Question 12 Marks
Write the following in the expanded form:
$(x+2 y+4 z)^2$
AnswerWe have,
$(x+2 y+4 z)^2=x^2+(2 y)^2+(4 z)^2+2 x \times 2 y+2 \times 2 y \times 4 z+2 x \times 4 z$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$(x+2 y+4 z)^2=x^2+4 y^2+16 z^2+4 x y+16 y z+8 x z$
View full question & answer→Question 22 Marks
Write the following in the expanded form: $(-3x + y + z)^2$
AnswerWe have, $(-3x + y + z)^2$
$= [(-3x)^2 + y^2 + z^2 + 2(-3x)y + 2yz + 2(-3x)z]$
$\big[\therefore$$(x + y + z)^2 = x^2+ y^2 + z^2 + 2xy + 2yz + 2xz$$\big]$
$9x^2 + y^2 + z^2 - 6xy + 2yz - 6xz (-3x + y + z)^{20}$
$= 9x^2 + y^2 + z^2 - 6xy + 2xy - 6xz$
View full question & answer→Question 32 Marks
Find the following:
$\left(x^3+1\right)\left(x^6-x^3+1\right)$
AnswerWe have,
$\left(x^3+1\right)\left(x^6-x^3+1\right)$
$=\left(x^3+1\right)\left[\left(x^3\right)^2-1 \times x^3+1^2\right]$
$=\left(x^3\right)^3+(1)^3\left[\because a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=x^9+1$
$\therefore\left(x^3+1\right)\left(x^6-x^3+1\right)=x^9+1$
View full question & answer→Question 42 Marks
Find the following products:
$\left(7 p^4+q\right)\left(49 p^8-7 p^4 q+q^2\right)$
Answerwe have,
$\left(7 p^4+q\right)\left(49 p^8-7 p^4 q+q^2\right)$
$=\left(7 p^4+q\right)\left[\left(7 p^4\right)^2-7 p^4 \times q+(q)^2\right]$
$=\left(7 p^4\right)^3+(q)^3\left[\because a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=343 p^{12}+q^3$
$\therefore\left(7 p^4+q\right)\left(49 p^8-7 p^4 q+q^2\right)=343 p^{12}+q^3$
View full question & answer→Question 52 Marks
Evaluate the following using identities:
$(0.98)^2$
AnswerWe have,
$(0.98)^2=(1-0.02)^2$
$=(1)^2+(0.02)^2-2 \times 1 \times 0.02$
$=1+0.0004-0.04[\text { Where, } a=1 \text { and } b=0.02]$
$=1.0004-0.04$
$=0.9604$
Therefore, $(0.98)^2=0.96041$
View full question & answer→Question 62 Marks
Find the following products:
$\Big(\frac{2}{\text{x}}+3\text{x}\Big)\Big(\frac{4}{\text{x}^2}+9\text{x}^2-6\Big)$
Answer We have,
$\Big(\frac{2}{\text{x}}+3\text{x}\Big)\Big(\frac{4}{\text{x}^2}+9\text{x}^2-6\Big)$
$\Big(\frac{2}{\text{x}}+3\text{x}\Big)\bigg[\Big(\frac{2}{\text{x}}\Big)^2+(3\text{x})^2-\frac{2}{\text{x}}\times3\text{x}\bigg]$
$=\Big(\frac{2}{\text{x}}\Big)^3+(3\text{x})^3$ $\big[\therefore\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\frac{8}{\text{x}^3}+27\text{x}^3$
$\therefore\Big(\frac{2}{\text{x}}+3\text{x}\Big)\Big(\frac{4}{\text{x}^2}+9\text{x}^2-6\Big)=\frac{8}{\text{x}^3}+27\text{x}^3$
View full question & answer→Question 72 Marks
Write the following in the expanded form:
$(2+x-2 y)^2$
AnswerWe have,
$(2+x-2 y)^2=2^2+x^2+(-2 y)^2+2(2)(x)+2(x)(-2 y)+2(2)(-2 y)$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$=4+x^2+4 y^2+4 x-4 x y-8 y$
$(2+x-2 y)^2=4+x^2+4 y^2+4 x-4 x y-8 y$
View full question & answer→Question 82 Marks
If $\text{x}+\frac{1}{\text{x}}=11$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
AnswerWe have $\text{x}+\frac{1}{\text{x}}=11$
Now, $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\big(\frac{1}{\text{x}}\big)^2+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(11)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow121=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=119.$
View full question & answer→Question 92 Marks
Simplify the following:
$322 \times 322-2 \times 322 \times 22+22 \times 22$
AnswerWe have,
$322 \times 322-2 \times 322 \times 22+22 \times 22$
$=(322-22)^2\left[a^2+b^2-2 a b=(a-b)^2\right]$
$=(300)^2[\text { Where } a=322 \text { and } b=22]$
$=90000$
Therefore, $322 \times 322-2 \times 322 \times 22+22 \times 22=90000$.
View full question & answer→Question 102 Marks
Find the following products:
$\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)$
Answer We have,
$\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)$
$=\Big(\frac{\text{x}}{2}+2\text{y}\Big)\bigg[\Big(\frac{\text{x}}{2}\Big)^2-\text{xy}-(2\text{y})^2\bigg]$
$=\Big(\frac{\text{x}}{2}\Big)^3+(2\text{y})^3$ $\big[\therefore\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\frac{\text{x}^3}{8}+8\text{y}^3$
$\therefore\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)=\frac{\text{x}^3}{8}+8\text{y}^3$
View full question & answer→Question 112 Marks
Evaluate:
$25^3-75^3+50^3$
AnswerLet $\mathrm{a}=25, \mathrm{~b}=-75$ and $\mathrm{c}=50$
Then,
$a+b+c=25-75+50$
$=0$
$\therefore a^3+b^3+c^3=3 a b c$
$\Rightarrow(25)^3+(-75)^3+(50)^3=3 \times 25 \times(-75) \times 50$
$=-75 \times 75 \times 50$
$=-5625 \times 50$
$=-281250$
$\therefore 25^3-75^3+50^3=-281250$
View full question & answer→Question 122 Marks
Write the following in the expanded form:
$(m+2 n-5 p)^2$
AnswerWe have,
$(m+2 n-5 p)^2=m^2+(2 n)^2+(-5 p)^2+2 m \times 2 n+(2 \times 2 n \times-5 p)+2 m \times-5 p$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$(m+2 n-5 p)^2=m^2+4 n^2+25 p^2+4 m n-20 n p-10 p m$
View full question & answer→Question 132 Marks
Write the following in the expanded form: $(2 a-3 b-c)^2$
AnswerWe have,
$(2 a-3 b-c)^2=\left[(2 a)+(-3 b)+(-c)^2\right.$
$(2 a)^2+(-3 b)^2+(-c)^2+2(2 a)(-3 b)+2(-3 b)(-c)+2(2 a)(-c)$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$4 a^2+9 b^2+c^2-12 a b+6 b c-4 c a$
$\therefore(2 a-3 b-c)^2=4 x^2+9 y^2+c^2-12 a b+6 b c-4 c a$
View full question & answer→Question 142 Marks
Find the following products: $\Big(\frac{3}{\text{x}}-2\text{x}^2\Big)\Big(\frac{9}{\text{x}^2}+4\text{x}^2-6\text{x}\Big)$
AnswerWe have, $\Big(\frac{3}{\text{x}}-2\text{x}^2\Big)\Big(\frac{9}{\text{x}^2}+4\text{x}^2-6\text{x}\Big)$
$=\Big(\frac{3}{\text{x}}-2\text{x}^2\Big)\bigg[\Big(\frac{3}{\text{x}}\Big)^2+(2\text{x})^2+\frac{3}{\text{x}}-\times2\text{x}^2\bigg]$
$=\Big(\frac{3}{\text{x}}\Big)^3-(2\text{x}^2)^3$
$\big[\therefore\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)\big]$
$=\frac{27}{\text{x}^3}-8\text{x}^6$
$\therefore\Big(\frac{3}{\text{x}}-2\text{x}^2\Big)\Big(\frac{9}{\text{x}^2}+4\text{x}^2-6\text{x}\Big)=\frac{27}{\text{x}^3}-8\text{x}^6$
View full question & answer→Question 152 Marks
Write the following in the expanded form: $\Big(\frac{\text{a}}{\text{bc}}+\frac{\text{b}}{\text{ca}}+\frac{\text{c}}{\text{ab}}\Big)^2$
AnswerWe have, $\Big(\frac{\text{a}}{\text{bc}}+\frac{\text{b}}{\text{ca}}+\frac{\text{c}}{\text{ab}}\Big)^2=\Big(\frac{\text{a}}{\text{bc}}\Big)^2+\Big(\frac{\text{b}}{\text{ca}}\Big)^2+\Big(\frac{\text{c}}{\text{ab}}\Big)^2$
$+2\Big(\frac{\text{a}}{\text{bc}}\Big)\Big(\frac{\text{b}}{\text{ca}}\Big)+2\Big(\frac{\text{b}}{\text{ca}}\Big)\Big(\frac{\text{c}}{\text{ab}}\Big)+2\Big(\frac{\text{a}}{\text{bc}}\Big)\Big(\frac{\text{c}}{\text{ab}}\Big)$
$\big[\therefore\big(\text{x}+\text{y}+\text{z}\big)=\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{xz}\big]$
$\Big(\frac{\text{a}}{\text{bc}}+\frac{\text{b}}{\text{ca}}+\frac{\text{c}}{\text{ab}}\Big)^2\Big(\frac{\text{a}}{\text{bc}}+\frac{\text{b}}{\text{ca}}+\frac{\text{c}}{\text{ab}}\Big)^2$
$=\Big(\frac{\text{a}^2}{\text{b}^2\text{c}^2}\Big)+\Big(\frac{\text{b}^2}{\text{c}^2\text{a}^2}\Big)+\Big(\frac{\text{c}^2}{\text{a}^2\text{b}^2}\Big)+2\frac{2}{\text{a}^2}+\frac{2}{\text{b}^2}+\frac{2}{\text{c}^2}$
View full question & answer→Question 162 Marks
Find the following products: $\Big(\frac{3}{\text{x}}-\frac{5}{\text{y}}\Big)\Big(\frac{9}{\text{x}^2}+\frac{25}{\text{y}^2}+\frac{15}{\text{xy}}\Big)$
AnswerWe have, $\Big(\frac{3}{\text{x}}-\frac{5}{\text{y}}\Big)\Big(\frac{9}{\text{x}^2}+\frac{25}{\text{y}^2}+\frac{15}{\text{xy}}\Big)$
$=\Big(\frac{3}{\text{x}}-\frac{5}{\text{y}}\Big)\bigg[\Big(\frac{3}{\text{x}}\Big)^2+\Big(\frac{5}{\text{y}}\Big)^2+\frac{3}{\text{x}}\times\frac{5}{\text{y}}\Big)\bigg]$
$=\Big(\frac{3}{\text{x}}\Big)^3-\Big(\frac{5}{\text{y}}\Big)^3$
$\big[\therefore\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)\big]$
$=\frac{27}{\text{x}^3}-\frac{125}{\text{y}^3}$
$\therefore\Big(\frac{3}{\text{x}}-\frac{5}{\text{y}}\Big)\Big(\frac{9}{\text{x}^2}+\frac{25}{\text{y}^2}+\frac{15}{\text{xy}}\Big)=\frac{27}{\text{x}^3}-\frac{125}{\text{y}^3}$
View full question & answer→Question 172 Marks
Find the following products:
$(1+x)\left(1-x+x^2\right)$
AnswerWe have,
$(1+x)\left(1-x+x^2\right)$
$=(1+x)\left[(1)^2-1 \times x+(x)^2\right]$
$=1^3+x^3\left[\because a^3-b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$\therefore(1+x)\left(1-+x+x^2\right)=1+x^3$
View full question & answer→Question 182 Marks
Find the following products: $\Big(3+\frac{5}{\text{x}}\Big)\Big(9-\frac{15}{\text{x}}+\frac{25}{\text{x}^2}\Big)$
AnswerWe have, $\Big(3+\frac{5}{\text{x}}\Big)\Big(9-\frac{15}{\text{x}}+\frac{25}{\text{x}^2}\Big)$
$\Big(3+\frac{5}{\text{x}}\Big)\Big[(3)^2-3\times\frac{5}{\text{x}}+\Big(\frac{5}{\text{x}}\Big)^2\Big]$
$=(3)^3+\Big(\frac{5}{\text{x}}\Big)^2$
$\big[\therefore\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=27+\frac{125}{\text{x}^3}$
$\therefore\Big(3+\frac{5}{\text{x}}\Big)\Big(9-\frac{15}{\text{x}}+\frac{25}{\text{x}^2}\Big)=27+\frac{125}{\text{x}^3}$
View full question & answer→Question 192 Marks
Simplify the following:
$\frac{7.83\times7.83-1.17\times1.17}{6.66}$
Answer We have,
$\frac{7.83\times7.83-1.17\times1.17}{6.66}$
$=\frac{(7.83+1.17)(7.83-1.17)}{6.66}$ $\big[\because(\text{a}-\text{b})^2=(\text{a}+\text{b})(\text{a}-\text{b})\big]$
$=\frac{(9.00)(6.66)}{6.66}=9$
$\therefore\frac{7.83\times7.83-1.17\times1.17}{6.66}=9$
View full question & answer→Question 202 Marks
Evaluate:
$(0.2)^3-(0.3)^3+(0.1)^3$
AnswerLet $\mathrm{a}=0.2, \mathrm{~b}=-0.3$ and $\mathrm{c}=0.1$
Then,
$a+b+c=0.2-0.3+0.1$
$=0.3-0.3$
$\Rightarrow a+b+c=0$
$\therefore a^3+b^3+c^3=3 a b c$
$\Rightarrow(0.2)^3+(-0.3)^3+(0.1)^3=3 \times(0.2) \times(-0.3) \times(0.1)$
$=-0.018$
View full question & answer→Question 212 Marks
Write the following in the expanded form:
$(-2 x+3 y+2 z)^2$
AnswerWe have,
$(-2 x+3 y+2 z)^2=(-2 x)^2+(3 y)^2+(2 z)^2+2(-2 x)(3 y)+2(3 y)(2 z)+2(-2 x)(2 z)$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$(-4 x+6 y+4 z)^2=4 x^2+9 y^2+4 z^2-12 x y+12 y z-8 x z$
View full question & answer→Question 222 Marks
Evaluate the following using identities: $\big(1.5\text{x}^2-0.3\text{y}^2\big)\big(1.5\text{x}+0.3\text{y}^2\big)$
AnswerWe have, $\big(1.5\text{x}^2-0.3\text{y}^2\big)\big(1.5\text{x}+0.3\text{y}^2\big)$
$=\big[1.5\text{x}^2\big]^2-\big[0.3\text{y}^2\big]^2$
$\begin{bmatrix}\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2$
$\text{a}=1.5\text{x}^2\ \text{and}\ \text{b}=0.3\text{y}^2\end{bmatrix}$
$=2.25\text{x}^4-0.09\text{y}^4$
$=\big[1.5\text{x}^2-0.3\text{y}^2\big]\big[1.5\text{x}^2+0.3\text{y}^2\big]$
$=2.25\text{x}^4-0.09\text{y}^4.$
View full question & answer→Question 232 Marks
Write the following in the expanded form: $\Big(\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{z}}+\frac{\text{z}}{\text{x}}\Big)^2$
AnswerWe have, $\Big(\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{z}}+\frac{\text{z}}{\text{x}}\Big)^2$
$=\Big(\frac{\text{x}}{\text{y}}\Big)^2+\Big(\frac{\text{y}}{\text{z}}\Big)+\Big(\frac{\text{z}}{\text{x}}\Big)^2$
$+2\frac{\text{x}}{\text{y}}\frac{\text{y}}{\text{z}}+2\frac{\text{y}}{\text{z}}\frac{\text{z}}{\text{x}}+2\frac{\text{z}}{\text{x}}\frac{\text{x}}{\text{y}}$
$\big[\therefore\big(\text{x}+\text{y}+\text{z}\big)$
$=\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{xz}\big]$
$\therefore\Big(\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{z}}+\frac{\text{z}}{\text{x}}\Big)^2$
$=\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\Big(\frac{\text{y}^2}{\text{z}^2}\Big)+\Big(\frac{\text{z}^2}{\text{x}^2}\Big)+2\frac{\text{x}}{\text{z}}+2\frac{\text{y}}{\text{x}}+2\frac{\text{x}}{\text{y}}$
View full question & answer→Question 242 Marks
Evaluate the following using identities: $\big(\text{a}^2\text{b}-\text{b}^2\text{a}\big)^2$
AnswerWe have, $\big(\text{a}^2\text{b}-\text{a}\text{b}^2\big)^2$
$=\big(\text{a}^2\text{b}\big)^2+\big(\text{ab}^2\big)^2-2\times\text{a}^2\times\text{ab}^2$
$\begin{bmatrix}\because\big(\text{a}-\text{b}\big)^2=\text{a}^2+\text{b}^2-2\text{ab} \text{Where}\ \text{a}=\text{a}^2\text{b}\ \text{and}\ \text{b}=\text{ab}^2\end{bmatrix}$
$=\text{a}^4\text{b}^2+\text{b}^4\text{a}^2-2\text{a}^3\text{b}^3$
$\therefore\big(\text{a}^2\text{b}-\text{b}^2\text{a}\big)^2=\text{a}^4\text{b}^2+\text{b}^4\text{a}^2-2\text{a}^3\text{b}^3$
View full question & answer→Question 252 Marks
If $\text{x}-\frac{1}{\text{x}}=-1$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
AnswerWe have, $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\big(\frac{1}{\text{x}}\big)^2-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Rightarrow(-1)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\big[\because\text{x}-\frac{1}{\text{x}}=-1\big]$
$\Rightarrow2+1=\text{x}^2+\frac{1}{\text{x}^2}$
$\therefore\ \text{x}^2+\frac{1}{\text{x}^2}=3.$
View full question & answer→Question 262 Marks
Write the following in the expanded form:
$(a b+b c+c a)^2$
AnswerWe have,
$(a b+b c+c a)^2=(a b)^2+(b c)^2+(c a)^2+2(a b)(b c)+2(b c)(c a)+2(a b)(c a)$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$=a^2 b^2+b^2 c^2+c^2 a^2+2(a c) b^2+2(a b)(c)^2+2(b c)(a)^2$
$(a b+b c+c a)^2=a^2 b^2+b^2 c^2+c^2 a^2+2 a c b^2+2 a b c^2+2 b c a^2$
View full question & answer→Question 272 Marks
Simplify the following expressions:
$(x+y-2 z)^2-x^2-y^2-3 z^2+4 x y$
AnswerExpanding, we get
$(x+y-2 z)^2-x^2-y^2-3 z^2+4 x y$
$=\left[x^2+y^2+4 z^2+2 x y+2 y(-2 z)+2 a(-2 c)\right]-x^2-y^2-3 z^2+4 x y$
$=z^2+6 x y-4 y z-4 z x$
$(x+y-2 z)^2-x^2-y^2-3 z^2+4 x y=z^2+6 x y-4 y z-4 z x$
View full question & answer→Question 282 Marks
Write the following in the expanded form:
$\left(a^2+b^2+c^2\right)^2$
AnswerWe have,
$\left(a^2+b^2+c^2\right)^2=\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2+2 a^2 b^2+2 b^2 c^2+2 a^2 c^2$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2 a^2 b^2+2 b^2 c^2+2 c^2 a^2$
View full question & answer→Question 292 Marks
Write the following in the expanded form:
$(2 x-y+z)^2$
AnswerWe have,
$(2 x-y+z)^2=(2 x)^2+(-y)^2+(z)^2+2(2 x)(-y)+2(-y)(z)+2(2 x)(z)$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$(2 x-y+z)^2=4 x^2+y^2+z^2-4 x y-2 y z+4 x z$
View full question & answer→Question 302 Marks
Write the following in the expanded form:
$(a+2 b+c)^2$
AnswerWe have,
$(a+2 b+c)^2=a^2+(2 b)^2+c^2+2 a(2 b)+2 a c+2(2 b) c$
${\left[\therefore(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 x z\right]}$
$\therefore(a+2 b+c)^2=a^2+4 b^2+c^2+4 a b+2 a c+4 b c$
View full question & answer→Question 312 Marks
Evaluate the following using identities:
$(2\text{x} +\text{y})(2\text{x} - \text{y})$
Answer We have,
$(2\text{x} +\text{y})(2\text{x} - \text{y})$
$= (2\text{x})^2 - (\text{y})^2 $ $\begin{bmatrix}\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2\\\text{Where}\ \text{a}=2\text{x}\ \text{and}\ \text{b}=\text{y}\end{bmatrix}$
$=4\text{x}^2-\text{y}^2$
$(2\text{x}+\text{y})(2\text{x}-\text{y})=4\text{x}^2-\text{y}^2$
View full question & answer→Question 322 Marks
Find the following products:
$(3 x+2 y)\left(9 x^2-6 x y+4 y^2\right)$
Answerwe have,
$(3 x+2 y)\left(9 x^2-6 x y+4 y^2\right)$
$=(3 x+2 y)\left[(3 x)^2-3 x \times 2 y+(2 y)^2\right]$
$=(3 x)^3+(2 y)^3\left[\because a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=27 x^3+8 y^3$
$\therefore(3 x+2 y)\left(9 x^2-6 x y+4 y^2\right)=27 x^3+8 y^3$
View full question & answer→Question 332 Marks
Find the following products:
$(4 x-5 y)\left(16 x^2+20 x y+25 y^2\right)$
Answerwe have,
$(4 x-5 y)\left(16 x^2+20 x y+25 y^2\right)$
$=(4 x-5 y)\left[(4 x)^2+4 x \times 5 y+(5 y)^2\right]$
$=(4 x)^3-(5 y)^3\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=64 x^3-125 y^3$
$\therefore(4 x-5 y)\left(16 x^2+20 x y+25 y^2\right)=64 x^3-125 y^3$
View full question & answer→Question 342 Marks
Simplify the following:
$175 \times 175+2 \times 175 \times 25+25 \times 25$
AnswerWe have,
$175 \times 175+2 \times 175 \times 25+25 \times 25=(175)^2+2(175)(25)+(25)^2$
$=(175+25)^2\left[a^2+b^2+2 a b=(a+b)^2\right]$
$=(200)^2[\text { Where } a=175 \text { and } b=25]$
$=40000$
Therefore, $175 \times 175+2 \times 175 \times 25+25 \times 25=40000$.
View full question & answer→Question 352 Marks
Simplify the following:
$0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24$
AnswerWe have,
$0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24$
$=(0.76+0.24)^2\left[a^2+b^2+2 a b=(a+b)^2\right]$
$=(1.00)^2[\text { Where } a=0.76 \text { and } b=0.24]$
$=1$
Therefore, $0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24=1$
View full question & answer→Question 362 Marks
Evaluate:
$48^3-30^3-18^3$
AnswerLet $\mathrm{a}=48, \mathrm{~b}=-30$ and $\mathrm{c}=-18$
Then,
$a+b+c=48-30-18$
$=0$
$\therefore a^3+b^3+c^3=3 a b c$
$\Rightarrow(48)^3+(-30)^3+(-18)^3=3 \times(48) \times(-30) \times(-18)$
$=144 \times 540$
$=77760$
$\therefore 48^3-30^3-18^3=77760$
View full question & answer→Question 372 Marks
Evaluate the following using identities: $\big(\text{a}-0.1\big)\big(\text{a}+0.1\big)$
AnswerWe have, $\big(\text{a}-0.1\big)\big(\text{a}+0.1\big)=\text{a}^2-(0.1)^2$
$\begin{bmatrix}\because(\text{a}-\text{b})(\text{a}+\text{b})\text{a}=\text{a}:\text{b}=0.1\end{bmatrix}$
$=\text{a}^2-0.01$
$(\text{a}-0.1)(\text{a}+0.1)=\text{a}^2-0.01$
View full question & answer→Question 382 Marks
Find the following products:
$(1-x)\left(1+x+x^2\right)$
AnswerWe have,
$(1-x)\left(1+x+x^2\right)$
$=(1-x)\left[(1)^2+1 \times x+(x)^2\right]$
$=1^3-x^3\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=1-x^3$
$\therefore(1-x)\left(1+x+x^2\right)=1-x^3$
View full question & answer→Question 392 Marks
Evaluate the following using identities: $\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2$
AnswerWe have, $\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=(2\text{x})^2+\big(\frac{1}{\text{x}}\big)^2-2.2\text{x}.\frac{1}{\text{x}}$
$\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=4\text{x}^2+\frac{1}{\text{x}^2}-4$
$\begin{bmatrix}\because(\text{a}-\text{b})^2=\text{a}^2+\text{b}^2-2\text{ab},\text{Where}\ \text{a}=2\text{x}\ \text{and}\ \text{b}=\frac{1}{\text{x}}\end{bmatrix}$
$\therefore\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=4\text{x}^2+\frac{1}{\text{x}^2}-4$
View full question & answer→Question 402 Marks
Find the following:
$\left(x^2-1\right)\left(x^4+x^2+1\right)$
AnswerWe have,
$\left(x^2-1\right)\left(x^4+x^2+1\right)$
$=\left(x^2-1\right)\left[\left(x^2\right) 2+1 \times x^2+1^2\right]$
$=\left(x^2\right)^3-(1)^3\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=x^6-1$
$\therefore\left(x^2-1\right)\left(x^4+x^2+1\right)=x^6-1$
View full question & answer→Question 412 Marks
Evaluate the following using identities:
$(399)^2$
AnswerWe have,
$399^2=(400-1)^2$
$=(400)^2+(1)^2-2 \times 400 \times 1\left[(a-b)^2=a^2+b^2-2 a b\right]$
Where, $a=400$ and $b=1$
$=160000+1-8000$
$=159201$
Therefore, $(399)^2=159201$
View full question & answer→