Question 12 Marks
Factorize:
$\text{x}^2-\sqrt{3}\text{x}-6$
Answer$\text{x}^2-\sqrt{3}\text{x}-6$
Splitting the middle term,
$=\text{x}^2-2\sqrt{3}\text{x}+\sqrt{3}\text{x}-6$
$\big[\therefore-\sqrt{3}=-2\sqrt{3}+\sqrt{3} \ \text{also} \ -2\sqrt{3}\times\sqrt{3}=-6\big]$
$=\text{x}\big(\text{x}-2\sqrt{3}\big)+\sqrt{3}\big(\text{x}-2\sqrt{3}\big)$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
$\therefore\text{x}^2-\sqrt{3}\text{x}-6$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
View full question & answer→Question 22 Marks
Factorize: $2(x + y)^2 - 9(x + y) - 5$
AnswerLet $x + y = z$
$= 2z^2 - 9z - 5$
Splitting the middle term,
$= 2z^2 - 10z + z - 5$
$= 2z(z - 5) + 1(z - 5)$
$= (z - 5)(2z + 1)$
Substituting $z = x + y$
$= (x + y - 5)(2(x + y) + 1)$
$= (x + y - 5)(2x + 2y + 1)$
$\therefore2(x + y)^2 - 9(x + y) - 5$
$ = (x + y - 5)(2x + 2y + 1)$
View full question & answer→Question 32 Marks
Factorize:
$\text{x}^2+2\sqrt{3}\text{x}-24$
Answer$\text{x}^2+2\sqrt{3}\text{x}-24$
Splitting the middle term,
$=\text{x}^2+4\sqrt{3}\text{x}-2\sqrt{3}\text{x}-24$
$\big[\therefore2\sqrt{3}=4\sqrt{3}-2\sqrt{3} \ \text{also} \ 4\sqrt{3}\big(-2\sqrt{3}\big)=-24\big]$
$=\text{x}\big(\text{x}+4\sqrt{3}\big)-2\sqrt{3}\big(\text{x}+4\sqrt{3}\big)$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
$\therefore\text{x}^2+2\sqrt{3}\text{x}-24$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
View full question & answer→Question 42 Marks
If $a + b + c = 0,$ then write the value of $a^3 + b^3 + c^3.$
AnswerRecall the formula
$a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
When$, (a + b + c) = 0,$ we have
$a^3 + b^3 + c^3 - 3abc $
$= 0.(a^2 + b^2 + c^2 - ab - bc - ca)$
$= 0$
$a^3 + b^3 + c^3 - 3abc = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$
View full question & answer→Question 52 Marks
Factorize: $a^3x^3 - 3a^2bx^2 + 3ab^2x - b^3$
Answer$a^3x^3 - 3a^2bx^2 + 3ab^2x - b^3$
$= (ax)^3 - 3(ax)^2 \times b + 3(ax)b^2 - b^3$
$= (ax - b)^3 \big[\because a^3 - 3a^2b + 3ab^2 - b^3= (a - b)^3\big]$
$= (ax - b)(ax - b)(ax - b)$
$\therefore a^3x^3 - 3a^2bx^2 + 3ab^2x - b^3$
$= (ax - b)(ax - b)(ax - b)$
View full question & answer→Question 62 Marks
Factorize the following expressions: $(2x - 3y)^3 + (4z - 2x)^3+ (3y - 4z)^3$
Answer$(2x - 3y)^3 + (4z - 2x)^3+ (3y - 4z)^3$
Let $2x - 3y = a, 4z - 2x = b, 3y - 4z = c$
$\therefore a + b + c = 2x - 3y + 4z - 2x + 3y - 4z = 0$
$\because a + b + c = 0$
$\therefore a^3 + b^3 + c^3 = 3abc$
$\therefore (2x - 3y)^3 + (4z - 2x)^3+ (3y - 4z)^3$
$= 3(2x - 3y)(4z - 2x)(3y - 4z)$
View full question & answer→Question 72 Marks
Factorize the following expressions:
$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$
Answer$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$$=\big(\sqrt{2}\text{a}\big)^3+\big(\sqrt{3}\text{b}\big)^3+\text{c}^3-3\times\sqrt{2}\text{a}\times\sqrt{3}\text{b}\times\text{c}$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\Big(\big(\sqrt{2}\text{a}\big)^2+\big(\sqrt{3}\text{b}\big)^2+\\\text{c}^2-(\sqrt{2}\text{a}\big)(\sqrt{3}\text{b}\big)-(\sqrt{3}\text{b}\big)\text{c}-(\sqrt{2}\text{a}\big)\text{c}\Big)$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6}\text{ab}-\sqrt{3}\text{bc}-\sqrt{2}\text{ac}\big)$
$\therefore2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6}\text{ab}-\sqrt{3}\text{bc}-\sqrt{2}\text{ac}\big)$
View full question & answer→Question 82 Marks
Give the possible expression for the length breadth of the rectangle having $35y^2 - 13y - 12$ as its area.
AnswerArea is given as $35y^2 - 13y - 12$
Splitting the middle term,
Area $= 35y^{2 }+ 218y - 15y - 12$
$= 7y(5y + 4) - 3(5y + 4)$
$= (5y + 4)(7y - 3)$
We also know that area of rectangle $=$ length $\times$ breadth
$\therefore$ Possible length $= (5y + 4)$ and breadth $= (7y - 3)$
Or possible length $= (7y - 3)$ and breadth $= (5y + 4)$
View full question & answer→Question 92 Marks
Factorize: $64a^3 + 125b^3 + 240a^2b + 300ab^2$
Answer$64a^3 + 125b^3 + 240a^2b + 300ab^2$
$= (4a)^3 + (5b)^3 + 3 \times (4a)^2 \times 5b + 3(4a)(5b)^2$
$= (4a + 5b)^3 \big[\because a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3\big]$
$= (4a + 5b)(4a + 5b)(4a + 5b)$
$\therefore 64a^3 + 125b^3 + 240a^2b + 300ab^2$
$= (4a + 5b)(4a + 5b)(4a + 5b)$
View full question & answer→Question 102 Marks
Multiply:$ (x^2 + 4y^2 + z^2 + 2xy + xz - 2yz) by (x - 2y - z)$
Answer$= (x - 2y - z)(x^2 + 4y^2 + z^2 + 2xy + xz - 2yz)$
$= (x + (-2y) + (-z))(x^2 + (-2y)^2 + (-z)^2 - x(-2y) - (-2y)(-z) - (-z)x)$
$= x^3 + (-2y)^3 + (-z)^3 - 3 \times x(-2y)(-z) \big[\because (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) $$= a^3 + b^3 + c^3 - 3abc\big]$
$= x^3 - 8y^3 - z^3 + 3 \times x \times 2yz$
$= x^3 - 8y^3 - z^3 - 6xyz$
View full question & answer→Question 112 Marks
Factorize: $8x^3 + 27y^3 + 36x^2y + 54xy^2$
Answer$8x^3 + 27y^3 + 36x^2y + 54xy^2$
$= (2x)^3 + (3y)^3 + 3 \times (2x)^2 \times 3y + 3 \times (2x)(3y)^2$
$= (2x + 3y)^3 \big[\because a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3\big]$
$= (2x + 3y)(2x + 3y)(2x + 3y)$
$\therefore 8x^3 + 27y^3 + 36x^2y + 54xy^2$
$= (2x + 3y)(2x + 3y)(2x + 3y)$
View full question & answer→Question 122 Marks
Factorize the following expressions: $1 - 27a^3$
Answer$1 - 27a^3$
$= (1)^3 - (3a)^3$
$= (1 - 3a)(1^2 + 1 \times 3a + (3a)^2)$
$\therefore [a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$= (1 - 3a)(1^2 + 3a + 9a^2)$
$\therefore 1 - 27a^3 $
$= (1 - 3a)(1^2 + 3a + 9a^2)$
View full question & answer→Question 132 Marks
Factorize the following expressions: $p^3 + 27$
Answer$p^3 + 27$
$= p^3 + 3^3$
$\therefore [a^3 + b^{3 }= (a + b)(a^2 - ab + b^2)]$
$= (p + 3)(p² - 3p - 9)$
$\therefore p^3 + 27 = (p + 3)(p² - 3p - 9)$
View full question & answer→Question 142 Marks
Factorize:
$\text{x}^2+5\sqrt{5}\text{x}+30$
Answer$\text{x}^2+5\sqrt{5}\text{x}+30$
Splitting the middle term,
$=\text{x}^2+2\sqrt{5}\text{x}+3\sqrt{5}\text{x}+30$
$\big[\therefore5\sqrt{5}=2\sqrt{5}+3\sqrt{5} \ \text{also} \ 2\sqrt{5}\times3\sqrt{5}=30\big]$
$=\text{x}\big(\text{x}+2\sqrt{5}\big)+3\sqrt{5}\big(\text{x}+2\sqrt{5}\big)$
$=\big(\text{x}+2\sqrt{5}\big)\big(\text{x}+3\sqrt{5}\big)$
$\therefore\text{x}^2+5\sqrt{5}\text{x}+30$
$=\big(\text{x}+2\sqrt{5}\big)\big(\text{x}+3\sqrt{5}\big)$
View full question & answer→Question 152 Marks
Factorize the following expressions:
$2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
Answer$2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
$=\big(\sqrt{2}\text{a}\big)^3+\big(2\sqrt{2}\text{b}\big)^3+(\text{c})^3-3\times\sqrt{2}\text{a}\times2\sqrt{2}\text{b}\times\text{c}$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b})+\text{c}\big)\Big(\big(\sqrt{2}\text{a}\big)^2+\big(2\sqrt{2}\text{b})^2+\text{c}^2\\\big(\sqrt{2}\text{a}\big)\big(2\sqrt{2}\text{b}\big)-\big(2\sqrt{2}\text{b}\big)\text{c}-\big(\sqrt{2}\text{a}\big)\text{c}\Big)$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2}\text{bc}-\sqrt{2}\text{ac}\big)$
$\therefore2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2}\text{bc}-\sqrt{2}\text{ac}\big)$
View full question & answer→Question 162 Marks
Factorize:$ 8a^3 + 27b^3 + 36a^2b + 54ab^2$
Answer$8a^3 + 27b^3 + 36a^2b + 54ab^2$
$= (2a)^3 + (3b)^3 + 3 \times (2a)^2 \times 3b + 3 \times (2a)(3b)^2$
$= (2a + 3b)^3 \big[\because a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3\big]$
$= (2a + 3b)(2a + 3b)(2a + 3b)$
$\therefore 8a^3 + 27b^3 + 36a^2b + 54ab^2$
$= (2a + 3b)(2a + 3b)(2a + 3b)$
View full question & answer→Question 172 Marks
Factorize: $8a^3 - 27b^3 - 36a^2b + 54ab^2$
Answer$8a^3 - 27b^3 - 36a^2b + 54ab^2$
$= (2a)^3 - (3b)^3 - 3 \times (2a)^2 \times 3b + 3 \times (2a)(3b)^2$
$= (2a - 3b)^3 \big[\because a^3 - b^3 - 3a^2b + 3ab^2 = (a - b)^3\big]$
$= (2a - 3b)(2a - 3b)(2a - 3b)$
$\therefore 8a^3 - 27b^3 - 36a^2b + 54ab^2$
$= (2a - 3b)(2a - 3b)(2a - 3b)$
View full question & answer→Question 182 Marks
Factorize the following expressions: $x^3 - 8y^3 + 27z^3 + 18xyz$
Answer$x^3 - 8y^3 + 27z^3 + 18xyz$
$= x^3 + (-2y)^3 + (3z)3 - 3 \times x \times (-2y)(3z)$
$= (x + (-2y) + 3z)(x^2 + (-2y)^2 +(3z)^2 - x(-2y) - (-2y)(3z) - 3z(x))$
$\big[\because a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= (x - 2y + 3z)(x^2 + 4y^2+ 9z^2 + 2xy + 6yz - 3zx)$
$\therefore x^3 - 8y^3 + 27z^3 + 18xyz$
$ = (x - 2y + 3z)(x^2 + 4y^2 + 9z^2 + 2xy + 6yz - 3zx)$
View full question & answer→Question 192 Marks
Factorize: $8x^3 + y^3 + 12x^2y + 6xy^2$
Answer$8x^3 + y^3 + 12x^2y + 6xy^2$
$= (2x)^3 + y^3 + 3 \times (2x)^2 \times y + 3(2\times ) \times y^2$
$= (2x + y)^3 \big[\because a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3\big]$
$= (2x + y)(2x + y)(2x + y)$
$\therefore 8x^3 + y^3 + 12x^2y + 6xy^2$
$= (2x + y)(2x + y)(2x + y)$
View full question & answer→Question 202 Marks
Factorize the following expressions: $27x^3 - y^3 - z^3 - 9xyz$
AnswerWe know that
$x^3 + y^3 + z^3 - 3xyz$
$= ( x + y + z)(x^2 + y^2 + z^2 - xy - yz -zx)$
$\therefore 27x^3 - y^3 - z^3 - 9xyz$
$= (3x)^3 + (-y)^3 + (-z)^3 - 3(3x)(-y)(-z)$
$= [3x +(-y) + (-z)][(3x)^2 + (-y)^2 + (-z)^2 - (3x)(-y)(-z)-(-z)(3x)]$
$= (3x - y - z)(9x^2 + y^2 + z^2 + 3xy - yz + 3zx)$
View full question & answer→Question 212 Marks
Factorize the following expressions: $10x^4y - 10xy^4$
Answer$10x^4y - 10xy^4$
$= 10xy(x^3 - y^3)$
$= 10xy(x - y)(x^2 + xy + y^2)$
$\therefore [x^3 - y^{3 }= (x - y)(x^2 + xy + y^2)]$
$\therefore 10x^4y - 10xy^4 $
$= 10xy(x - y)(x^2 + xy + y^2)$
View full question & answer→Question 222 Marks
Factorize:
$\text{x}^2+6\sqrt{2}\text{x}+10$
Answer$\text{x}^2+6\sqrt{2}\text{x}+10$
Splitting the middle term,
$=\text{x}^2+5\sqrt{2}\text{x}+\sqrt{2}\text{x}+10$
$\big[\therefore6\sqrt{2}=5\sqrt{2}+\sqrt{2} \ \text{and} \ 5\sqrt{2}\times\sqrt{2}=10\big]$
$=\text{x}\big(\text{x}+5\sqrt{2}\big)+\sqrt{2}\big(\text{x}+5\sqrt{2}\big)$
$=\big(\text{x}+5\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)$
$\therefore\text{x}^2+6\sqrt{2}\text{x}+10$
$=\big(\text{x}+5\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)$
View full question & answer→Question 232 Marks
Factorize: $125x^3 - 27y^3 - 225x^2y + 135xy^2$
Answer$125x^3 - 27y^3 - 225x^2y + 135xy^2$
$= (5x)^3 - (3y)^3 - 3 \times (5x)^2 \times 3y + 3 \times (5x)(3y)^2$
$= (5x - 3y)^3 \big[\because a^3 - b^3 - 3a^2b + 3ab^2 = (a - b)^3\big]$
$= (5x - 3y)(5x - 3y)(5x - 3y)$
$\therefore 125x^3 - 27y^3 - 225a^2y + 135xy^2$
$= (5x - 3y)(5x - 3y)(5x - 3y)$
View full question & answer→Question 242 Marks
Multiply: $(x^2 + 4y^2 + 2xy - 3x + 6y + 9)$ by $(x - 2y + 3)$
Answer$= (x - 2y + 3)(x^2 + 4y^2 + 9 + 2xy + 6y - 3x)$
$= (x + (-2y) + 3) (x^2 + (-2y)^2 + 3^2 - x(-2y) - (-2y)3 - 3x$)
$= x^3 + (-2y)^3 + 3^3 - 3 \times x (-2y)^3\big[\because (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc\big]$
$= x^3 - 8y^3 + 27 + 18xy$
View full question & answer→Question 252 Marks
Factorize the following expressions: $(3x - 2y)^3 + (2y - 4z)^3+ (4z - 3x)^3$
Answer$(3x - 2y)^3 + (2y - 4z)^3 + (4z - 3x)^3$
Let $(3x - 2y) = a, (2y - 4z) = b, (4z - 3x) = c$
$\therefore a + b + c = 3x - 2y + 2y - 4z + 4z -3x = 0$
$\because a + b + c = 0$
$\therefore a^3 + b^3 + c^3 + 3abc$
$\therefore (3x - 2y)^3 + (2y - 4z)^3 + (4z - 3x)^3$
$= 3(3x - 2y)(2y - 4z)(4z - 3x)$
View full question & answer→Question 262 Marks
Factorize the following expressions:$ (a - 2b)^3 - 512b^3$
Answer$(a - 2b)^3 - 512b^3$
$= (a - 2b)^3 - (8b)^3$
$= (a - 2b - 8b)((a - 2b)^2 + (a - 2b)8b + (8b)^2)$
$\therefore [a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$= (a - 10b)(a^2 + 4b^2 - 4ab + 8ab - 16b^2 + 64b^2)$
$= (a - 10b)(a^2 + 52b^2 + 4ab)$
$\therefore (a - 2b)^3 − 512b^3$
$ = (a - 10b)(a^2 + 52b^2 + 4ab)$
View full question & answer→Question 272 Marks
Factorize the following expressions:
$\frac{\text{x}^3}{216}=8\text{y}^3$
Answer$\frac{\text{x}^3}{216}-8\text{y}^3$
$=\frac{\text{x}^3}{6}-(2\text{y})^3$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\Big(\frac{\text{x}}{6}\Big)^2+\frac{\text{x}}{6}\times2\text{y}+(2\text{y})^2\Big)$
$\therefore\big[\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2)\big]$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}^2}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
$\therefore\frac{\text{x} ^3}{216}-8\text{y} ^3=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
View full question & answer→Question 282 Marks
If$ a + b + c = 9$ and a$b + bc + ca = 40,$ find $a^2 + b^2 +c^2.$
AnswerRecall the formula $(a + b + c)^2$
$= a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given that$(a + b + c) $
$= 9, ab + bc + ca $
$= 40,$
Then we have $(a + b + c)^2 $
$= a^2 + b^2 + c^2 + 2(ab + bc + ca) (9)^2$
$= a^2 + b^2 + c^2 + 2.(40) a^2 + b^2 + c^2 + 80$
$= 81 a^2 + b^2 + c^2$
$= 81 - 80 a^2 + b^2 + c^2$
$ = 1$
View full question & answer→Question 292 Marks
Factorize the following expressions: $y^3 + 125$
Answer$y^3 + 125$
$= y^3 + 5^3$
$\therefore [a^3 + b^3 = (a + b)(a^2 - ab + b^2)]$
$= (y + 5)(y^{2 }- 5y + 5^2)$
$= (y + 5)(y^2 - 5y + 25)$
$\therefore y^3 + 125$
$= (y + 5)(y^2 - 5y + 25)$
View full question & answer→Question 302 Marks
Factorize the following expressions:
$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
Answer$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\times\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
$=\big(\sqrt{3}\text{a}+(-\text{b})+\big(-\sqrt{5}\text{c}\big)\big)\Big(\big(\sqrt{3}\text{a}\big)^2+(-\text{b})^2+\big(-\sqrt{5}\text{c}\big)^2\\-\sqrt{3}\text{a}(-\text{b})-(-\text{b})\big(-\sqrt{5}\text{c}\big)-\big(-\sqrt{5}\text{c}\big)\sqrt{3}\text{a}\Big)$
$=\big(\sqrt{3}\text{a}+(-\text{b})-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{ac})$
$\therefore3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}+(-\text{b})-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{ac})$
View full question & answer→Question 312 Marks
Factorize: $x^4 + x^2y^2 + y^4$
Answer$x^4 + x^2y^2 + y^4$
Adding $x^2y^2$ and subtracting $x^2y^2$ to the given equation
$= x^4 + x^2y^2 + y^4 + x^2y^2 - x^2y^2$
$= x^4 + 2x^2y^{2 }+ y^4 - x^2y^2$
$= (x^2)^2 + 2 \times x^2 \times y^2 + (y^2)^2 - (xy)^2$
Using the identity$ (p + q)^2 = p^2 + q^2 + 2pq$
$= (x^2 + y^2)^2 - (xy)^2$
Using the identity $p^2 - q^2 = (p + q)(p - q)$
$= (x^2 + y^2 + xy)(x^2 + y^2 - xy)$
$\therefore x^4 + x^2y^2 + y^4$
$ = (x^2 + y^{2 }+ xy)(x^2 + y^2 - xy)$
View full question & answer→Question 322 Marks
Factorize: $a^2 + b^2 + 2(ab + bc + ca)$
Answer$= a^2 + b^2 + 2ab + 2bc + 2ca$
Using the identity $(p + q)^2 $
$= p^2 + q^2 + 2pq$
We get,
$= (a + b)^2 + 2bc + 2ca$
$= (a + b)^2 + 2c(b + a)$
Or$ (a + b)^2 + 2c(a + b)$
Taking (a + b) common
$= (a + b)(a + b + 2c)$
$\therefore a^{2 }+ b^2 + 2(ab + bc + ca)$
$ = (a + b)(a + b + 2c)$
View full question & answer→Question 332 Marks
Factorize: $x^3 - 12x(x - 4) - 64$
Answer$x^3 - 12x(x - 4) - 64$
$= x^3 - 12x^2 + 48x - 64$
$= (x)^3 - 3 \times x^2 \times 4 + 3 \times 4^2 \times x - 4^3 \big[\because a^3 - 3a^2b + 3ab^2 - b^3= (a - b)^3\big]$
$= (x - 4)(x - 4)(x - 4)$
$\therefore x^3 - 12x(x - 4) - 64$
$ = (x - 4)(x - 4)(x - 4)$
View full question & answer→Question 342 Marks
Factorize: $a^2 - b^2 + 2bc - c^2$
Answer$a^2 - b^2 + 2bc - c^2$
$a^2 - (b^2 - 2bc + c^2)$
Using the identity $(a - b)^2 $
$= a^2 + b^2 - 2ab$
$= a^2 - (b - c)^2$
Using the identity $a^2 - b^2 $
$= (a + b)(a - b)$
$= (a + b - c)(a - (b - c))$
$= (a + b - c)(a - b + c)$
$\therefore a^2 - b^2 + 2bc - c^2$
$= (a + b - c)(a - b + c)$
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What are the possible expression for the cuboid having volume $3x^2 - 12x.$
AnswerVolume $= 3x^2 - 12x$
$= 3x(x - 4)$
$= 3 \times x(x - 4)$
Also volume $=$ Length \times Breadth $\times$ Height
$\therefore$ Possible expression for dimensions of cuboid are $= 3, x, (x - 4)$
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If $a^2 + b^2 + c^2 = 20$ and $a + b + c = 0,$ find $ab + bc + ca.$
AnswerRecall the formula
$(a + b + c)^2 $
$= a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given that
$a^2 + b^2 + c^2 = 20$
$(a + b + c) = 0$
Then we have
$(a + b + c)^2 $
$= a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$(0)^2 = 20 + 2(ab + bc + ca)$
$20 + 2(ab + bc + ca) = 0$
$2(ab + bc + ca) = -20$
$(ab + bc + ca) = -10$
View full question & answer→Question 372 Marks
Factorize the following expressions: $64a^{3 }- b^3$
Answer$64a^{3 }- b^3$
$= (4a)^3 - b^3$
$= (4a - b)((4a)^2 + 4a \times b + b^2)$
$\therefore [a^{3 }- b^3 = (a - b)(a^2 + ab + b^2)]$
$= (4a − b)(16a^2 + 4ab + b^2)$
$\therefore 64a^3 - b^3$
$ = (4a - b)(16a^{2 }+ 4ab + b^2)$
View full question & answer→Question 382 Marks
Factorize the following expressions: $x^6 + y^6$
Answer$= (x^2)^3 + (y^2)^3$
$= (x^2 + y^2)((x^2)^2 - x^2y^2 + (y^2)^2)$
$= (x^2 + y^2)(x^4 - x^2y^2 + y^4)$
$\big[\therefore a^3 + b^3= (a + b)(a^2 - ab + b^2)\big]$
$\therefore x^6 + y^6 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)$
View full question & answer→Question 392 Marks
Factorize the following expressions: $a^3 + 8b^3 + 64c^3 - 24abc$
Answer$a^3 + 8b^3 + 64c^3 - 24abc$
$= (a)^3 + (2b)^3 + (4c)^3 - 3 \times 2b \times 4c$
$= (a + 2b + 4c)(a^2 + (2b)^2 + (4c)^2 - a \times 2b - 2b \times 4c - 4c \times a)$
$\big[\because a^3 + b^3 + c^3 - 3ab$
$c = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= (a + 2b + 4c)(a^2 + 4b^2+ 16c^2 - 2ab - 8bc - 4ac)$
$\therefore a^3 + 8b^3 + 64c^3 - 24abc$
$ = (a + 2b + 4c)(a^2 + 4b^2 + 16c^2 - 2ab - 8bc - 4ac)$
View full question & answer→Question 402 Marks
Factorize:
$\text{x}^2-2\sqrt{2}\text{x}-30$
Answer$\text{x}^2-2\sqrt{2}\text{x}-30$
Splitting the middle term,
$=\text{x}^2=5\sqrt{2}\text{x}+3\sqrt{2}\text{x}-30$
$\big[\therefore-2\sqrt{2}=-5\sqrt{2}+3\sqrt{2} \ \text{also} \ -5\sqrt{2}\times3\sqrt{2}=-30\big]$
$=\text{x}\big(\text{x}-5\sqrt{2}\big)+3\sqrt{2}\big(\text{x}-5\sqrt{2}\big)$
$=\big(\text{x}-5\sqrt{2}\big)\big(\text{x}+3\sqrt{2}\big)$
$\therefore\text{x}^2-2\sqrt{2}\text{x}-30$
$=\big(\text{x}-5\sqrt{2}\big)\big(\text{x}+3\sqrt{2}\big)$
View full question & answer→Question 412 Marks
Multiply: $(x^2 + y^2 + z^2 - xy + xz + yz)$ by $(x + y -z)$
Answer$(x^2 + y^2 + z^2 - xy + xz + yz)$ by $(x + y -z)$
$= (x + y - z)(x^2 + y^2 + z^2 - xy + xz + yz)$
$= (x + y + (-z))(x^2 + y^2 + (-z)^2 - xy - y(-z) - (-z)x)$
$= x^3 + y^3 + (-z)^3 - 3xyz(-z) \big[\because (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc\big]$
$= x^3 + y^3 - z^3 + 3xyz$
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Factorize the following expressions: $a^{12} + b^{12}$
Answer$= (a^4)^3 + (b^4)^3$
$= (a^4 + b^4)((a^4)^2 - a^4 \times b^4 + (b^4)^2)$
$\big[\therefore a^3 + b^3 = (a + b)(a^2 - ab + b^2)\big]$
$= (a^4 + b^4)(a^8 - a^4b^{4 }+ b^8)$
$\therefore a^{12} + b^{12}$
$ = (a^4 + b^4)(a^8 - a^4b^4 + b^8)$
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Factorize: $x^3 + 8y^3 + 6x^2y + 12xy^2$
Answer$x^3 + 8y^3 + 6x^2y + 12xy^2$
$= (x)^3 + (2y)^3 + 3 \times x^2 \times 2y + 3 \times x \times (2y)^2$
$= (x + 2y)^3 \big[\because a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3\big]$
$= (x + 2y)(x + 2y)(x + 2y)$
$\therefore x^3 + 8y^3 + 6x^2y + 12xy^2$
$= (x + 2y)(x + 2y)(x + 2y)$
View full question & answer→Question 442 Marks
Factorize the following expressions: $8x^3 + 27y^3 - 216z^3 + 108xyz$
Answer$8x^3 + 27y^3 - 216z^3 + 108xyz$
$= (2x)^3 + (3y)^3 + (-6z)^3 - 3(2x)(3y)(-6z)$
$= (2x + 3y - 6z)((2x)^2 + (3y)^2 + (-6z)^2 - 2x \times 3y - 3y(-6z)-(-6z)2x)$
$= (2x + 3y - 6z)(4x^2 + 9y^2 + 36z^2 - 6xy + 18yz + 12zx)$
$\therefore 8x^3 + 27y^3 - 216z^3 + 108xyz$
$= (2x + 3y - 6z)(4x^2 + 9y^2 + 36z^2 - 6xy + 18yz + 12zx)$
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If $a^2 + b^2 + c^2 = 250$ and $ab + bc + ca = 3,$ find $a + b + c.$
AnswerRecall the formula
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given that
$a^2 + b^2 + c^2 = 250, ab + bc + ca = 3$
Then we have
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$(a + b + c)^2 = 250 + 2.(3)$
$(a + b + c)^2 = 256$
$(a + b + c) = ±16$
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Factorize the following expressions: $8x^3y^3 + 27a^3$
Answer$8x^3y^3 + 27a^3$
$= (2xy)^{3 }+ (3a)^3$
$= (2xy + 3a)((2xy)^2 - 2xy \times 3a + (3a)^2)$
$\therefore [a^3 + b^3 = (a + b)(a^2 - ab + b^2)]$
$= (2xy + 3a)(4x^2y^2 - 6xya + 9a^2)$
$\therefore 8x^3y^3 + 27a^3$
$= (2xy + 3a)(4x^2y^2 - 6xya + 9a^2)$
View full question & answer→Question 472 Marks
Factorize the following expressions: $(a - 3b)^3 + (3b - c)^3 + (c - a)^3$
Answer$(a - 3b)^3 + (3b - c)^3 + (c - a)^3$Let $(a - 3b) = x, (3b - c) = y, (c - a) = z$
$x + y + z = a - 3b + 3b - c + c - a = 0$
$\because x + y + z = 0$
$\therefore x^3 + y^3 + z^3 = 3xyz$
$\therefore (a - 3b)^3 + (3b - c)^3 + (c - a)^3$
$= 3(a - 3b)(3b - c)(c - a)$
View full question & answer→Question 482 Marks
Factorize: $\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$
Answer$\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$$=\Big(\frac{2}{3}\text{x}\Big)^3+(1)^3+3\times\Big(\frac{2}{3}\text{x}\Big)^2\times1+3(1)^2\times\Big(\frac{2}{3}\text{x}\Big)$
$=\Big(\frac{2}{3}\text{x}+1\Big)^3$$\big[\because a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3\big]$
$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$
$\therefore\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$
$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$
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