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Maths 2 Marks Questions

Factorisation of Polynomials · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 158 questions.

Questions · Page 2 of 4

2 Marks Questions

Question 512 Marks
Factorise:
125a3 + b3 + 64c3 - 60abc
Answer
125a3 + b3 + 64c3 - 60abc
= (5a)3 + (b)3 + (4c)3 - 3 × 5a × b × 4c
= (5a + b + 4c)[(5a)2 + (b)2 + (4c)2 - 5a × b - b × 4c - 5a × 4c]
= (5a + b + 4c)(25a2 + b2 + 16c2 - 5ab - 4bc - 20ac
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Question 522 Marks
Expand:
$\Big(1+\frac{2}{3}{\text{a}}\Big)^3$
Answer
$\Big(1+\frac{2}{3}{\text{a}}\Big)^3$

$=\Big(\frac{2}{3}\text{a}\Big)^3+3\times\Big(\frac{2}{3}\text{a}\Big)^2\times1+3\text{a}\frac{2}{3}\text{a}\times(1)^2+(1)^3$

$=\frac{8}{27}\text{a}^3+\frac{4}{3}\text{a}^2+2\text{a}+1$

 

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Question 532 Marks
Factorise:
3a3b - 243ab3
Answer
3a3b - 243ab3
= 3ab(a2 - 81b2)
= 3ab[(a)2 - (9b)2]
= 3ab(a + 9b)(a - 9b) $\big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
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Question 542 Marks
If a, b, c are all nonzero and a + b + c = 0, prove that:
$\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3$
Answer
a + b + c = 0
⇒ a3 + b3 + c3 = 3abc
Thus,
We have:
$\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}}$
$=3$
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Question 552 Marks
Factorise:
$2\text{x}^2+3\sqrt{3}\text{x}+3$
Answer
$2\text{x}^2+3\sqrt{3}\text{x}+3$
$2\text{x}^2+2\sqrt{3}\text{x}+\sqrt{3}\text{x}+3$
$=2\text{x}\big(\text{x}+\sqrt{3}\big)+\sqrt{3}\big(\text{x}+\sqrt{3}\big)$
$=\big(\text{x}+\sqrt{3}\big)\big(2\text{x}+\sqrt{3}\big)$
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Question 562 Marks
Evaluate:
(107)2
Answer
(107)2 = (100 + 7)2
= (100)2 + 2 × (100) × (7) + (7)2
= 10000 + 1400 + 49
= 11449
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Question 572 Marks
Factorise:
x2 - 32x - 105
Answer
x2 - 32x - 105
= x2 - 35x + 3x - 105
= x(x - 35) + 3(x - 35)
= x(x - 35)(x + 3)
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Question 582 Marks
Factorise:
$6\sqrt{3}\text{x}^2-47\text{x}+5\sqrt{3}$
Answer
$6\sqrt{3}\text{x}^2-47\text{x}+5\sqrt{3}$
$=6\sqrt{3}\text{x}^2-45\text{x}-2\text{x}+5\sqrt{3}$
$=3\sqrt{3}\text{x}\big(2\text{x}-5\sqrt{3}\big)-1\big(2\text{x}-5\sqrt{3}\big)$
$=\big(2\text{x}-5\sqrt{3}\big)\big(3\sqrt{3}\text{x}-1\big)$
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Question 592 Marks
Factorise:
(a + b)3 - a - b
Answer
(a + b)3 - a - b
= (a + b)3 - (a + b)
= (a + b)[(a + b)2 - 12] $\big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
= (a + b)(a + b + 1)(a + b - 1)
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Question 602 Marks
Factorise:
$2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
Answer
$x =2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
$=\big(\sqrt{2}\text{a}\big)^3+\big(2\sqrt{2}\text{b}\big)^3+\text{c}^3-3\times\big(\sqrt{2}\text{a}\big)\times\big(2\sqrt{2}\text{b}\big)\times(\text{c})$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\Big[\big(\sqrt{2}\text{a}\big)^2+(2\sqrt{2}\text{b})^2+\text{c}^2\\-\big(\sqrt{2}\text{a}\big)\times\big(2\sqrt{2}\text{b}\big)-\big(2\sqrt{2}\text{b}\big)\times(\text{c})-\big(\sqrt{2}\text{a}\big)\times(\text{c})\Big]$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2\text{bc}}-\sqrt{2}\text{ac}\big)$
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Question 612 Marks
Factorise:
x2 - x - 156
Answer
x2 - x - 156
= x2 - 13x + 12x - 156
= x(x - 13) + 12(x - 13)
= (x - 13)(x + 12)
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Question 622 Marks
Factorise:
$\text{x}^2+\sqrt{2}\text{x}-24$
Answer
$\text{x}^2+\sqrt{2}\text{x}-24$
$=\text{x}^2+4\sqrt{2}\text{x}-3\sqrt{2}\text{x}-24$
$=\text{x}(\text{x}+4\sqrt{2})-3\sqrt{2}(\text{x}+4\sqrt{2})$
$=(\text{x}-4\sqrt{2})(\text{x}-3\sqrt{2})$
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Question 632 Marks
Factorise:
$\frac{3}{2}\text{x}^2+16\text{x}+10$
Answer
$\frac{3}{2}\text{x}^2+16\text{x}+10$
$=\frac{3}{2}\text{x}^2+\text{x}+15\text{x}+10$
$=\frac{\text{x}}{2}(3\text{x}+2)+5(3\text{x}+2)$
$(3\text{x}+2)\Big(\frac{\text{x}}{2}+5\Big)$
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Question 642 Marks
Factorise:
x - 8xy3
Answer
x - 8xy3
= x(1 - 8y3)
= x[(1)3 - (2y)3]
= x(1 - 2y)[(1)2 + 1 × 2y + (2y)2] Since a3 - b3 = (a - b)(a2 + a × b + b2)
= x (1 - 2y)(1 + 2y + 4y2)
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Question 652 Marks
Factorise:
$\text{x}^4+\frac{4}{\text{x}^4}$
Answer
$\text{x}^4+\frac{4}{\text{x}^4}$
$=\text{x}^4+\frac{4}{\text{x}^4}+4-4$
$=\big(\text{x}^2\big)^2+\Big(\frac{2}{\text{x}^2}\Big)^2+2\big(\text{x}^2\big)\Big(\frac{2}{\text{x}^2}\Big)-2^2$
$=\bigg[\big(\text{x}^2\big)^2+\Big(\frac{2}{\text{x}^2}\Big)^2+2\big(\text{x}^2\big)\Big(\frac{2}{\text{x}^2}\Big)\bigg]-2^2$
$=\Big[\text{x}^2+\frac{2}{\text{x}^2}\Big]^2-2^2$
$=\Big(\text{x}^2+\frac{2}{\text{x}^2}+2\Big)\Big(\text{x}^2+\frac{2}{\text{x}^2}-2\Big)$
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Question 662 Marks
Factorise:
x6 - 729
Answer
x6 - 729
= (x2)3 - (9)3
= (x2 - 9)[(x2)2 + x2 × 9 + (9)2] Since a3 - b3 = (a - b)(a2 + a × b + b2)
= (x2 - 9)(x4 + 9x2 + 81)
= (x + 3)(x - 3)[(x2 + 9)2 - (3x)2]
= (x + 3)(x - 3)(x2 + 3x + 9)(x2 - 3x + 9)
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Question 672 Marks
Factorise:
27a3 - b3 + 8c3 - 18abc
Answer
27a3 - b3 + 8c3 - 18abc
= (3a)3 + (-b)3 + (2c)3 - 3 × (3a) × (-b) × (2c)
= [3a + (-b) + 2c][(3a)2 + (-b)2 + (2c)2 - 3a(-b)2c - 3a × 2c]
= (3a - b + 2c)(9a2 + b2 + 4c2 + 3ab + 2bc - 6ac)
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Question 682 Marks
Factorise:
$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
Answer
$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
$=(1)^3+\Big(\frac{3}{5}\text{a}\Big)^3+3(1)^2\Big(\frac{3}{5}\text{a}\Big)+3(1)\Big(\frac{3}{5}\text{a}\Big)^2$
$=\Big(1+\frac{3}{5}\text{a}\Big)^3$
Hence, factorisation of $1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$ is $=\Big(1+\frac{3}{5}\text{a}\Big)^3$
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Question 692 Marks
Factorise:
$21\text{x}^2-2\text{x}+\frac{1}{21}$
Answer
$21\text{x}^2-2\text{x}+\frac{1}{21}$
$21\text{x}^2-\text{x}-\text{x}+\frac{1}{21}$
$=21\text{x}\Big(\text{x}-\frac{1}{21}\Big)-1\Big(\text{x}-\frac{1}{21}\Big)$
$=\Big(\text{x}-\frac{1}{21}\Big)(21\text{x}-1)$
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Question 702 Marks
Factorise:
a3 + 0.008
Answer
a3 + 0.008
= (a)3 + (0.2)3
= (a + 0.2)[(a)2 - a × 0.2 + (0.2)2] Since a3 + b3 = (a + b)(a2 - a × b + b2)
= (a + 0.2)(a2 - 0.2a + 0.04)
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Question 712 Marks
Expand:
(2b - b + c)2
Answer
(2b - b + c)2 = [(2a) + (-b) + (c)]2
= (2a)2 + (-b)2 + (c)2 + 2(2a)(-b) + 2(-b)(c) + 4(a)(c)
= 14a2 + b2 + c2 - 4ab - 2bc + 4ac
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Question 722 Marks
Factorise:
$\sqrt{3}\text{x}^2-10\text{x}+8\sqrt{3}$
Answer
$\sqrt{3}\text{x}^2-10\text{x}+8\sqrt{3}$
$=\sqrt{3}\text{x}^2+4\text{x}+6\text{x}+8\sqrt{3}$
$=\text{x}\big(\sqrt{3}\text{x}+4\big)+2\sqrt{3}\big(\sqrt{3}\text{x}+4\big)$
$=\big(\sqrt{3}\text{x}+4\big)\big(\text{x}+2\sqrt{3}\big)$
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Question 732 Marks
Factorise:
$\text{x}^2+5\sqrt{5}\text{x}+30$
Answer
$\text{x}^2+5\sqrt{5}\text{x}+30$
$=\text{x}^2+2\sqrt{5}\text{x}+3\sqrt{5}\text{x}+30$
$=\text{x}(\text{x}+2\sqrt{5})+3\sqrt{5}(\text{x}+2\sqrt{5})$
$=(\text{x}+2\sqrt{5})(\text{x}+3\sqrt{5})$
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Question 742 Marks
Factorise:
125 - 8x3 - 27y3 - 90xy
Answer
125 - 8x3 - 27y3 - 90xy
= 53 + (-2x)3 + (-3y)3 - 3 × 5 × (-2x) × (-3y)
= [5 + (-2x) + (-3y)][52 + (-2x)2 + (-3y)2 - 5 × (-2x) - (-2x)(-3y) - 5 × (-3y)]
= (5 - 2x - 3y)(25 + 4x2 + 9y2 + 10x - 6xy + 15y)
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Question 752 Marks
Factorise:
25x2 - 10x + 1 - 36y2
Answer
25x2 - 10x + 1 - 36y2
= (25x2 - 10x + 1) - 36y2
= [(5x)2 - 2(5x)(1) + (1)2] - (6y)2
= (5x - 1)2 - (6y)2
= (5x - 1 - 6y)(5x - 1 + 6y)
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Question 762 Marks
Factorise:
4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
Answer
We have:
4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
= [(2x) + (3y) + (-4z)]2
= (2x + 3y - 4z)2
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Question 772 Marks
Find the product.
(x + y - z)(x2 + y2 + z2 - xy + yz + zx)
Answer
(x + y - z)(x2 + y2 + z2 - xy + yz + zx)
= [x + y + (-z)][x2 + y2 + (-z)2 - xy - y × (-z) - [-z] × x]
= x3 + y3 + (-z)3 - 3x × y × (-z)
= x3 + y3 + (-z)3 - 3x × y × (-z)
= x3 + y3 - z3 + 3xyz
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Question 782 Marks
Factorise:
$125\text{a}^3+\frac{1}{8}$
Answer
$125\text{a}^3+\frac{1}{8}$
We know that:
Since a2 + b3 = (a + b)(a2 - a × b + b2)
Let us rewrite
$125\text{a}^3+\frac{1}{8}$
$=(5\text{a})^3+\Big(\frac{1}{2}\Big)^3$
$=\Big(5\text{a}+\frac{1}{2}\Big)\bigg[(5\text{a})^2-5\text{a}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\bigg]$
$=\Big(5\text{a}+\frac{1}{2}\Big)\Big(25\text{a}^2-\frac{5\text{a}}{2}+\frac{1}{4}\Big)$
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Question 792 Marks
Factorise:
x4 - 625
Answer
x4 - 625
= (x2)2 - (25)2
= (x2 + 25)(x2 - 25)
= (x2 + 25)(x2 - 52) $\Big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\Big]$
= (x2 + 25)(x + 5)(x - 5)
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Question 802 Marks
Factorise:
9a2 + 3a - 8b - 64b2
Answer
9a2 + 3a - 8b - 64b2
= 9a2 - 64b2 + 3a - 8b
= (3a)2 - (8b)2 + (3a - 8b) $\big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
= (3a + 8b)(3a - 8b) + (3a - 8b)
= (3a - 8b)(3a + 8b + 1)
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Question 812 Marks
Factorise:
1029 - 3x3
Answer
1029 - 3x3
= 3(343 - x3)
= 3[(7)3 - x3]
= 3[(7 - x)(72 + 7x + x2)]
= 3(7 - x)(49 + 7x + x2)
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Question 822 Marks
Expand:
(-3a + 4b - 5c)2
Answer
(-3a + 4b - 5c)2 = [(-3a) + (4b) + (-5c)]2
= (-3a)2 + (4b)2 + (-5c)2 + 2(-3a)(4b) + 2(4b)(-5c) + 2(-3a)(-5c)
=9a2 + 16b2 + 25c2 - 24ab - 40bc + 30ac
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Question 832 Marks
Factorise:
a12 - b12
Answer
a12 - b12
= (a6)2 - (b6)2
= (a6 - b6)(a6 + b6)
= [(a3)2 - (b3)2][(a2)3 + (b2)3]
= (a3 - b3)(a3 + b3)[(a2 + b2)(a4 - a2b2 + b4)]
= (a - b)(a2 + ab + b2)(a + b)(a2 - ab + b2)(a2 + b2)(a4 - a2b2 + b4)
= (a - b)(a + b)(a2 + b2)(a2 + ab + b2)(a2 - ab + b2)(a4 - a2b2 + b4)
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Question 842 Marks
Factorise:
$\frac{2}{3}\text{x}^2-\frac{17}{3}\text{x}-28$
Answer
$\frac{2}{3}\text{x}^2-\frac{17}{3}\text{x}-28$
$=\frac{2}{3}\text{x}^2+\frac{7}{3}\text{x}-8\text{x}-28$
$=\frac{\text{x}}{2}(2\text{x}+7)-4(2\text{x}+7)$
$=\Big(\frac{\text{x}}{3}-4\Big)(2\text{x}+7)$
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Question 852 Marks
Evaluate:
(99)3
Answer
(99)3
= (100 - 1)3
= (100)3 - (1)3 - 3(100)2 × (1) + 3(100)(1)2
= 1000000 - 1 - 30000 + 300
= 1000300 - 30001
= 970299
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Question 862 Marks
Factorise:
24x2 - 41x + 12
Answer
24x2 - 41x + 12
= 24x2 - 32x - 9x + 12
= 8x(3x - 4) - 3(3x - 4)
= (3x - 4)(8x - 3)
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Question 872 Marks
Factorise:
27x3 - y3 - z3 - 9xyz
Answer
27x3 - y3 - z3 - 9xyz
= (3x)3 - y3 - z3 - 3 × (3x) × (-y) × (-z)
We know,
a3 + b3 + c3 - 3abc
= (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
a = 3x, b = -y, c = -z
(3x)3 - y3 - z3 - 3 × (3x) × (-y) × (-z)
= (3x - y - z)(9x2 + y2 + z2 + 3xy - yz + 3xz)
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Question 882 Marks
Factorise:
9 - a2 + 2ab - b2
Answer
9 - a2 + 2ab - b2
= 9 - (a2 - 2ab + b2)
= 32 - (a - b)2 $\big[\therefore\ \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
= (3 + a - b)(3 - a + b)
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Question 892 Marks
Evaluate:
(103)3
Answer
(103)3
= (100 + 3)3
= (100)3 + (3)3 + 3(100)2 × (3) + 3(100)(3)2
= 1000000 + 27 + 90000 + 2700
= 1092727
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Question 902 Marks
Factorise:
x9 - y9
Answer
x9 - y9
= (x3)3 - (y3)3
= [(x3 - y3)][(x3)2 + x3y3 + (y3)2]
= [(x - y)(x2 + xy + y2)(x6 + x3y3 + y6)
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Question 912 Marks
Factorise:
(a - b)3 + (b - c)3 + (c - a)3
Answer
(a - b)3 + (b - c)3 + (c - a)3
Putting (a - b) = x, (b - c) = y and (c - a) = z
We get: (a - b)3 + (b - c)3 + (c - a)3
= x3 + y3 + z3 [(x + y + z) = (a - b) + (b - c) + (c - a) = 0]
= 3xyz [(x + y + z) = 0 ⇒ x3 + y3 + z3 = 3xyz]
= 3(a - b)(b - c)(c - a)
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Question 922 Marks
Factorise:
(ax + by)2 + (bx - ay)2
Answer
(ax + by)2 + (bx - ay)2
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 - 2abxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + b2x2 + b2y2 + a2y2
= x2(a2 + b2) + y2(a2 + b2)
= (a2 + b2)(x2 + y2)
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Question 932 Marks
Factorise:
$\text{x}^2+3\sqrt{3}\text{x}+6$
Answer
$\text{x}^2+3\sqrt{3}\text{x}+6$
$=\text{x}^2+2\sqrt{3}\text{x}+\sqrt{3}\text{x}+6$
$=\text{x}(\text{x}+2\sqrt{3})+\sqrt{3}(\text{x}+2\sqrt{3})$
$=(\text{x}+2\sqrt{3})(\text{x}+\sqrt{3})$
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Question 942 Marks
Factorise:
a3 - 0.064
Answer
a3 - 0.064
= (a)3 - (0.4)3
= (a - 0.4)[(a)2 + a × 0.4 + (0.4)2] Since a3 - b3 = (a - b)(a2 + a × b + b2)
= (a - 0.4)(a2 + 0.4a + 0.16)
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Question 952 Marks
Factorise:
x2 + 11x + 30
Answer
x2 + 11x + 30
= x2 + 6x + 5x + 30
= x(x + 6) + 5(x + 6)
= (x + 6)(x + 5)
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Question 962 Marks
Factorise:
81x4 - y4
Answer
81x4 - y4
= (9x2)2 - (y2)2
= (9x2 - y2)(9x2 + y2)
= [(3x)2 - y2](9x2 + y2)
= (3x - y)(3x + y)(9x2 + y2)
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Question 972 Marks
Factorise:
8a3 + 125b3 - 64c3 + 120abc
Answer
8a3 + 125b3 - 64c3 + 120abc
= (2a)3 + (5b)3 + (-4c)3 - 3 × (2a) × (5b) × (-4c)
= (2a + 5b - 4c)[(2a)2 + (5b)2 + (-4c)2 - (2a)(5b) - (5b)(-4c) - (2a) × (-4c)]
= (2a + 5b - 4c)(4a2 + 25b2 + 16c2 - 10ab + 20bc + 8ac)
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Question 982 Marks
Factorise:
(x + 2)3 + (x - 2)3
Answer
(x + 2)3 + (x - 2)3
= [(x + 2) + (x - 2)][(x + 2)2 - (x + 2)(x - 2) + (x - 2)2]
= (2x)(x2 + 4x + 4 - x2 + 4 + x2 - 4x + 4)
= 2x(x2 + 12)
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Question 992 Marks
Factorise:
8a3 - b3 - 4ax + 2bx
Answer
8a3 - b3 - 4ax + 2bx
= 8a3 - b3 - 2x(2a - b)
= (2a)3 - (b)3 - 2x(2a - b) Since a3 - b3 = (a - b)(a2 + a × b + b2)
= (2a - b)[(2a)2 + 2a × b + (b)2] - 2x(2a - b)
= (2a - b)(4a2 + 2ab + b2) - 2x(2a - b)
= (2a - b)(4a2 + 2ab + b2 - 2x)
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Question 1002 Marks
Factorise:
a2x2 + (ax2 + 1)x + a
Answer
a2x2 + (ax2 + 1)x + a
= a2x2 + ax3 + x + a
= ax2(a + x) + 1(x + a)
= (ax2 + 1)(a + x)
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2 Marks Questions - Page 2 - Maths STD 9 Questions - Vidyadip