4 Marks Questions

Question 14 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 24 Marks

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

Answer

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

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Question 34 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 44 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 54 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 64 Marks

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

Answer

$=(a+b)^3-[2(a-b)]^3=(a+b)^3-[2 a-2 b]^3$
$=(a+b-(2 a-2 b))\left((a+b)^2+(a+b)(2 a-2 b)+(2 a-2 b)^2\right)$
$\therefore\left[a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=(a+b-2 a+2 b)\left(a^2+b^2+2 a b+(a+b)(2 a-2 b)+(2 a-2 b)^2\right)$
$=(a+b-2 a+2 b)\left(a^2+b^2+2 a b+2 a^2-2 a b+2 a b-2 b^2+(2 a-2 b)^2\right)$
$=(3 b-a)\left(3 a^2+2 a b-b^2+(2 a-2 b)^2\right)$
$=(3 b-a)\left(3 a^2+2 a b-b^2+4 a^2+4 b^2-8 a b)=(3 b-a)(3 a^2+4 a^2-b^2+4 b^2-8 a b+2 a b\right)$
$=(3 b-a)\left(7 a^2+3 b^2-6 a b\right)$
$\therefore(a+b)^3-8(a-b)^3=(3 b-a)\left(7 a^2+3 b^2-6 a b\right)$

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Question 74 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 84 Marks

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

Answer

$=x^3+6 x^2+12 x+8+8 $
$=x^3+3 \times x^2 \times 2+3 \times x \times 2^2+2^3+8 $
$=(x+2)^3+8\left[\therefore a^3+3 a^2 b+3 a b^2+b^3=(a+b)^3\right] $
$=(x+2)^3+23 $
$=(x+2+2)\left((x+2)^2-2(x+2)+2^2\right)\left[\therefore a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right] $
$=(x+2+2)\left(x^2+4+4 x-2 x-4+4\right)\left[\therefore(a+b)^2=a^2+b^2+2 a b\right] $
$=(x+4)\left(x^2+4+2 x\right) $
$\therefore x^3+6 x^2+12 x+16=(x+4)\left(x^2+4+2 x\right)$

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