3 Mark Question

Question 13 Marks

Factorise:
$2(x+y)^2-9(x+y)-5$

Answer

Let $x+y=z$
Then, $2(x+y)^2-9(x+y)-5$
$=2 z^2-9 z-5$
$=2 z^2-10 z+z-5$
$=2 z(z-5)+1(z-5)$
$=(z-5)(2 z+1)$
Now, replacing $z$ by $(x+y)$, we get
$2(x+y)^2-9(x+y)-5$
$=[(x+y)-5][(2(x+y)+1)]$
$=(x+y-5)(2 x+2 y+1)$

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Question 23 Marks

Factorise:
$4 x^4+7 x^2-2$

Answer

Let $x^2=y$
Then, $4 x^4+7 x^2-2$
$=4 y^2+7 y-2$
$=4 y^2+8 y-y-2$
$=4 y(y+2)-1(y+2)$
$=(y+2)(4 y-1)$
Now replacing y by $x^2$, we get
$4 x^4+7 x^2-2$
$=\left(x^2+2\right)\left(4 x^2-1\right) \text { Since } a^2-b^2=(a-b)(a+b)$
$=\left(x^2+2\right)(2 x+1)(2 x-1)$

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Question 33 Marks

Factorise:$10\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3$

Answer

Given equation:
$10\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3$ Let $3\text{x}+\frac{1}{\text{x}}=\text{a}$
Then, we have $= 10a^2 - a - 3 = 10a^2 - 6a + 5a - 3$
$= 2a(5a - 3) + 1(5a - 3) = (5a - 3)(2a + 1)$
$=\bigg[5\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3\bigg]\bigg[2\Big(3\text{x}+\frac{1}{\text{x}}\Big)+1\bigg]$
$=\Big(15\text{x}+\frac{5}{\text{x}}-3\Big)\Big(6\text{x}+\frac{2}{\text{x}}+1\Big)$

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Question 43 Marks

Factorise:
Prove that $\frac{0.85\times0.85\times0.85+0.15\times0.15\times0.15}{0.85\times0.85-0.85\times0.15+0.15\times0.15}=1$

Answer

Let 0.85 = a and 0.15 = b Then, we have$\text{L.H.S.}$
$=\frac{0.85\times0.85\times0.85+0.15\times0.15\times0.15}{0.85\times0.85-0.85\times0.15+0.15\times0.15}$
$=\frac{\text{a}\times\text{a}\times\text{a}+\text{b}\times\text{b}\times\text{b}}{\text{a}\times\text{a}-\text{a}\times\text{b}+\text{b}\times\text{b}}$
$=\frac{\text{a}^3+\text{b}^3}{\text{a}^2-\text{ab}+\text{b}^2}$
$=\frac{(\text{a}+\text{b})\big(\text{a}^2-\text{ab}+\text{b}^2\big)}{\big(\text{a}^2-\text{ab}+\text{b}^2\big)}$
$=\text{a}+\text{b}$
$=0.85+0.15$
$=1$
$=\text{R.H.S.}$

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Question 53 Marks

Factorise:
$(a-3 b)^3+(3 b-c)^3+(c-a)^3$

Answer

We know
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$x^3+y^3+z^3=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)+3 x y z$
Here, $x=(a-3 b), y=(3 b-c), z=(c-a)$
$(a-3 b)^3+(3 b-c)^3+(c-a)^3$
$=(a-3 b+3 b-c+c-a)$
${\left[(a-3 b)^2+(3 b-c)^2+(c-a)^2-(a-3 b)(3 b-c)-(3 b-c)(c-a)-(c-a)(a-3 b)\right]}$
$+3(a-3 b)(3 b-c)(c-a)$
$=0+3(a-3 b)(3 b-c)(c-a)$
$=3(a-3 b)(3 b-c)(c-a)$

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Question 63 Marks

If $a+b+c=9$ and $a^2+b^2+c^2=35$, find the value of $\left(a^3+b^3+c^3-3 a b c\right)$.

Answer

$a+b+c=9$
$\Rightarrow(a+b+c)^2=9^2=81$
$\Rightarrow a^2+b^2+c^2+2(a b+b c+c a)=81$
$\Rightarrow 35+2(a b+b c+c a)=81$
$\Rightarrow(a b+b c+c a)=23$
We have,
$\left(a^3+b^3+c^3-3 a b c\right)=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=(9)(35-23)$
$=108$

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Question 73 Marks

Factorise:$\text{x}^2+\frac{1}{\text{x}^2}-3$

Answer

$\text{x}^2+\frac{1}{\text{x}^2}-3$$=\text{x}^2+\frac{1}{\text{x}^2}-2-1$
$=\Big(\text{x}^2+\frac{1}{\text{x}^2}-2\Big)-1$
$=\bigg(\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x}^2)\Big(\frac{1}{\text{x}^2}\Big)\bigg)-1^2$
$=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2-1^2$
$=\Big(\text{x}-\frac{1}{\text{x}}-1\Big)\Big(\text{x}-\frac{1}{\text{x}}+1\Big)$

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Question 83 Marks

Factorise:
Prove that $=\frac{59\times59\times59-9\times9\times9}{59\times59+59\times9+9\times9}=50$

Answer

Let 59 = a and 9 = b Then, we have$\text{L.H.S.}$
$=\frac{59\times59\times59-9\times9\times9}{59\times59+59\times9+9\times9}$
$=\frac{\text{a}\times\text{a}\times\text{a}-\text{b}\times\text{b}\times\text{b}}{\text{a}\times\text{a}+\text{a}\times\text{b}+\text{b}\times\text{b}}$
$=\frac{\text{a}^3-\text{b}^3}{\text{a}^2+\text{ab}+\text{b}^2}$
$=\frac{(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)}{\big(\text{a}^2+\text{ab}+\text{b}^2\big)}$
$=\text{a}-\text{b}$
$=59-9$
$=50$
$=\text{R.H.S.}$

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Question 93 Marks

Factorise:
$9(2 a-b)^2-4(2 a-b)-13$

Answer

Let $2 a - b = c$
Then, $9(2 a-b)^2-4(2 a-b)-13$
$=9 c^2-4 c-13$
$=9 c^2-13 c+9 c-13$
$=c(9 c-13)+1(9 c-13)$
$=(c+1)(9 c-13)$
Now, replacing $c$ by ( $2 a - b )$, we get
$9(2 a-b)^2-4(2 a-b)-13$
$=(2 a-b+1)[9(2 a-b)-13]$
$=(2 a-b+1)(18 a-9 b-13)$

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Question 103 Marks

Factorise:$6\Big(2\text{x}-\frac{3}{\text{x}}\Big)^2+7\Big(2\text{x}-\frac{3}{\text{x}}\Big)-20$

Answer

Given equation: $6\Big(2\text{x}-\frac{3}{\text{x}}\Big)^2+7\Big(2\text{x}-\frac{3}{\text{x}}\Big)-20$ Let $2\text{x}-\frac{3}{\text{x}}=\text{a}$
Then, we have $= 6a^2 + 7a - 20 = 6a^2 + 15a - 8a - 20$
$= 3a(2a + 5) - 4(2a + 5)$
$= (2a + 5)(3a - 4)$
$=\bigg[2\Big(2\text{x}-\frac{3}{\text{x}}\Big)+5\bigg]\bigg[3\Big(2\text{x}-\frac{3}{\text{x}}\Big)-4\bigg]$
$=\Big(4\text{x}-\frac{6}{\text{x}}+5\Big)\Big(6\text{x}-\frac{9}{\text{x}}-4\Big)$

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Question 113 Marks

Find the product.
$(3 x-5 y+4)\left(9 x^2+25 y^2+15 x y-20 y+12 x+16\right)$

Answer

$(3 x-5 y+4)\left(9 x^2+25 y^2+15 x y-20 y+12 x+16\right)$
$=(3 x+(-5 y)+4)\left(9 x^2+25 y^2+16+15 x y-20 y+12 x\right)$
$(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c$
Here, $a=3 x , b =-5 y , c =4$
$=(3 x+(-5 y)+4)\left(9 x^2+25 y^2+16+15 x y-20 y+12 x\right)$
$=(3 x)^3+(-5 y)^3+4^3-3 \times 3 x(-5 y)(4)$
$=27 x^3-125 y^3+64+180 x y$

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Question 123 Marks

Find the product.
$(x - y - z)(x^2 + y^2 + z^2 + xy - yz + xz)$

Answer

$(x-y-z)\left(x^2+y^2+z^2+x y-y z+x z\right)$
$=\left(x+(-y)+(-z)\left(x^2+y^2+z^2+x y-y z+x z\right)\right.$
$=\left(x+(-y)+(-z)\left(x^2+y^2+z^2+x y-y z+x z\right)\right.$
We have
$(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c$
Here, $a=x, b=-y, c=-z$
$\left(x+(-y)+(-z)\left(x^2+y^2+z^2+x y-y z+x z\right)=x^3-y^3-z^3-3 x y z\right.$

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Question 133 Marks

Factorise:
$x^8- 1$

Answer

$x^8- 1 = (x^4)^2 - (1)^2$
$= (x^4 - 1)(x^4 + 1)$
$= [(x^2)^2 - (1)^2)(x^4 + 1)$
$= (x^2 - 1)(x^2 + 1)(x^4 + 1)$
$= (x - 1)(x + 1)(x^2 + 1)[(x^2)^2 + (1)^2 + 2x^2 - 2x^2$
$= (x - 1)(x + 1)(x^2 + 1)[(x^2)^2 + (1)^2 + 2x^2) - 2x^2$
$=(\text{x}-1)(\text{x}+1)\big(\text{x}^2+1\big)\\\ \ \ \ \ \Big[\big(\text{x}^2+1) -\big(\sqrt{2}\text{x}\big)^2\Big] $
$=(\text{x}-1)(\text{x}+1)\big(\text{x}^2+1\big)\big(\text{x}^2+1-\sqrt{2}\text{x}\big)\\\ \ \ \ \big(\text{x}^2+1+\sqrt{2}\text{x}\big) $

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Question 143 Marks

Factorise:$\frac{1}{3}\text{x}^2-2\text{x}-9$

Answer

$\frac{1}{3}\text{x}^2-2\text{x}-9$$=\frac{1}{3}\text{x}^2-3\text{x}+\text{x}-9$
$=\text{x}\Big(\frac{\text{x}}{3}-3\Big)+(\text{x}-9)$
$=\frac{\text{x}}{3}(\text{x}-9)+(\text{x}-9)$
$=(\text{x}-9)\Big(\frac{\text{x}}{3}+1\Big)$
$=(\text{x}-9)\frac{(\text{x}+3)}{3}$
$=\frac{1}{3}(\text{x}-9)(\text{x}+3)$

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Question 153 Marks

Factorise:$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$

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\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
We know
$\text{x}^3+\text{y}^3+\text{z}^3-3\text{xyz}=(\text{x}+\text{y}+\text{z})\$\text{x}^2+\text{y}^2+\text{z}^2-\text{xy}-\text{yz}-\text{zx})$
Here, $\text{x}=\big(\sqrt{3}\text{a}\big),\ \text{y}=(-\text{b}),\ \text{z}=\big(-\sqrt{5}\text{c}\big)$
$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
$=\big(\sqrt{3}\text{a}-\text{b}-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{c}\big)$

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Question 163 Marks

Factorise:$\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+\frac{2}{\text{a}}$

Answer

We know that, Since $a^3 - b^3 = (a - b)(a^2 + a \times b + b^2)$$\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+\frac{2}{\text{a}}$
$=\text{a}^3-\frac{1}{\text{a}^3}-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\text{a}\times\frac{1}{\text{a}}+\frac{1}{\text{a}^2}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}-2\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$

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Question 173 Marks

Factorise:$\sqrt{5}\text{x}^2+2\text{x}-3\sqrt{5}$

Answer

We have: We have to split 2 into numbers such that their sum is 2 and product is (-15), i.e., $\sqrt{5}\times\big(-3\sqrt{5}\big)$ Clearly, 5 + (-3) = 2 and 5 × (-3) = -15.$\therefore\ \sqrt{5}\text{x}^2+2\text{x}-3\sqrt{5}$
$=\sqrt{5}\text{x}^2+5\text{x}-3\text{x}-3\sqrt{5}$
$=\sqrt{5}\text{x}\big(\text{x}+\sqrt{5}\big)-3\big(\text{x}+\sqrt{5}\big)$
$=\big(\text{x}+\sqrt{5}\big)\big(\sqrt{5}\text{x}-3\big)$

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Question 183 Marks

Factorise:$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$

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\big(\sqrt{2}\text{a}\big)\big(\sqrt{3}\text{b}\big)\text{c}$
We know
$\text{x}^3+\text{y}^3+\text{z}^3-3\text{xyz}=(\text{x}+\text{y}+\text{z})\\\big(\text{x}^2+\text{y}^2+\text{z}^2-\text{xy}-\text{yz}-\text{zx}\big)$
$\text{x}=\sqrt{2}\text{a},\ \text{y}=\sqrt{3}\text{b},\ \text{z}=\text{c}$
$\big(\sqrt{2}\text{a}\big)^3+\big(\sqrt{3}\text{b}\big)^3+\text{c}^3-3\big(\sqrt{2}\text{a}\big)\big(\sqrt{3}\text{b}\big)\text{c}$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6\text{ab}}-\sqrt{3\text{bc}}-\sqrt{2}\text{ac}\big)$

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Question 193 Marks

Factorise:
$7(x - 2y)^2 - 25(x - 2y) + 12$

Answer

Let $x-2 y=z$
Then, $7( x -2 y )^2-25( x -2 y )+12$
$=7 z^2-25 z+12$
$=7 z^2-21 z-4 z+12$
$=7 z(z-3)-4(z-3)$
$=(z-3)(7 z-4)$
Now replace $z$ by ( $x-2 y$ ), we get
$7(x-2 y)^2-25(x-2 y)+12$
$=(x-2 y-3)[7(x-2 y)-4]$
$=(x-2 y-3)(7 x-14 y-4)$

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Question 203 Marks

Factorise:
$(5 a-7 b)^3+(7 b-9 c)^3+(9 c-5 a)^3$

Answer

Put $(5 a -7 b)= x ,(7 b-9 c )= y ,(9 c -5 a )= z$.
Here,
$x+y+z=5 a-7 b+9 c-5 a+7 b-9 c=0$
Thus,
We have:
$(5 a-7 b)^3+(9 c-5 a)^3+(7 b-9 c)^3=x^3+z^3+y^3$
$=3 x y z\left[\text { When } x+y+z=0, x^3+y^3+z^3=3 x y z\right]$
$=3(5 a-7 b)(9 c-5 a)(7 b-9 c)$

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Question 213 Marks

Factorise:$\text{x}^2-2\text{x}+\frac{7}{16}$

Answer

$\text{x}^2-2\text{x}+\frac{7}{16}$$=\frac{1}{16}\big(16\text{x}^2-32\text{x}+7\big)$
$=\frac{1}{16}\big(16\text{x}^2-4\text{x}-28\text{x}+7\big)$
$=\frac{1}{16}\Big[4\text{x}(4\text{x}-1)-7(4\text{x}-1)\Big]$
$=\frac{1}{16}(4\text{x}-1)(4\text{x}-7)$
$=(4\text{x}-1)\Big(\frac{\text{x}}{4}-\frac{7}{16}\Big)$

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