4 Mark Question

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

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

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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 - [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 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

$= x^3 + 6x^2 + 12x + 8 + 8$
$= x^3 + 3 \times x^2 \times 2 + 3 \times x \times 2^2 + 2^3 + 8$
$= (x + 2)^3 + 8$
$\big[\therefore$ $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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Maths 4 Mark Question

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