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