Chapter 02 Polynomials Exercise-04
EXERCISE 2.4
1. Use suitable identities to find the following products:
(i) $(x+4)(x+10)$
(ii) $(x+8)(x-10)$
(iii) $(3 x+4)(3 x-5)$
(iv) $\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)$
(v) $(3-2 x)(3+2 x)$
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Solution
(i) By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$, $(x+4)(x+10)=x^{2}+(4+10) x+4 \times 10$
$ =x^{2}+14 x+40 $
(ii) By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$,
$ \begin{aligned} (x+8)(x-10) & =x^{2}+(8-10) x+(8)(-10) \\ & =x^{2}-2 x-80 \end{aligned} $
(iii)
$ (3 x+4)(3 x-5)=9(x+\frac{4}{3})(x-\frac{5}{3}) \text{. } $
By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$;
$ \begin{aligned} 9(x+\frac{4}{3})(x-\frac{5}{3}) & =9[x^{2}+(\frac{4}{3}-\frac{5}{3}) x+(\frac{4}{3})(-\frac{5}{3})] \\ & =9[x^{2}-\frac{1}{3} x-\frac{20}{9}] \\ & =9 x^{2}-3 x-20 \end{aligned} $
(iv) Bv usina the identitv $(\overline{x+y})(\overline{x-y})=x^{2}-y^{2}$,
$ \begin{aligned} (y^{2}+\frac{3}{2})(y^{2}-\frac{3}{2}) & =(y^{2})^{2}-(\frac{3}{2})^{2} \\ & =y^{4}-\frac{9}{4} \end{aligned} $
(v) By using the identity $(x+y)(x-y)=x^{2}-y^{2}$
$ \begin{aligned} (3-2 x)(3+2 x) & =(3)^{2}-(2 x)^{2} \\ & =9-4 x^{2} \end{aligned} $
2. Evaluate the following products without multiplying directly:
(i) $103 \times 107$
(ii) $95 \times 96$
(iii) $104 \times 96$
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Solution
(i) $103 \times 107=(100+3)(100+7)$
$=(100)^{2}+(3+7) 100+(3)(7)$
[By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$, where $x=$
$100, a=3$, and $b=7]$
$=10000+1000+21$
$=11021$ (ii) $95 \times 96=(100-5)(100-4)$
$=(100)^{2}+(-5-4) 100+(-5)(-4)$
[By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$, where $x=$ $100, a=-5$, and $b=-4]$
$=10000-900+20$
$=9120$
(iii) $104 \times 96=(100+4)(100-4)$
$=(100)^{2}-(4)^{2}[(x+y)(x-y)=x^{2}-y^{2}]$
$=10000-16$
$=9984$
3. Factorise the following using appropriate identities:
(i) $9 x^{2}+6 x y+y^{2}$
(ii) $4 y^{2}-4 y+1$
(iii) $x^{2}-\frac{y^{2}}{100}$
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Solution
(i) $9 x^{2}+6 x y+y^{2}=(3 x)^{2}+2(3 x)(y)+(y)^{2}$
$=(3 x+y)(3 x+y) \quad[x^{2}+2 x y+y^{2}=(x+y)^{2}]$
(ii) $4 y^{2}-4 y+1=(2 y)^{2}-2(2 y)(1)+(1)^{2}$
$=(2 y-1)(2 y-1) \quad[x^{2}-2 x y+y^{2}=(x-y)^{2}]$
(iii) $x^{2}-\frac{y^{2}}{100}=x^{2}-(\frac{y}{10})^{2}$
$=(x+\frac{y}{10})(x-\frac{y}{10}) \quad[x^{2}-y^{2}=(x+y)(x-y)]$
4. Expand each of the following, using suitable identities:
(i) $(x+2 y+4 z)^{2}$
(ii) $(2 x-y+z)^{2}$
(iii) $(-2 x+3 y+2 z)^{2}$
(iv) $(3 a-7 b-c)^{2}$
(v) $(-2 x+5 y-3 z)^{2}$
(vi) $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$
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Solution
It is known that,
$ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x $
$ \begin{aligned} & \text{ (i) }(x+2 y+4 z)^{2}=x^{2}+(2 y)^{2}+(4 z)^{2}+2(x)(2 y)+2(2 y)(4 z)+2(4 z)(x) \\ & =x^{2}+4 y^{2}+16 z^{2}+4 x y+16 y z+8 x z \\ & \text{ (ii) }(2 x-y+z)^{2}=(2 x)^{2}+(-y)^{2}+(z)^{2}+2(2 x)(-y)+2(-y)(z)+2(z)(2 x) \\ & =4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z \\ & \text{ (iii) }(-2 x+3 y+2 z)^{2} \\ & =(-2 x)^{2}+(3 y)^{2}+(2 z)^{2}+2(-2 x)(3 y)+2(3 y)(2 z)+2(2 z)(-2 x) \\ & =4 x^{2}+9 y^{2}+4 z^{2}-12 x y+12 y z-8 x z \\ & \text{ (iv) }(3 a-7 b-c)^{2} \\ & =(3 a)^{2}+(-7 b)^{2}+(-c)^{2}+2(3 a)(-7 b)+2(-7 b)(-c)+2(-c)(3 a) \\ & =9 a^{2}+49 b^{2}+c^{2}-42 a b+14 b c-6 a c \\ & (-2 x+5 y-3 z)^{2} \\ & \text{ (v) }(-2 x)^{2}+(5 y)^{2}+(-3 z)^{2}+2(-2 x)(5 y)+2(5 y)(-3 z)+2(-3 z)(-2 x) \\ & =4 x^{2}+25 y^{2}+9 z^{2}-20 x y-30 y z+12 x z \\ & {[\frac{1}{4} a-\frac{1}{2} b+1]^{2}} \\ & \text{ (vi) } \\ & =(\frac{1}{4} a)^{2}+(-\frac{1}{2} b)^{2}+(1)^{2}+2(\frac{1}{4} a)(-\frac{1}{2} b)+2(-\frac{1}{2} b)(1)+2(\frac{1}{4} a)(1) \\ & =\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a \end{aligned} $
5. Factorise:
(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
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Solution
It is known that,
$ \begin{aligned} & (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x \\ & \text{ (i) } 4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z \\ & =(2 x)^{2}+(3 y)^{2}+(-4 z)^{2}+2(2 x)(3 y)+2(3 y)(-4 z)+2(2 x)(-4 z) \\ & =(2 x+3 y-4 z)^{2} \\ & =(2 x+3 y-4 z)(2 x+3 y-4 z) \\ & \text{ (ii) } 2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z \\ & =(-\sqrt{2} x)^{2}+(y)^{2}+(2 \sqrt{2} z)^{2}+2(-\sqrt{2} x)(y)+2(y)(2 \sqrt{2} z)+2(-\sqrt{2} x)(2 \sqrt{2} z) \\ & =(-\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) \end{aligned} $
6. Write the following cubes in expanded form:
(i) $(2 x+1)^{3}$
(ii) $(2 a-3 b)^{3}$
(iii) $\left[\frac{3}{2} x+1\right]^{3}$
(iv) $\left[x-\frac{2}{3} y\right]^{3}$
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Solution
It is known that,
$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
and $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
(i) $(2 x+1)^{3}=(2 x)^{3}+(1)^{3}+3(2 x)(1)(2 x+1)$
$ \begin{aligned} & =8 x^{3}+1+6 x(2 x+1) \\ & =8 x^{3}+1+12 x^{2}+6 x \\ & =8 x^{3}+12 x^{2}+6 x+1 \\ & \text{ (ii) }(2 a-3 b)^{3}=(2 a)^{3}-(3 b)^{3}-3(2 a)(3 b)(2 a-3 b) \\ & =8 a^{3}-27 b^{3}-18 a b(2 a-3 b) \\ & =8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2} \\ & \text{ (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}+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 \\ & \\ & \text{ (vi) }[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}-2 x y(x-\frac{2}{3} y) \\ & =x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2} \end{aligned} $
7. Evaluate the following using suitable identities:
(i) $(99)^{3}$
(ii) $(102)^{3}$
(iii) $(998)^{3}$
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Solution
It is known that,
(i) $(99)^{3}=(100-1)^{3}$
$ \begin{aligned} & (a+b)^{3}=a^{3}+b^{3}+3 a b(a+b) \\ & \text{ and }(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b) \end{aligned} $
$ \begin{aligned} & =(100)^{3}-(1)^{3}-3(100)(1)(100-1) \\ & =1000000-1-300(99) \\ & =1000000-1-29700 \\ & =970299 \end{aligned} $
$ \begin{aligned} & \text{ (ii) }(102)^{3}=(100+2)^{3} \\ & =(100)^{3}+(2)^{3}+3(100)(2)(100+2) \\ & =1000000+8+600(102) \\ & =1000000+8+61200=1061208 \\ & \text{ (iii) }(998)^{3}=(1000-2)^{3} \\ & =(1000)^{3}-(2)^{3}-3(1000)(2)(1000-2) \\ & =1000000000-8-6000(998) \\ & =1000000000-8-5988000 \\ & =1000000000-5988008 \\ & =994011992 \text{ Question } \\ & \text{ 8: } \end{aligned} $
8. Factorise each of the following:
(i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
(ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$
(iii) $27-125 a^{3}-135 a+225 a^{2}$
(iv) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$
(v) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p$
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Solution
It is known that, $(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$
and $(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$
(i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
$ \begin{aligned} & =(2 a)^{3}+(b)^{3}+3(2 a)^{2} b+3(2 a)(b)^{2} \\ & =(2 a+b)^{3} \\ & =(2 a+b)(2 a+b)(2 a+b) \\ & \text{ (ii) } 8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2} \\ & =(2 a)^{3}-(b)^{3}-3(2 a)^{2} b+3(2 a)(b)^{2} \\ & =(2 a-b)^{3} \\ & =(2 a-b)(2 a-b)(2 a-b) \\ & \text{ (iii) } 27-125 a^{3}-135 a+225 a^{2} \\ & =(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2} \\ & =(3-5 a)^{3} \\ & =(3-5 a)(3-5 a)(3-5 a) \\ & (\text{ iv) } 64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}. \\ & =(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2} \\ & =(4 a-3 b)^{3} \\ & =(4 a-3 b)(4 a-3 b)(4 a-3 b) \\ & 27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p \\ & \text{ (v) } \\ & =(3 p)^{3}-(\frac{1}{6})^{3}-3(3 p)^{2}(\frac{1}{6})+3(3 p)(\frac{1}{6})^{2} \\ & =(3 p-\frac{1}{6})^{3} \\ & =(3 p-\frac{1}{6})(3 p-\frac{1}{6})(3 p-\frac{1}{6}) \end{aligned} $
9. Verify : (i) $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$
(ii) $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$
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Solution
(i) It is known that,
$ \begin{aligned} (x+y)^{3} & =x^{3}+y^{3}+3 x y(x+y) \\ x^{3}+y^{3} & =(x+y)^{3}-3 x y(x+y) \\ & =(x+y)[(x+y)^{2}-3 x y] \\ & =(x+y)(x^{2}+y^{2}+2 x y-3 x y) \\ & =(x+y)(x^{2}+y^{2}-x y) \\ & =(x+y)(x^{2}-x y+y^{2}) \end{aligned} $
(ii) It is known that,
$ \begin{aligned} (x-y)^{3} & =x^{3}-y^{3}-3 x y(x-y) \\ x^{3}-y^{3} & =(x-y)^{3}+3 x y(x-y) \\ & =(x-y)[(x-y)^{2}+3 x y] \\ & =(x-y)(x^{2}+y^{2}-2 x y+3 x y) \\ & =(x-y)(x^{2}+y^{2}+x y) \\ & =(x-y)(x^{2}+x y+y^{2}) \end{aligned} $
10. Factorise each of the following:
(i) $27 y^{3}+125 z^{3}$
(ii) $64 m^{3}-343 n^{3}$
[Hint : See Question 9.]
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Solution
$ \begin{aligned} & \text{ (i) } 27 y^{3}+125 z^{3} \\ & =(3 y)^{3}+(5 z)^{3} \\ & =(3 y+5 z)[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)] \quad[\because a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)] \\ & =(3 y+5 z)[9 y^{2}+25 z^{2}-15 y z] \\ & \text{ (ii) } 64 m^{3}-343 n^{3} \\ & =(4 m)^{3}-(7 n)^{3} \\ & =(4 m-7 n)[(4 m)^{2}+(7 n)^{2}+(4 m)(7 n)] \quad[\because a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)] \\ & =(4 m-7 n)[16 m^{2}+49 n^{2}+28 m n] \end{aligned} $
11. Factorise : $27 x^{3}+y^{3}+z^{3}-9 x y z$
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Solution
It is known that,
$ \begin{aligned} & x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)(x^{2}+y^{2}+z^{2}-x y-y z-z x) \\ & \therefore 27 x^{3}+y^{3}+z^{3}-9 x y z \\ & =(3 x)^{3}+(y)^{3}+(z)^{3}-3(3 x)(y)(z) \\ & =(3 x+y+z)[(3 x)^{2}+y^{2}+z^{2}-(3 x)(y)-(y)(z)-z(3 x)] \\ & =(3 x+y+z)[9 x^{2}+y^{2}+z^{2}-3 x y-y z-3 x z] \end{aligned} $
12. Verify that $x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$
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Solution
It is known that,
$ \begin{aligned} & x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)(x^{2}+y^{2}+z^{2}-x y-y z-z x) \\ = & \frac{1}{2}(x+y+z)[2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x] \\ = & \frac{1}{2}(x+y+z)[(x^{2}+y^{2}-2 x y)+(y^{2}+z^{2}-2 y z)+(x^{2}+z^{2}-2 z x)] \\ = & \frac{1}{2}(x+y+z)[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}] \end{aligned} $
13. If $x+y+z=0$, show that $x^{3}+y^{3}+z^{3}=3 x y z$.
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Solution
It is known that,
$ x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)(x^{2}+y^{2}+z^{2}-x y-y z-z x) $
Put $x+y+z=0$,
$ \begin{aligned} & x^{3}+y^{3}+z^{3}-3 x y z=(0)(x^{2}+y^{2}+z^{2}-x y-y z-z x) \\ & x^{3}+y^{3}+z^{3}-3 x y z=0 \\ & x^{3}+y^{3}+z^{3}=3 x y z \end{aligned} $
14. Without actually calculating the cubes, find the value of each of the following:
(i) $(-12)^{3}+(7)^{3}+(5)^{3}$
(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$
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Solution
(i)
$ (-12)^{3}+(7)^{3}+(5)^{3} $
Let $x=-12, y=7$, and $z=5$
It can be observed that, $x+y$
$+z=-12+7+5=0$
It is known that if $x+y+z=0$, then $x^{3}+y^{3}+z^{3}=3 x y z$
$\therefore \quad(-12)^{3}+(7)^{3}+(5)^{3}=3(-12)(7)(5)$
$=-1260$
(ii)
$ (28)^{3}+(-15)^{3}+(-13)^{3} $
Let $x=28, y=-15$, and $z=-13$
It can be observed that,
$x+y+z=28+(-15)+(-13)=28-28=0$
It is known that if $x+y+z=0$, then
$x^{3}+y^{3}+z^{3}=3 x y z$
$\begin{aligned} \therefore(28)^{3}+(-15)^{3}+(-13)^{3} & =3(28)(-15)(-13) \\ & =16380\end{aligned}$
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
$$ \text { Area : } 25 a^{2}-35 a+12 $$
$$ \text { Area: } 35 y^{2}+13 y-12 $$
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Solution
Area $=$ Length $\times$ Breadth
The expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth.
$ \begin{aligned} & \text{ (i) } 25 a^{2}-35 a+12=25 a^{2}-15 a-20 a+12 \\ & =5 a(5 a-3)-4(5 a-3) \\ & =(5 a-3)(5 a-4) \end{aligned} $
Therefore, possible length $=5 a-3$
And, possible breadth $=5 a-4$
$ \begin{aligned} & \text{ (ii) } 35 y^{2}+13 y-12=35 y^{2}+28 y-15 y-12 \\ & =7 y(5 y+4)-3(5 y+4) \\ & =(5 y+4)(7 y-3) \end{aligned} $
Therefore, possible length $=5 y+4$
And, possible breadth $=7 y-3$
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) $\boxed{\text { Volume :} 3 x^2-12 x}$
(ii) $ \boxed{\text{ Volume : } 12 k y^{2}+6 k y-20 k} $
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Solution
Volume of cuboid $=$ Length $\times$ Breadth $\times$ Height
The expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height.
(i) $3 x^{2}-12 x=3 x(x-4)$
One of the possible solutions is as follows.
Length $=3$, Breadth $=x$, Height $=x-4$
(ii) $12 k y^{2}+8 k y-20 k=4 k(3 y^{2}+2 y-5)$
$ \begin{aligned} & =4 k[3 y^{2}+5 y-3 y-5] \\ & =4 k[y(3 y+5)-1(3 y+5)] \\ & =4 k(3 y+5)(y-1) \end{aligned} $
One of the possible solutions is as follows.
Length $=4 k$, Breadth $=3 y+5$, Height $=y-1$
