4 Marks Questions

Question 14 Marks

Factorize the following expressions: $(a + b)^3 - 8(a - b)^3$

Answer

$= (a + b)^3 - [2(a - b)]^3$
$= (a + b)^3 - [2a - 2b]^3$
$= (a + b - (2a - 2b))((a + b)^2 + (a + b)(2a - 2b) + (2a - 2b)^2)$
$\therefore [a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$= (a + b - 2a + 2b)(a^2 + b^2 + 2ab + (a + b)(2a - 2b) + (2a - 2b)^2)$
$= (a + b - 2a + 2b)(a^2 + b^2 + 2ab + 2a^2 - 2ab + 2ab - 2b^2 + (2a - 2b)^2)$
$= (3b - a)(3a^2 + 2ab - b^2 + (2a - 2b)^2)$
$= (3b - a)(3a^2 + 2ab - b^2 + 4a^2 + 4b^2 - 8ab)$
$= (3b - a)(3a^2 + 4a^2 - b^2 + 4b^2 - 8ab + 2ab)$
$= (3b - a)(7a^2 +3b^{2 }- 6ab)$
$\therefore (a + b)^3 - 8(a - b)^3 $
$= (3b - a)(7a^2 + 3b^2 - 6ab)$

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Question 24 Marks

Factorize:
$\text{x}^2+\frac{12}{35}\text{x}+\frac{1}{35}$

Answer

$\text{x}^2+\frac{12}{35}\text{x}+\frac{1}{35}$
Splitting the middle term,
$=\text{x}^2+\frac{5}{35}\text{x}+\frac{7}{35}\text{x}+\frac{1}{35}$
$\Big[\therefore\frac{12}{35}=\frac{5}{35}+\frac{7}{35} \ \text{and} \ \frac{5}{35}\times\frac{7}{35}=\frac{1}{35}\Big]$
$=\text{x}^2+\frac{\text{x}}{7}+\frac{\text{x}}{5}+\frac{1}{35}$
$=\text{x}\Big(\text{x}+\frac{1}{7}\Big)+\frac{1}{5}\Big(\text{x}+\frac{1}{7}\Big)$
$=\Big(\text{x}+\frac{1}{7}\Big)\Big(\text{x}+\frac{1}{5}\Big)$
$\therefore\text{x}^2+\frac{12}{35}\text{x}+\frac{1}{35}=\Big(\text{x}+\frac{1}{7}\Big)\Big(\text{x}+\frac{1}{5}\Big)$

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Question 34 Marks

Factorize: $a^2 + 2ab + b^2 - c^2$

Answer

$a^2 + 2ab + b^2 - c^2$
Using the identity $(p + q)^2 $
$= p^2 + q^2 + 2pq$
$= (a + b)^2 - c^2$
Using the identity $p^2 - q^2 $
$= (p + q)(p - q)$
$= (a + b + c)(a + b - c)$
$\therefore a^2 + 2ab + b^2 - c^2 $
$= (a + b + c)(a + b - c)$

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Question 44 Marks

Factorize: $x(x - 2)(x - 4) + 4x - 8$

Answer

$x(x - 2)(x - 4) + 4x - 8$
$= x(x - 2)(x - 4) + 4(x - 2)$
Taking $(x - 2) $common in both the terms
$=(x - 2){x(x - 4) + 4}$
$=(x - 2){x^2 - 4x + 4}$
Now splitting the middle term of $x^2 - 4x + 4$
$= (x - 2){x^2 - 2x - 2x + 4}$
$= (x - 2){x( x - 2) -2(x - 2)}$
$= (x - 2){(x - 2)(x - 2)}$
$= (x - 2)(x - 2)(x - 2)$
$= (x - 2)^3$
$\therefore x(x - 2)(x - 4) + 4x - 8$
$ = (x - 2)^3$

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Question 54 Marks

Factorize: $a(a + b)^3 - 3a^2b(a + b)$

Answer

$a(a + b)^3 - 3a^2b(a + b)$
Taking $(a + b)$ common in the two terms
$= (a + b){a(a + b)^2 - 3a^2b}$
Now, using $(a + b)^2 = a^2 + b^2 + 2ab$
$= (a + b){a(a^2 + b^2 + 2ab) - 3a^2b}$
$= (a + b){a^3 + ab^2 + 2a^2b - 3a^2b}$
$= (a + b){a^3 + ab^2 - a^2b}$
$= (a + b)p{a^2 + b^2 - ab}$
$= p(a + b)(a^2 + b^2 - ab)$
$\therefore a(a + b)^3 - 3a^2b(a + b)$
$= a(a + b)(a^2 + b^2 - ab)$

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Question 64 Marks

Factorize: $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-4\Big(\text{x}+\frac{1}{\text{x}}\Big)+6$

Answer

$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-4\Big(\text{x}+\frac{1}{\text{x}}\Big)+6$
$=\text{x}^2+\frac{1}{\text{x}^2}-4\text{x}-\frac{4}{\text{x}}+4+2$
$=\text{x}^2+\frac{1}{\text{x}^2}+4+2-4\text{x}-4\text{x}$
$=\big(\text{x}^2\big)+\Big(\frac{1}{\text{x}}\Big)^2+(-2)^2+2\times\text{x}\times\frac{1}{\text{x}}+2\times\frac{1}{\text{x}}\times(-2)+2(-2)\text{x}$
Using identity
$x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x + y + z)^2$
We get,
$=\Big[\text{x}+\frac{1}{\text{x}}+(-2)\Big]^2$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]^2$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]\Big[\text{x}+\frac{1}{\text{x}}-2\Big]$
$\therefore\Big[\text{x}^2+\frac{1}{\text{x}^2}\Big]-4\Big[\text{x}+\frac{1}{\text{x}}\Big]+6$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]\Big[\text{x}+\frac{1}{\text{x}}-2\Big]$

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Question 74 Marks

Factorize the following expressions:
$\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$

Answer

$\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$
Let $\Big(\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big)=\text{a},\Big(\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big)=\text{b},\Big(-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big)=\text{c}$
$\text{a}+\text{b}+\text{c}=\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}+\frac{\text{x}}{3}-\frac{\text{2y}}{3}+\text{z}-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=\Big(\frac{\text{x}}{2}+\frac{\text{x}}{3}-\frac{5\text{x}}{6}\Big)+\Big(\text{y}-\frac{\text{2y}}{3}-\frac{\text{y}}{3}\Big)+\Big(\frac{\text{z}}{3}+\text{z}-\frac{4\text{z}}{3}\Big)$
$\text{a}+\text{b}+\text{c}=\frac{3\text{x}}{6}+\frac{2\text{x}}{6}-\frac{5\text{x}}{6}+\frac{3\text{y}}{3}-\frac{2\text{y}}{3}-\frac{\text{y}}{3}+\frac{\text{z}}{3}+\frac{3\text{z}}{3}-\frac{4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=\frac{5\text{x}-5\text{x}}{6}+\frac{3\text{y}-3\text{y}}{3}+\frac{4\text{z}-4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=0$
$\because \ \text{a}+\text{b}+\text{c}=0$
$\therefore \ \text{a}^3+\text{b}^3+\text{c}^3=3\text{abc}$
$\therefore\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$
$=3\Big(\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big)\Big(\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big)\Big(-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big)$

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Question 84 Marks

Factorize the following expressions: $x^3 + 6x^2 + 12x + 16$

Answer

$= x^3 + 6x^2 + 12x + 8 + 8$
$= x^3 + 3 \times x^2 \times 2 + 3 \times x \times 2^2 + 2^3 + 8$
$= (x + 2)^3 + 8$
$\big[\therefore a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3\big]$
$= (x + 2)^3 + 23$
$= (x + 2 + 2)((x + 2)^2 - 2(x + 2) + 2^2)$
$\big[\therefore a^3 + b^3 = (a + b)(a^{2 }- ab + b^2)\big]$
$= (x + 2 + 2)(x^2 + 4 + 4x - 2x - 4 + 4)$
$\big[\therefore (a + b)^2 = a^2 + b^2 + 2ab\big]$
$= (x + 4)(x^2 + 4 + 2x)$
$\therefore x^3 + 6x^2 + 12x + 16$
$ = (x + 4)(x^2 + 4 + 2x)$

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