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MATHS 5 Marks Questions

Factorization Of Polynomials · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 15 questions.

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MCQ 15 Marks
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 - 3AB(A - B)$
$\Rightarrow [(x + y) - (x - y)]^3$
 $= (x + y)^3 - (x - y)^3 - 3(x + y)(x - y)[(x + y) - (x - y)]$
$= (x + y)^3 - (x - y)^3 - 3(x^2 - y^2)(2y)$
$= (x + y)^3 - (x - y)^3 - 6y(x^2 - y^2)$
But$, (x + y)^3 - (x - y)^3 - 6y(x^2 - y^2) = ky^3$
$\Rightarrow [(x + y) - (x - y)]^3 = (2y)^3 = k8y^3$
$\Rightarrow (2y)^3 = ky^3$
$\Rightarrow 8y^3 = ky^3$
$\Rightarrow k = 8$
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MCQ 25 Marks
The expression $(a - b)^3 + (b - c)^3 + (c - a)^3$ can be factorized as:
  • A
    $(a - b)(b - c)(c - a)$
  • B
    $3(a - b)(b - c)(c - a)$
  • ✓
    $-3(a - b)(b - c)(c - a)$
  • D
    $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
Answer
Correct option: C.
$-3(a - b)(b - c)(c - a)$
By we know that $a^3 + b^3 + c^3 - 3abc$
$ = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0$, then
$a^3 + b^3 + c^3 = 3abc$
In given expression,
Let $a - b = A, b - c = B, c - a = C$
Now, $a - b + b - c + c - a = 0$
$i.e. A + B + C = 0$
$\Rightarrow A^3 + B^3 + C^3 = 3ABC$
$\Rightarrow (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a)$
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MCQ 35 Marks
The factors of$ x^3 - 7x + 6$ are:
  • A
    $x(x - 6)(x - 1)$
  • B
    $(x^2 - 6)(x - 1)$
  • C
    $(x + 1)(x + 2)(x + 3)$
  • ✓
    $(x - 1)(x + 3)(x - 2)$
Answer
Correct option: D.
$(x - 1)(x + 3)(x - 2)$
$x^3 - 7x + 6 = x^3 - 7x + 6 + 1 - 1 ($by adding $+1 -1$ to $\text{R.H.S})$
$= x^3 - 7x + 7 - 1$
$= (x^3 - 1) - 7(x - 1)$
Now by identity $a^3 - b^3 = (a - b)(a^2 + b^2 + ab),$ we get
$x^3 - 7x + 6 = (x^3 - 1) - 7(x - 1)$
$= (x - 1)(x^2 + x + 1) - 7(x - 1)$
$= (x - 1)(x^2 + x + 1 - 7)$
$= (x - 1)(x^2 + x - 6)$
$= (x - 1)(x + 3)(x - 2)$
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MCQ 45 Marks
The expression $x^4 + 4$ can be factorized as:
  • ✓
    $(x^2 + 2x + 2)(x^2 - 2x + 2)$
  • B
    $(x^2 + 2x + 2)(x^2 + 2x - 2)$
  • C
    $(x^2 - 2x - 2)(x^2{ }- 2x + 2)$
  • D
    $(x^2 + 2)(x^2 - 2)$
Answer
Correct option: A.
$(x^2 + 2x + 2)(x^2 - 2x + 2)$
$x^4 + 4$
$= x^4 + 4 + 4x^2 - 4x^2$
$= (x^4 + 4x^2 + 4) - 4x^2$
$= (x^2 + 2)^2 - (2x)^2$
$= (x^2 + 2 - 2x)(x^2 + 2 + 2x)$
$= (x^2 + 2x + 2)(x^2 - 2x + 2)$
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MCQ 55 Marks
The factors of $x^2 + 4y^2 + 4y - 4xy - 2x - 8,$ are:
  • ✓
    $(x - 2y - 4)(x - 2y + 2)$
  • B
    $(x - y + 2)(x - 4y - 4)$
  • C
    $(x + 2y - 4)(x + 2y + 2)$
  • D
    None of these.
Answer
Correct option: A.
$(x - 2y - 4)(x - 2y + 2)$
$x^2 + 4y^2 + 4y - 4xy - 2x - 8$
$= x^2 +(2y)^2 - 2 \times x(2y) + 4y - 2x - 8$
$= (x - 2y)^2 + 4y - 2x - 8 ...(1)$
Now making $eq(1)$ a perfect square by adding $1$ and $-1$
$(x - 2y)^2 + 4y - 2x - 8 = (x - 2y)^2 + 4y - 2x - 8 + 1 - 1$
$= (x - 2y)^2 + (1)^2 - 2 \times (1) \times (x - 2y) - 9$
$= (x - 2y - 1)^2 - (3)^2$
$= [(x - 2y - 1) - 3][x - 2y - 1 + 3]$
$= (x - 2y - 4)(x - 2y + 2)$
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MCQ 65 Marks
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
$\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2}$
$=\frac{\{(0.013)+(0.007)\}(0.013)^2-(0.013)(0.007)+(0.007)^2}{(0.013)^2-(0.013)(0.007)+(0.007)^2}$
$=0.013+0.007$
$=0.020$
$=0.02$
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MCQ 75 Marks
The factors of $a^2 - 1 - 2x - x^2$ are:
  • A
    $(a - x + 1)(a - x - 1)$
  • B
    $(a + x - 1)(a - x + 1)$
  • C
    $(a + x +1)(a - x + 1)$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
$a^2 - 1 - 2x - x^2$
$= a^2 - (1 + 2x + x^2)$
$= a^2 - (1 + x)^2$
$= [a - (1 + x)][a + (1 + x)]$
$= (a - x - 1)(a + x + 1)$
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Question 85 Marks
The value of $\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09},$ is:
  1. 2
  2. 3
  3. 2.327
  4. 2.273
Answer
  1. 2
Solution:
$\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09}$
$=\frac{(2.3)^3-(0.3)^3}{(2.3)^2+(0.3)^3+(2.3)(0.3)}$
$=\frac{(2.3 - 0.3)\{(2.3)^2+(0.3)^2+(2.3)(0.3)\}}{((2.3)^2+(0.3)^2+(2.3)(0.3))}$
$=2.3 - 0.3$
$=2$
Hence, correct option is (a).
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MCQ 95 Marks
The factors of $x^4 + x^2 + 25$ are:
  • ✓
    $(x^2 + 3x + 5)(x^2 - 3x + 5)$
  • B
    $(x^2{ }+ 3x + 5)(x^2 + 3x − 5)$
  • C
    $(x^2 + x +5)(x^2 - x + 5)$
  • D
    None of these.
Answer
Correct option: A.
$(x^2 + 3x + 5)(x^2 - 3x + 5)$
For making perfect square to $x^4 + x^2 + 25$
We add $+10x^2$ and -$10x^2$ to it.
$= x^4 + x^2 + 25$
$= x^4 + x^2 + 25 + 10x^2 - 10x^2$
$= [x^4 + 10x^2 + 25] - 9x^2$
$= (x^2 + 5)^2 + (3x)^2$
$= [(x^2 + 5) + 3x][(x^2 + 5) - 3x]$
$= (x^2 + 3x + 5)(x^2 - 3x + 5)$
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MCQ 105 Marks
$(x + y)^3 - (x - y)^3$ can be factorized as:
  • ✓
    $2y(3x^2 + y^2)$
  • B
    $2x(3x^2 + y^2)$
  • C
    $2y(3y^2 + x^2)$
  • D
    $2x(x^2+ 3y^2)$
Answer
Correct option: A.
$2y(3x^2 + y^2)$
We know the identity
$a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$
Let $x + y = a$ and $x - y = b$
Then,
$a^3 - b^3$
$= (x + y)^3 - (x - y)^3$
$= [(x + y) - (x - y)][(x + y)^2 + (x - y)^2 + (x + y)(x - y)]$
$= 2y[x^2 + y^2 + 2xy + x^2 + y^2 - 2xy + x^2 - y^2]$
$= 2y(3x^2 + y^2)$
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MCQ 115 Marks
If $3x = a + b + c,$ 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.
$0$
$3x = a + b + c$
$\Rightarrow a + b + c - 3x = 0$
$\Rightarrow 3x - (a + b + c) = 0$
$\Rightarrow (x - a) + (x - b) + (x - c) = 0 ...(1)$
Using identity if $a + b + c = 0$ then, $a^3 + b^3 + c^3 - 3abc = 0$
If we take $x - a = A, x - b = B, x - c = C$ in equation $(1),$ we get
$A + B + C = 0$
$\Rightarrow A^3 + B^3 + C^3 - 3ABC= 0$
$\Rightarrow (x - a)^3 + (x - b)^3 + (x - c)^3 - 3(x - a) (x - b) (x - c) = 0$
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MCQ 125 Marks
The factors of $x^3 - 1 + y^3 + 3xy$ are:
  • ✓
    $(x - 1 + y)(x^2 + 1 + y^2 + x + y - xy)$
  • B
    $(x + y + 1)(x^2 + y^2 + 1 - xy - x - y)$
  • C
    $(x - 1 + y)(x^2 - 1 - y^2 + x + y + xy)$
  • D
    $3(x + y - 1)(x^2 + y^2 - 1)$
Answer
Correct option: A.
$(x - 1 + y)(x^2 + 1 + y^2 + x + y - xy)$
By using identity
$a^3 + b^3 + c^3 - 3abc$
$ = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
We can write,
$x^3 - 1 + y^3 + 3xy$
$= (x^3) + (-1)^3 + (y^3) - 3(-1)(x)(y)$
$= [x + (-1) + y][x^2 + (-1)^2 + y^2 - x(-1) - y(-1) - xy]$
$= (x - 1 + y)(x^2 + 1 + y^2+ x + y - xy)$
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MCQ 135 Marks
The factors of $8a^3 + b^3 - 6ab + 1$ are:
  • A
    $(2a + b - 1)(4a^2 + b^2 + 1 - 3ab - 2a)$
  • B
    $(2a - b + 1)(4a^2 + b^2 - 4ab + 1 - 2a + b)$
  • ✓
    $(2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$
  • D
    $(2a - 1 + b)(4a^2 + 1 - 4a - b - 2ab)$
Answer
Correct option: C.
$(2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$
We know the identity
$a^3 + b^3 + c^3 - 3abc$
$ = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
So by using identity, we can write given expression as
$(2a)^3 + (b)^3 + (1)^3 - 3(2a)(b)(1)$
$= (2a + b + 1)[(2a)^2 + b^2 + 1^2 -2a \times b - b \times 1 - 2a \times 1]$
$= (2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$
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MCQ 145 Marks
If $x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c),$ then $a + b + c =$
  • A
    $4$
  • B
    $12$
  • ✓
    $-10$
  • D
    $3$
Answer
Correct option: C.
$-10$
The given equation is
x$^3 - 3x^2 + 3x - 7$
$ = (x + 1)(ax^2 + bx + c)$
This can be written as
$x^3 - 3x^2 + 3x - 7$
$ = (x + 1)(ax^2 + bx + c)$
$= x^3 - 3x^2 + 3x - 7 $
$= ax^3 + bx^2 + cx + ax^2 + bx + c$
$= x^3 - 3x^2 + 3x - 7 $
$= ax^3 + (a + b)x^2 + (b + c)x + c$
Comparing the cofficients on both sides of the equation.
We get,
$a = 1 ...(1)$
$a + b = 3 ...(2)$
$b + c = 3 ...(3)$
$c = -7 ...(4)$
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,
$a + b + c = 1 - 4 - 7 = -10$
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MCQ 155 Marks
The factors of $x^3 - x^2y - xy^2{ }+ y^3$ are:
  • A
    $(x + y)(x^2 - xy + y^2)$
  • B
    $(x + y)(x^2 + xy + y^2)$
  • C
    $(x + y)^2(x - y)$
  • ✓
    $(x - y)^2(x + y)$
Answer
Correct option: D.
$(x - y)^2(x + y)$
$x^3 - x^2y - xy^2 + y^3$
 $= x^3 + y^3 - xy(x + y)$
Now by identity$  x^3 + y^3 = (x + y)(x^2 + y^2 - xy),$ we have
$x^3 - x^2y - xy^2 + y^3$
$ = (x + y)(x^2 + y^2 - xy) - xy(x + y)$
$= (x + y)(x^2 + y^2 - xy - xy)$
$= (x + y)(x^2 + y^2 - 2xy)$
$= (x + y)(x - y)^2$
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