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Maths MCQ(1M)

Factorization Of Algebraic Expressions · Maharashtra Board STD 9 (MSBSHSE - English Medium). 15 questions.

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MCQ(1M)

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15 questions · auto-graded multiple-choice test.

MCQ 11 Mark
If $(x + y)^3 - (x - y)^3 - 6y(x^2 - y^2) = ky^2$, then $k =$
  • A
    $1$
  • B
    $2$
  • C
    $4$
  • ✓
    $8$
Answer
Correct option: D.
$8$
Let $x + y = A$ and $x - y = B$
Now, $(A-B)^3=A^3-B^3-3 A B(A-B)$
$\Rightarrow[(x+y)-(x-y)]^3=(x+y)^3-(x-y)^3-3(x+y)(x-y)[(x+y)-(x-y)]$
$=(x+y)^3-(x-y)^3-3\left(x^2-y^2\right)(2 y)$
$=(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)$
But, $(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)=k y^3$
$\Rightarrow[(x+y)-(x-y)]^3=(2 y)^3=k 8 y^3$
$\Rightarrow(2 y)^3=k y^3$
$\Rightarrow 8 y^3=k y^3$
$\Rightarrow k=8$
Hence, correct option is $(d).$
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MCQ 21 Mark
The expression $(a-b)^3+(b-c)^3+(c-a)^3$ can be factorized as:
  • A
    $(a-b)(b-c)(c-a)$
  • ✓
    $3(a-b)(b-c)(c-a)$
  • C
    $-3(a-b)(b-c)(c-a)$
  • D
    $(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
Answer
Correct option: B.
$3(a-b)(b-c)(c-a)$
By we know that $a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
If $a+b+c=0$, then
$a^3+b^3+c^3=3 a b c$
In given expression,
Let $a - b = A , b - c = B , c - a = C$
Now, $a-b+b-c+c-a=0$
i.e. $A+B+C=0$
$\Rightarrow A^3+B^3+C^3=3 A B C$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3$
$=3(a-b)(b-c)(c-a)$
Hence, correct option is $(b).$
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MCQ 31 Mark
The value of $\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09},$ is:
  • ✓
    $2$
  • B
    $3$
  • C
    $2.327$
  • D
    $2.273$
Answer
Correct option: A.
$2$
$=\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09}$
$=\frac{(2.3)^3-(0.3)^3}{(2.3)^2+(0.3)^3+(2.3)(0.3)}$
$=\frac{(2.3 - 0.3)\{(2.3)^2+(0.3)^2+(2.3)(0.3)\}}{((2.3)^2+(0.3)^2+(2.3)(0.3))}$
$=2.3 - 0.3$
$=2$
Hence, correct option is $(a)$.
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MCQ 41 Mark
The factors of $x^3-7 x+6$ are:
  • A
    $x(x-6)(x-1)$
  • B
    $\left(x^2-6\right)(x-1)$
  • C
    $(x+1)(x+2)(x+3)$
  • ✓
    $(x-1)(x+3)(x-2)$
Answer
Correct option: D.
$(x-1)(x+3)(x-2)$
$x^3-7 x+6=x^3-7 x+6+1-1 ($by adding $+1\ \&\ -1 \text { to R.H.S) }$
$=x^3-7 x+7-1$
$=\left(x^3-1\right)-7(x-1)$
Now by identity $a^3-b^3=(a-b)\left(a^2+b^2+a b\right)$, we get
$x^3-7 x+6$
$=\left(x^3-1\right)-7(x-1)$
$=(x-1)\left(x^2+x+1\right)-7(x-1)$
$=(x-1)\left(x^2+x+1-7\right)$
$=(x-1)\left(x^2+x-6\right)$
$=(x-1)(x+3)(x-2)$
Hence, correct option is $(d).$
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MCQ 51 Mark
The expression $x^4+4$ can be factorized as:
  • ✓
    $\left(x^2+2 x+2\right)\left(x^2-2 x+2\right)$
     
  • B
     $\left(x^2+2 x+2\right)\left(x^2+2 x-2\right)$
     
  • C
    $\left(x^2-2 x-2\right)\left(x^2-2 x+2\right)$
     
  • D
    $\left(x^2+2\right)\left(x^2-2\right)$
Answer
Correct option: A.
$\left(x^2+2 x+2\right)\left(x^2-2 x+2\right)$
 
$x^4+4$
$=x^4+4+4 x^2-4 x^2$
$=\left(x^4+4 x^2+4\right)-4 x^2$
$=\left(x^2+2\right)^2-(2 x)^2$
$=\left(x^2+2-2 x\right)\left(x^2+2+2 x\right)$
$=\left(x^2+2 x+2\right)\left(x^2-2 x+2\right)$
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MCQ 61 Mark
The factors of $x^2+4 y^2+4 y-4 x y-2 x-8$, are:
  • ✓
    $(x-2 y-4)(x-2 y+2)$
  • B
    $(x-y+2)(x-4 y-4)$
  • C
    $(x+2 y-4)(x+2 y+2)$
  • D
    None of these.
Answer
Correct option: A.
$(x-2 y-4)(x-2 y+2)$
$x^2+4 y^2+4 y-4 x y-2 x-8$
$=x^2+(2 y)^2-2 \times x(2 y)+4 y-2 x-8$
$=(x-2 y)^2+4 y-2 x-8 \ldots(1)$
Now making eq(1) a perfect square by adding 1 and -1
$(x-2 y)^2+4 y-2 x-8=(x-2 y)^2+4 y-2 x-8+1-1$
$=(x-2 y)^2+(1)^2-2 x(1) \times(x-2 y)-9$
$=(x-2 y-1)^2-(3)^2$
$=[(x-2 y-1)-3][x-2 y-1+3]$
$=(x-2 y-4)(x-2 y+2)$
Hence, correct option is $(a).$
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MCQ 71 Mark
The value of $\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2},$ is:
  • A
    $0.006$
  • ✓
    $0.02$
  • C
    $0.0091$
  • D
    $0.00185$
Answer
Correct option: B.
$0.02$
By using identity $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$, we have
$\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2}$
$=\frac{\{(0.013)+(0.007)\}(0.013)^2-(0.013)(0.007)+(0.007)^2}{(0.013)^2-(0.013)(0.007)+(0.007)^2}$
$=0.013+0.007$
$=0.020$
$=0.02$
Hence, correct option is $(b).$
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MCQ 81 Mark
The factors of $a^2-1-2 x-x^2$ are:
  • A
    $(a-x+1)(a-x-1)$
  • B
    $(a+x-1)(a-x+1)$
  • ✓
    $(a+x+1)(a-x+1)$
  • D
    None of these.
Answer
Correct option: C.
$(a+x+1)(a-x+1)$
$a^2-1-2 x-x^2$
$=a^2-\left(1+2 x+x^2\right)$
$=a^2-(1+x)^2$
$=[a-(1+x)][a+(1+x)]$
$=(a-x-1)(a+x+1)$
Hence, correct option is $(c).$
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MCQ 91 Mark
The factors of $x ^4+ x ^2+25$ are:
  • ✓
    $\left(x^2+3 x+5\right)\left(x^2-3 x+5\right)$
  • B
    $\left(x^2+3 x+5\right)\left(x^2+3 x-5\right)$
  • C
    $\left(x^2+x+5\right)\left(x^2-x+5\right)$
  • D
    None of these.
Answer
Correct option: A.
$\left(x^2+3 x+5\right)\left(x^2-3 x+5\right)$
For making perfect square to $x^4+x^2+25$
We add $+10 x^2$ and $-10 x^2$ to it.
$=x^4+x^2+25$
$=x^4+x^2+25+10 x^2-10 x^2$
$=\left[x^4+10 x^2+25\right]-9 x^2$
$=\left(x^2+5\right)^2+(3 x)^2$
$=\left[\left(x^2+5\right)+3 x\right]\left[\left(x^2+5\right)-3 x\right]$
$=\left(x^2+3 x+5\right)\left(x^2-3 x+5\right)$
Hence, correct option is $(a).$
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MCQ 101 Mark
The factors of $x^3-x^2 y-x y^2+y^3$ are:
  • A
    $(x+y)\left(x^2-x y+y^2\right)$
  • B
    $(x+y)\left(x^2+x y+y^2\right)$
  • C
    $(x+y)^2(x-y)$
  • ✓
    $(x-y)^2(x+y)$
Answer
Correct option: D.
$(x-y)^2(x+y)$
$x^3-x^2 y-x y^2+y^3=x^3+y^3-x y(x+y)$
Now by identity $x^3+y^3$
$=(x+y)\left(x^2+y^2-x y\right)$, we have
$x^3-x^2 y-x y^2+y^3$
$=(x+y)\left(x^2+y^2-x y\right)-x y(x+y)$
$=(x+y)\left(x^2+y^2-x y-x y\right)$
$=(x+y)\left(x^2+y^2-2 x y\right)$
$=(x+y)(x-y)^2$
Hence, correct option is $(d).$
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MCQ 111 Mark
$(x+y)^3-(x-y)^3$ can be factorized as:
  • ✓
    $2 y\left(3 x^2+y^2\right)$
  • B
    $2 x\left(3 x^2+y^2\right)$
  • C
    $2 y\left(3 y^2+x^2\right)$
  • D
    $2 x\left(x^2+3 y^2\right)$
Answer
Correct option: A.
$2 y\left(3 x^2+y^2\right)$
We know the identity
$a^3-b^3=(a-b)\left(a^2+b^2+a b\right)$
Let $x+y=a$ and $x-y=b$
Then,
$a^3-b^3$
$=(x+y)^3-(x-y)^3$
$=[(x+y)-(x-y)]\left[(x+y)^2+(x-y)^2+(x+y)(x-y)\right]$
$=2 y\left[x^2+y^2+2 x y+x^2+y^2-2 x y+x^2-y^2\right]$
$=2 y\left(3 x^2+y^2\right)$
Hence, correct option is $(a).$
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MCQ 121 Mark
If $x^3-3 x^2+3 x-7=(x+1)\left(a x^2+b x+c\right)$, then $a+b+c=$
  • A
    $4$
  • B
    $12$
  • ✓
    $-10$
  • D
    $3$
Answer
Correct option: C.
$-10$
The given equation is
$x^3-3 x^2+3 x-7=(x+1)\left(a x^2+b x+c\right)$
This can be written as
$x^3-3 x^2+3 x-7=(x+1)\left(a x^2+b x+c\right)$
$=x^3-3 x^2+3 x-7=a x^3+b x^2+c x+a x^2+b x+c$
$=x^3-3 x^2+3 x-7=a x^3+(a+b) x^2+(b+c) x+c$
Comparing the cofficients on both sides of the equation.
$\text { We get, }$
$\text { a = } 1 \ldots(1)$
$a+b=3 \ldots$
$b+c=3 \ldots$
$c=-7 \ldots(4)$
Putting the value of a form $(1)$ in $(2)$
We get,
$1+b=3$
$b=-3-1$
$b=-4$
So the value of $a, b$ and $c$ is $1,-4$ and $-7$ respectively.
Therefore,
$a+b+c=1-4-7=-10$
Hence, correct option is $(c).$

 
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MCQ 131 Mark
If $3 x=a+b+c$, then the value of $(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)$ is:
  • A
    $a+b+c$
  • B
    $(a-b)(b-c)(c-a)$
  • ✓
    $0$
  • D
    None of these.
Answer
Correct option: C.
$0$
$3 x=a+b+c$
$\Rightarrow a+b+c-3 x=0$
$\Rightarrow 3 x-(a+b+c)=0$
$\Rightarrow(x-a)+(x-b)+(x-c)=0$
Using identity if $a+b+c=0$ then, $a^3+b^3+c^3-3 a b c=0$
If we take $x - a = A , x - b = B , x - c = C$ in equation $(1),$ we get
$A+B+C=0$
$\Rightarrow A^3+B^3+C^3-3 A B C=0$
$\Rightarrow(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)=0$
Hence, correct option is $(c).$
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MCQ 141 Mark
The factors of $x^3-1+y^3+3 x y$ are:
  • ✓
    $(x-1+y)\left(x^2+1+y^2+x+y-x y\right)$
  • B
    $(x+y+1)\left(x^2+y^2+1-x y-x-y\right)$
  • C
    $(x-1+y)\left(x^2-1-y^2+x+y+x y\right)$
  • D
    $3(x+y-1)\left(x^2+y^2-1\right)$
Answer
Correct option: A.
$(x-1+y)\left(x^2+1+y^2+x+y-x y\right)$
By using identity
$a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
We can write,
$x^3-1+y^3+3 x y$
$=\left(x^3\right)+(-1)^3+\left(y^3\right)-3(-1)(x)(y)$
$=[x+(-1)+y]\left[x^2+(-1)^2+y^2-x(-1)-y(-1)-x y\right]$
$=(x-1+y)\left(x^2+1+y^2+x+y-x y\right)$
Hence, correct option is $(a).$
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MCQ 151 Mark
The factors of $8 a^3+b^3-6 a b+1$ are:
  • A
    $(2 a+b-1)\left(4 a^2+b^2+1-3 a b-2 a\right)$
  • B
    $(2 a-b+1)\left(4 a^2+b^2-4 a b+1-2 a+b\right)$
  • ✓
    $(2 a+b+1)\left(4 a^2+b^2+1-2 a b-b-2 a\right)$
  • D
    $(2 a-1+b)\left(4 a^2+1-4 a-b-2 a b\right)$
Answer
Correct option: C.
$(2 a+b+1)\left(4 a^2+b^2+1-2 a b-b-2 a\right)$
We know the identity
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
So by using identity, we can write given expression as
$(2 a)^3+(b)^3+(1)^3-3(2 a)(b)(1)$
$=(2 a+b+1)\left[(2 a)^2+b^2+1^2-2 a \times b-b \times 1-2 a \times 1\right]$
$=(2 a+b+1)\left(4 a^2+b^2+1-2 a b-b-2 a\right)$
Hence, correct option is $(c).$
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