Factorise: $3x(x – 4)$
Dimensions are factors: $3$, $x$, and $(x – 4)$
Factorise: $3x(x – 4)$
Dimensions are factors: $3$, $x$, and $(x – 4)$
(ii) Volume: $12ky^2 + 8ky – 20k$
$= 4k(3y^2 + 2y – 5)$
Factorise quadratic: $3y^2 + 5y – 3y – 5 = y(3y + 5) – 1(3y + 5) = (3y + 5)(y – 1)$
Dimensions: $4k$, $(3y + 5)$, and $(y – 1)$
$= 4k(3y^2 + 2y – 5)$
Factorise quadratic: $3y^2 + 5y – 3y – 5 = y(3y + 5) – 1(3y + 5) = (3y + 5)(y – 1)$
Dimensions: $4k$, $(3y + 5)$, and $(y – 1)$