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Maths 3 Marks Question

Factorisation of Polynomials · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 21 questions.

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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
= 2z2 - 9z - 5
= 2z2 - 10z + z - 5
= 2z(z - 5) + 1(z - 5)
= (z - 5)(2z + 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)(2x + 2y + 1)
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Question 23 Marks
Factorise:
4x4 + 7x2 - 2
Answer
Let x2 = y
Then, 4x4 + 7x2 - 2
= 4y2 + 7y - 2
= 4y2 + 8y - y - 2
= 4y(y + 2) - 1(y + 2)
= (y + 2)(4y - 1)
Now replacing y by x2, we get
4x4 + 7x2 - 2
= (x2 + 2)(4x2 - 1) Since a2 - b2 = (a - b)(a + b)
= (x2 + 2)(2x + 1)(2x - 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
= 10a2 - a - 3
= 10a2 - 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:
(a - 3b)3 + (3b - c)3 + (c - a)3
Answer
We know
x3 + y3 + z3 - 3xyz = (x + y +z)(x2 + y2 + z2 - xy - yz - zx)
x3 + y3 + z3 = (x + y +z)(x2 + y2 + z2 - xy - yz - zx) + 3xyz
Here, x = (a - 3b), y = (3b - c), z = (c - a)
(a - 3b)3 + (3b - c)3 + (c - a)3
= (a - 3b + 3b - c + c - a)
[(a - 3b)2 + (3b - c)2 + (c - a)2 - (a - 3b)(3b - c) - (3b - c)(c - a) - (c - a)(a - 3b)]
+ 3(a - 3b)(3b - c)(c - a)
= 0 + 3(a - 3b)(3b - c)(c - a)
= 3(a - 3b)(3b - c)(c - a)
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Question 53 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 63 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 73 Marks
Factorise:
9(2a - b)2 - 4(2a - b) - 13
Answer
Let 2a - b = c
Then, 9(2a - b)2 - 4(2a - b) - 13
= 9c2 - 4c - 13
= 9c2 - 13c + 9c - 13
= c(9c - 13) + 1(9c - 13)
= (c + 1)(9c - 13)
Now, replacing c by (2a - b), we get
9(2a - b)2 - 4(2a - b) - 13
= (2a - b + 1)[9(2a - b) - 13]
= (2a - b + 1)(18a - 9b - 13)
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Question 83 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
= 6a2 + 7a - 20
= 6a2 + 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 93 Marks
Find the product.
(3x - 5y + 4)(9x2 + 25y2 + 15xy - 20y + 12x + 16)
Answer
(3x - 5y + 4)(9x2 + 25y2 + 15xy - 20y + 12x + 16)
= (3x + (-5y) + 4)(9x2 + 25y2 + 16 + 15xy - 20y + 12x)
(a + b + c)(a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc
Here, a = 3x, b = -5y, c = 4
= (3x + (-5y) + 4)(9x2 + 25y2 + 16 + 15xy - 20y + 12x)
= (3x)3 + (-5y)3 + 43 - 3 × 3x (-5y)(4)
= 27x3 - 125y3 + 64 + 180xy
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Question 103 Marks
Find the product.
(x - y - z)(x2 + y2 + z2 + xy - yz + xz)
Answer
(x - y - z)(x2 + y2 + z2 + xy - yz + xz)
= (x + (-y) + (-z)(x2 + y2 + z2 + xy - yz + xz)
= (x + (-y) + (-z)(x2 + y2 + z2 + xy - yz + xz)
We have
(a + b + c)(a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc
Here, a = x, b = -y, c = -z
(x + (-y) + (-z)(x2 + y2 + z2 + xy - yz + xz) = x3 - y3 - z3 - 3xyz
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Question 113 Marks
Factorise:
x8 - 1
Answer
x8 - 1
= (x4)2 - (1)2
= (x4 - 1)(x4 + 1)
= [(x2)2 - (1)2)(x4 + 1)
= (x2 - 1)(x2 + 1)(x4 + 1)
= (x - 1)(x + 1)(x2 + 1)[(x2)2 + (1)2 + 2x2 - 2x2
= (x - 1)(x + 1)(x2 + 1)[(x2)2 + (1)2 + 2x2) - 2x2
$=(\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 123 Marks
If a + b + c = 9 and a2 + b2 + c2 = 35, find the value of (a3 + b3 + c3 - 3abc).
Answer
a + b + c = 9
⇒ (a + b + c)2 = 92 = 81
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81
⇒ 35 + 2(ab + bc + ca) = 81
⇒ (ab + bc + ca) = 23
We have,
(a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
= (9)(35 - 23)
= 108
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Question 133 Marks
Factorise:
$\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+\frac{2}{\text{a}}$
Answer
We know that, Since a3 - b3 = (a - b)(a2 + a × b + b2)
$\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+\frac{2}{\text{a}}$
$=\text{a}^3-\frac{1}{\text{a}^3}-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\text{a}\times\frac{1}{\text{a}}+\frac{1}{\text{a}^2}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}-2\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
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Question 143 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 153 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 163 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 173 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 183 Marks
Factorise:
7(x - 2y)2 - 25(x - 2y) + 12
Answer
Let x - 2y = z
Then, 7(x - 2y)2 - 25(x - 2y) + 12
= 7z2 - 25z + 12
= 7z2 - 21z - 4z + 12
= 7z(z - 3) - 4(z - 3)
= (z - 3)(7z - 4)
Now replace z by (x - 2y), we get
7(x - 2y)2 - 25(x - 2y) + 12
= (x - 2y - 3)[7(x - 2y) - 4]
= (x - 2y - 3)(7x - 14y - 4)
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Question 193 Marks
Factorise:
(5a - 7b)3 + (7b - 9c)3 + (9c - 5a)3
Answer
Put (5a - 7b) = x, (7b - 9c) = y, (9c - 5a) = z.
Here,
x + y + z = 5a - 7b + 9c - 5a + 7b - 9c = 0
Thus,
We have:
(5a - 7b)3 + (9c - 5a)3 + (7b - 9c)3 = x3 + z3 + y3
= 3xyz [When x + y + z = 0, x3 + y3 + z3 = 3xyz]
= 3(5a - 7b)(9c - 5a)(7b - 9c)
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Question 203 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 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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