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Maths 1 Mark Question

Factorisation of Polynomials · Maharashtra Board STD 9 (MSBSHSE - English Medium). 49 questions.

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Question 21 Mark
Factorise:
$(x+1)^3+(x-1)^3$
Answer
$(x+1)^3+(x-1)^3$
$=(x+1+x-1)\left[(x+1)^2-(x+1)(x-1)+(x-1)^2\right]$
$=2 x\left(x^2+2 x+1-x^2+1+x^2-2 x+1\right)$
$=2 x\left(x^2+3\right)$
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Question 31 Mark
Factorise:
$a(a-2 b-c)+2 b c$
Answer
$a(a-2 b-c)+2 b c$
$=a^2-2 a b-a c+2 b c$
$=a(a-2 b)-c(a-2 b)$
$=(a-2 b)(a-c)$
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Question 41 Mark
Factorise:
2x + 4y - 8xy - 1
Answer
2x + 4y - 8xy - 1
= 2x - 1 - 8xy + 4y
= (2x - 1) - 4y(2x - 1)
= (2x - 1)(1 - 4y)
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Question 51 Mark
Factorise:
$a^2+2 a b+b^2-9 c^2$
Answer
$a^2+2 a b+b^2-9 c^2$
$=(a+b)^2-(3 c)^2$
$=(a+b+3 c)(a+b-3 c)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
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Question 61 Mark
Factorise:
$20 x^2-45$
Answer
$20 x^2-45$
$=5\left(4 x^2-9\right)$
$=5\left[(2 x)^2-(3)^2\right]$
$=2(1+5 x)(1-5 x)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=5(2 x+3)(2 x-3)$
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Question 71 Mark
Factorise:
$2 x\left(p^2+q^2\right)+4 y\left(p^2+q^2\right)$
Answer
$2 x\left(p^2+q^2\right)+4 y\left(p^2+q^2\right)$
$=(2 x+4 y)\left(p^2+q^2\right)$
$=2(x+2 y)\left(p^2+q^2\right)$
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Question 81 Mark
Factorise:
$8 a b^2-18 a^3$
Answer
$8 a b^2-18 a^3$
$=2 a\left(4 b^2-9 a^2\right)$
$=2 a\left[(2 b)^2-(3 a)^2\right]$
$=2 a(2 b+3 a)(2 b-3 a)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
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Question 91 Mark
Factorise:
$(3 a+5 b)^2-4 c^2$
Answer
$(3 a+5 b)^2-4 c^2$
$=(3 a+5 b)^2-(2 c)^2$
$=(3 a+5 b-2 c)(3 a+5 b+2 c)$
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Question 111 Mark
Factorise:
$(3 a-1)^2-6 a+2$
Answer
$(3 a-1)^2-6 a+2$
$=(3 a-1)^2-2(3 a-1)$
$=(3 a-1)[(3 a-1)-2]$
$=(3 a-1)(3 a-3)$
$=3(3 a-1)(a-1)$
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Question 121 Mark
Factorise:
$x^3+27$
Answer
$x^3+27$
$=x^3+3^3$
$=(x+3)\left(x^2-3 x+9\right) \text { Since } a^3+b=(a+b)\left(a^2-a b+b^2\right)$
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Question 131 Mark
Factorise:
$5-20 x^2$
Answer
$5-20 x^2$
$=5\left(1-4 x^2\right)$
$=5\left[(1)^2-(2 x)^2\right]$
$=5[(1-2 x)(1+2 x)]$
$=5(1-2 x)(1+2 x)$
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Question 161 Mark
Factorise:
$2-50 x^2$
Answer
$2-50 x^2$
$=2\left(1-25 x^2\right)$
$=2\left[(1)^2-(5 x)^2\right]$
$=2(1+5 x)(1-5 x)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
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Question 171 Mark
Factorise:
$64 a^3-27 b^3-144 a^2 b+108 a b^2$
Answer
$64 a^3-27 b^3-144 a^2 b+108 a b^2$
$=(4 a)^3-(3 b)^3-3(4 a)^2(3 b)+3(4 a)(3 b)^2$
$=(4 a-3 b)^3$
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Question 181 Mark
Factorise:
$x^3-2 x^2 y+3 x y^2-6 y^3$
Answer
$x^3-2 x^2 y+3 x y^2-6 y^3$
$=x^2(x-2 y)+3 y^2(x-2 y)$
$=(x-2 y)\left(x^2+3 y^2\right)$
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Question 191 Mark
Expand:
$(a+2 b+5 c)^2$
Answer
$(a+2 b+5 c)^2$
$=(a)^2+(2 b)^2+(5 c)^2+2(a)+2(2 b)(5 c)+2(5 c)(a)$
$=a^2+4 b^2+25 c^2+4 a b+20 b c+10 a c$
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Question 201 Mark
Factorise:
$27 a^2-48 b^2$
Answer
$27 a^2-48 b^2$
$=3\left(9 a^2-16 b^2\right)$
$=3\left[(3 a)^2-(4 b)^2\right]$
$=3(3 a+4 b)(3 a-4 b)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
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Question 211 Mark
Factorise: $a^3 b-a^2 b+5 a b-5 b$
Answer
$a^3 b-a^2 b+5 a b-5 b$
$=a^2 b(a-1)+5 b(a-1)$
$=(a-1)\left(a^2 b+5 b\right)$
$=(a-1) b\left(a^2+5\right)$
$=b(a-1)\left(a^2+5\right)$
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Question 241 Mark
Factorise:
$3 x^3-48 x$
Answer
$3 x^3-48 x$
$=3 x\left(x^2-16\right)$
$=3 x\left[(x)^2-(4)^2\right]$
$=3 x(x+4)(x-4)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
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Question 251 Mark
Factorise:
$a^3 x^3-3 a^2 b x^2+3 a b^2 x-b^3$
Answer
$a^3 x^3-3 a^2 b x^2+3 a b^2 x-b^3$
$(a x)^3-(b)^3-3(a x)^2(b)+(a x)(b)^2$
$=(a x-b)^3$
Hence, factorisation of $a^3 x^3-3 a^2 b x^2+3 a b^2 x-b^3$ is $=(a x-b)^3$
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Question 261 Mark
Factorise:
$a^3-12 a(a-4)-64$
Answer
$a^3-12 a(a-4)-64$
$a^3-12 a^2+48 a-64$
$=(a)^3-(4)^3-3(a)^2(4)+3(a)(4)^2$
$=(a-4)^3$
Hence, factorisation of $a^3-12 a(a-4)-64$ is $=(a-4)^3$
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Question 271 Mark
Factorise:$\Big(\frac{25}{4}\text{x}^2-\frac{1}{9}\text{y}^2\Big)$
Answer
$\Big(\frac{25}{4}\text{x}^2-\frac{1}{9}\text{y}^2\Big)$$\Big(\frac{5}{2}\text{x}\Big)^2-\Big(\frac{1}{3}\text{y}\Big)^2$
$=\Big(\frac{5}{2}\text{x}-\frac{1}{3}\text{y}\Big)\Big(\frac{5}{2}\text{x}+\frac{1}{3}\text{y}\Big)$
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Question 301 Mark
Factorise:
$150-6 x^2$
Answer
$150-6 x^2$
$=6\left(25-x^2\right)$
$=6\left(5^2-x^2\right)$
$=6(5+x)(5-x)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
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Question 311 Mark
Factorise:
$a^3+a b(1-2 a)-2 b^2$
Answer
$a^3+a b(1-2 a)-2 b^2$
$=a^3+a b-2 a^2 b-2 b^2$
$=a\left(a^2+b\right)-2 b\left(a^2+b\right)$
$=\left(a^2+b\right)(a-2 b)$
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Question 321 Mark
Factorise:
$x^3-5 x^2-x+5$
Answer
$x^3-5 x^2-x+5$
$=x^2(x-5)-1(x-5)$
$=(x-5)\left(x^2-1\right)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(x-5)(x+1)(x-1)$
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Question 341 Mark
Factorise:
$27 a^3+64 b^3$
Answer
$27 a^3+64 b^3$
$=(3 a)^3+(4 b)^3$
$=(3 a+4 b)\left[(3 a)^2-(3 a)(4 b)+(4 b)^2\right]$
$=(3 a+4 b)\left(9 a^2-12 a b+16 b^2\right)$
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Question 351 Mark
Factorise:
$125 x^3-27 y^3-225 x^2 y+135 x y^2$
Answer
$125 x^3-27 y^3-225 x^2 y+135 x y^2$
$(5 x)^3-(3 y)^3-3(5 x)^2(3 y)+3(5 x)(3 y)^2$
$=(5 x-3 y)^3$
Hence, factorisation of $125 x^3-27 y^3-225 x^2 y+135 x y^2$ is $(5 x-3 y)^3$
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Question 361 Mark
Factorise:
$(2 x-3)^2-8 x+12$
Answer
$(2 x-3)^2-8 x+12$
$=(2 x-3)^2-4(2 x-3)$
$=(2 x-3)(2 x-3-4)$
$=(2 x-3)(2 x-7)$
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Question 391 Mark
Factorise:
$a^6+b^6$
Answer
$a^6+b^6$
$=\left(a^2\right)^3+\left(b^2\right)^3$
$=\left(a^2+b^2\right)\left[\left(a^2\right)^2-\left(a^2 b^2\right)+\left(b^2\right)^2\right]$
$=\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
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Question 401 Mark
Factorise:
$2 a^2+b c-2 a b-a c$
Answer
$2 a^2+b c-2 a b-a c$
$=2 a^2-2 a b-a c+b c$
$=2 a(a-b)-c(a-b)$
$=(a-b)(2 a-c)$
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Question 411 Mark
Factorise:3ax - 6ay - 8by + 4bx
Answer
3ax - 6ay - 8by + 4bx
= 3a(x - 2y) + 4b(x - 2y)
= (x - 2y)(3a + 4b)
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Question 431 Mark
Factorise:
$8 a^3+27 b^3+36 a^2 b+54 a b^2$
Answer
$8 a^3+27 b^3+36 a^2 b+54 a b^2$
$=(2 a)^3+(3 b)^3+3(2 a)^2(3 b)+3(2 a)(3 b)^2$
$=(2 a+3 b)^3$
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Question 441 Mark
Factorise:
$a^3+a-3 a^2-3$
Answer
$a^3+a-3 a^2-3$
$=a\left(a^2+1\right)-3\left(a^2+1\right)$
$=(a-3)\left(a^2+1\right)$
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Question 461 Mark
Factorise:$\frac{64}{125}\text{a}^3-\frac{96}{25}\text{a}^2+\frac{48}{5}\text{a}-8$
Answer
$\frac{64}{125}\text{a}^3-\frac{96}{25}\text{a}^2+\frac{48}{5}\text{a}-8$$\Big(\frac{4}{5}\text{a}\Big)^3-(2)^3-3\Big(\frac{4}{5}\text{a}\Big)^2(2)+3\Big(\frac{4}{5}\text{a}\Big)(2)^2$
$=\Big(\frac{4}{5}\text{a}-2\Big)^3$
Hence, factorisatoin of $\frac{64}{125}\text{a}^3-\frac{96}{25}\text{a}^2+\frac{48}{5}\text{a}-8$ is $\Big(\frac{4}{5}\text{a}-2\Big)^3$
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Question 471 Mark
Factorise:
$x-64 x^3$
Answer
$x-64 x^3$
$=x\left(1-64 x^2\right)$
$=x\left[(1)^2-(8 x)^2\right]$
$=x(1+8 x)(1-8 x)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
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Question 491 Mark
Factorise:
$a b x^2+a^2 x+b^2 x+a b$
Answer
$a b x^2+a^2 x+b^2 x+a b$
$=a x(b x+a)+b(b x+a)$
$=(b x+a)(a x+b)$
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