Download App

Maths 3 Marks Question

Factorisation of Polynomials · Haryana Board (HBSE) STD 9 (Haryana - English Medium). 21 questions.

Questions

3 Marks Question

🎯

Test yourself on this topic

21 questions · timed · auto-graded

Question 13 Marks
Factorise: $2(x + y)^2 - 9(x + y) - 5$
Answer
Let $x + y = z$
Then, $2(x + y)^2 - 9(x + y) - 5$
$= 2z^2 - 9z - 5$
$= 2z^2 - 10z + z - 5$
$= 2z(z - 5) + 1(z - 5)$
$= (z - 5)(2z + 1)$
Now, replacing $z$ by $(x + y)$, we get $2(x + y)^2 - 9(x + y) - 5$
$= [(x + y) - 5][(2(x + y) + 1)] = (x + y - 5)(2x + 2y + 1)$
View full question & answer→
Question 23 Marks
Factorise: $4 x^4+7 x^2-2$
Answer
Let $x^2 = y$ Then, $4x^4 + 7x^2 - 2$
$= 4y^2 + 7y - 2$
$= 4y^2 + 8y - y - 2$
$= 4y(y + 2) - 1(y + 2)$
$= (y + 2)(4y - 1)$
Now replacing $y$ by $x^2,$
we get $4x^4 + 7x^2 - 2 = (x^2 + 2)(4x^2 - 1)$
Since $a^2 - b^2 = (a - b)(a + b) = (x^2 + 2)(2x + 1)(2x - 1)$
View full question & answer→
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 $
$= 10a^2 - 6a + 5a - 3$
$= 2a(5a - 3) + 1(5a - 3)$
$= (5a - 3)(2a + 1)$
$=\bigg[5\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3\bigg]\bigg[2\Big(3\text{x}+\frac{1}{\text{x}}\Big)+1\bigg]$
$=\Big(15\text{x}+\frac{5}{\text{x}}-3\Big)\Big(6\text{x}+\frac{2}{\text{x}}+1\Big)$
View full question & answer→
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 $\text{L.H.S.}$
$=\frac{0.85\times0.85\times0.85+0.15\times0.15\times0.15}{0.85\times0.85-0.85\times0.15+0.15\times0.15}$
$=\frac{\text{a}\times\text{a}\times\text{a}+\text{b}\times\text{b}\times\text{b}}{\text{a}\times\text{a}-\text{a}\times\text{b}+\text{b}\times\text{b}}$
$=\frac{\text{a}^3+\text{b}^3}{\text{a}^2-\text{ab}+\text{b}^2}$
$=\frac{(\text{a}+\text{b})\big(\text{a}^2-\text{ab}+\text{b}^2\big)}{\big(\text{a}^2-\text{ab}+\text{b}^2\big)}$
$=\text{a}+\text{b}$
$=0.85+0.15$
$=1$
$=\text{R.H.S.}$
View full question & answer→
Question 53 Marks
Factorise: $(a-3 b)^3+(3 b-c)^3+(c-a)^3$
Answer
We know $x^3 + y^3 + z^3 - 3xyz$
$= (x + y +z)(x^2 + y^2 + z^2 - xy - yz - zx) x^3 + y^3 + z^3$
$= (x + y +z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz$
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)$
$= 0 + 3(a - 3b)(3b - c)(c - a)$
$= 3(a - 3b)(3b - c)(c - a)$
View full question & answer→
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
$a + b + c = 9$
$\Rightarrow (a + b + c)^2 = 9^2 = 81$
$\Rightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 81$
$\Rightarrow 35 + 2(ab + bc + ca) = 81$
$\Rightarrow (ab + bc + ca) = 23$
We have, $(a^3 + b^3 + c^3 - 3abc)$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= (9)(35 - 23)$
$= 108$
View full question & answer→
Question 73 Marks
Factorise: $\text{x}^2+\frac{1}{\text{x}^2}-3$
Answer
$\text{x}^2+\frac{1}{\text{x}^2}-3$
$=\text{x}^2+\frac{1}{\text{x}^2}-2-1$
$=\Big(\text{x}^2+\frac{1}{\text{x}^2}-2\Big)-1$
$=\bigg(\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x}^2)\Big(\frac{1}{\text{x}^2}\Big)\bigg)-1^2$
$=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2-1^2$
$=\Big(\text{x}-\frac{1}{\text{x}}-1\Big)\Big(\text{x}-\frac{1}{\text{x}}+1\Big)$
View full question & answer→
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 $\text{L.H.S.}$
$=\frac{59\times59\times59-9\times9\times9}{59\times59+59\times9+9\times9}$
$=\frac{\text{a}\times\text{a}\times\text{a}-\text{b}\times\text{b}\times\text{b}}{\text{a}\times\text{a}+\text{a}\times\text{b}+\text{b}\times\text{b}}$
$=\frac{\text{a}^3-\text{b}^3}{\text{a}^2+\text{ab}+\text{b}^2}$
$=\frac{(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)}{\big(\text{a}^2+\text{ab}+\text{b}^2\big)}$
$=\text{a}-\text{b}$
$=59-9$
$=50$
$=\text{R.H.S.}$
View full question & answer→
Question 93 Marks
Factorise:$ 9(2a - b)^2 - 4(2a - b) - 13$
Answer
Let $2a - b = c$
Then, $9(2a - b)^2 - 4(2a - b) - 13$
$= 9c^2 - 4c - 13 = 9c^2 - 13c + 9c - 13$
$= c(9c - 13) + 1(9c - 13)$
$= (c + 1)(9c - 13)$
Now, replacing $c$ by $(2a - b),$
we get $9(2a - b)^2 - 4(2a - b) - 13$
$= (2a - b + 1)[9(2a - b) - 13]$
$= (2a - b + 1)(18a - 9b - 13)$
View full question & answer→
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 $
$= 6a^2 + 15a - 8a - 20$
$= 3a(2a + 5) - 4(2a + 5)$
$= (2a + 5)(3a - 4)$
$=\bigg[2\Big(2\text{x}-\frac{3}{\text{x}}\Big)+5\bigg]\bigg[3\Big(2\text{x}-\frac{3}{\text{x}}\Big)-4\bigg]$
$=\Big(4\text{x}-\frac{6}{\text{x}}+5\Big)\Big(6\text{x}-\frac{9}{\text{x}}-4\Big)$
View full question & answer→
Question 113 Marks
Find the product. $(3x - 5y + 4)(9x^2 + 25y^2 + 15xy - 20y + 12x + 16)$
Answer
$(3x - 5y + 4)(9x^2 + 25y^2 + 15xy - 20y + 12x + 16)$
$= (3x + (-5y) + 4)(9x^2 + 25y^2 + 16 + 15xy - 20y + 12x) $
$(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= a^3 + b^3 + c^3 - 3abc$
Here, $a = 3x, b = -5y, c = 4$
$​​​​​​​= (3x + (-5y) + 4)(9x^2 + 25y^2 + 16 + 15xy - 20y + 12x)$
$= (3x)^3 + (-5y)^3 + 4^3 - 3 \times 3x (-5y)(4)$
$= 27x^3 - 125y^3 + 64 + 180xy$
View full question & answer→
Question 123 Marks
Find the product. $(x - y - z)(x^2 + y^2 + z^2 + xy - yz + xz)$
Answer
$(x - y - z)(x^2 + y^2 + z^2 + xy - yz + xz)$
$= (x + (-y) + (-z)(x^2 + y^2 + z^2 + xy - yz + xz)$
$= (x + (-y) + (-z)(x^2 + y^2 + z^2 + xy - yz + xz)$
We have $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= a^3 + b^3 + c^3 - 3abc$
Here, $a = x, b = -y, c = -z (x + (-y) + (-z)
(x^2 + y^2 + z^2 + xy - yz + xz)$
$= x^3 - y^3 - z^3 - 3xyz$
View full question & answer→
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) $
View full question & answer→
Question 143 Marks
Factorise: $\frac{1}{3}\text{x}^2-2\text{x}-9$
Answer
$\frac{1}{3}\text{x}^2-2\text{x}-9$
$=\frac{1}{3}\text{x}^2-3\text{x}+\text{x}-9$
$=\text{x}\Big(\frac{\text{x}}{3}-3\Big)+(\text{x}-9)$
$=\frac{\text{x}}{3}(\text{x}-9)+(\text{x}-9)$
$=(\text{x}-9)\Big(\frac{\text{x}}{3}+1\Big)$
$=(\text{x}-9)\frac{(\text{x}+3)}{3}$
$=\frac{1}{3}(\text{x}-9)(\text{x}+3)$
View full question & answer→
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, $\text{x}=\big(\sqrt{3}\text{a}\big),\ \text{y}=(-\text{b}),\ \text{z}=\big(-\sqrt{5}\text{c}\big)$
$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
$=\big(\sqrt{3}\text{a}-\text{b}-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{c}\big)$
View full question & answer→
Question 163 Marks
Factorise: $\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+\frac{2}{\text{a}}$
Answer
We know that, Since $a^3 - b^3 = (a - b)(a^2 + a \times b + b^2)$
$\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+\frac{2}{\text{a}}$
$=\text{a}^3-\frac{1}{\text{a}^3}-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\text{a}\times\frac{1}{\text{a}}+\frac{1}{\text{a}^2}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}-2\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
View full question & answer→
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 $5 \times (-3) = -15$.
$\therefore\ \sqrt{5}\text{x}^2+2\text{x}-3\sqrt{5}$
$=\sqrt{5}\text{x}^2+5\text{x}-3\text{x}-3\sqrt{5}$
$=\sqrt{5}\text{x}\big(\text{x}+\sqrt{5}\big)-3\big(\text{x}+\sqrt{5}\big)$
$=\big(\text{x}+\sqrt{5}\big)\big(\sqrt{5}\text{x}-3\big)$
View full question & answer→
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
$\text{x}^3+\text{y}^3+\text{z}^3-3\text{xyz}=(\text{x}+\text{y}+\text{z})\\\big(\text{x}^2+\text{y}^2+\text{z}^2-\text{xy}-\text{yz}-\text{zx}\big)$
$\text{x}=\sqrt{2}\text{a},\ \text{y}=\sqrt{3}\text{b},\ \text{z}=\text{c}$
$\big(\sqrt{2}\text{a}\big)^3+\big(\sqrt{3}\text{b}\big)^3+\text{c}^3-3\big(\sqrt{2}\text{a}\big)\big(\sqrt{3}\text{b}\big)\text{c}$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6\text{ab}}-\sqrt{3\text{bc}}-\sqrt{2}\text{ac}\big)$
View full question & answer→
Question 193 Marks
Factorise: $7(x - 2y)^2 - 25(x - 2y) + 12$
Answer
Let $\mathrm{x}-2 \mathrm{y}=\mathrm{z}$ Then, $7(\mathrm{x}-2 \mathrm{y})^2-25(\mathrm{x}-2 \mathrm{y})+12=7 \mathrm{z}^2-25 \mathrm{z}+12=7 \mathrm{z}^2-21 \mathrm{z}-4 \mathrm{z}+12=7 \mathrm{z}(\mathrm{z}-3)-4(z-3)=(z-3)(7 z-4)$
Now replace $z$ by $(x-2 y)$, we get $7(x-2 y)^2-25(x-2 y)+12=(x-2 y-3)[7(x-2 y)-4]=(x-2 y-3)(7 x-14 y-4)$
View full question & answer→
Question 203 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: $(5a - 7b)^3 + (9c - 5a)^3 + (7b - 9c)^3 = x^3 + z^3 + y^3 = 3xyz$
[When $x + y + z = 0, x^3 + y^3 + z^3 = 3xyz] $
$= 3(5a - 7b)(9c - 5a)(7b - 9c)$
View full question & answer→
Question 213 Marks
Factorise: $\text{x}^2-2\text{x}+\frac{7}{16}$
Answer
$\text{x}^2-2\text{x}+\frac{7}{16}$
$=\frac{1}{16}\big(16\text{x}^2-32\text{x}+7\big)$
$=\frac{1}{16}\big(16\text{x}^2-4\text{x}-28\text{x}+7\big)$
$=\frac{1}{16}\Big[4\text{x}(4\text{x}-1)-7(4\text{x}-1)\Big]$
$=\frac{1}{16}(4\text{x}-1)(4\text{x}-7)$
$=(4\text{x}-1)\Big(\frac{\text{x}}{4}-\frac{7}{16}\Big)$
View full question & answer→
Generate Paper FreeAll Factorisation of Polynomials topics
3 Marks Question - Maths STD 9 Questions - Vidyadip