2 Marks Questions - Page 2 - MATHS STD 8 Questions - Vidyadip

Questions · Page 2 of 2

2 Marks Questions

Question 512 Marks

Find the common factors of the given term: $3 x^2 y^3, 10 x^3 y^2, 6 x^2 y^2 z$

Answer

$3 x^2 y^3=$ $3\times x\times x\times y\times y\times y$
$10x^3y^2=$ $2\times 5\times x\times x\times x\times y\times y$
$6x^2y^2z =$ $2\times 3\times x\times x\times y\times y\times z$
Hence the common factors are $x, x, y, y$
and $x\times x\times y\times y$ $= x^2y^2$

View full question & answer→

Question 522 Marks

Find the common factors of the given term: $10pq, 20qr, 30rp$

Answer

$10pq$ $ = \underline 2 \times \underline{\underline 5} \times p \times q$
$20qr$ $ = \underline 2 \times 2 \times \underline{\underline 5} \times q \times r$
$30rp$ $ = \underline 2 \times 3 \times \underline{\underline 5} \times r \times p$
Common prime factors are $2$ and $5$
$\therefore H.C.F. = 2 \times 5 = 10$

View full question & answer→

Question 532 Marks

Find the common factors of the given term: $16 x^{3},-4 x^{2}, 32 x$

Answer

The given terms $16 x^{3},-4 x^{2}, 32 x$ can be written as:
$16x^3=$ $2 \times 2 \times 2 \times 2 \times x \times x \times x$
$-4x^2=$ $-1 \times 2 \times 2 \times x \times x$
$32x =$ $2 \times 2 \times 2 \times 2 \times 2 \times x$
The common factors are $2, 2$ and $x$ = $2 \times 2 \times x$ = 4x

View full question & answer→

Question 542 Marks

Find the common factors of the given term: $6abc, 24ab^2, 12a^2b$

Answer

6abc $= 2 × 3 × a × b × c$
$24ab^2 = 2 × 2 × 2 × 3 × a × b × b$
$12a^2b = 2 × 2 × 3 × a × a × b$
Common prime factors are $2, 3, a$ and $b$
$\therefore H.C.F. = 2 \times 3 \times a \times b$
$= 6ab$

View full question & answer→

Question 552 Marks

Find the common factors of the given term: $2x, 3x^2, 4$

Answer

$2x, 3x^2, 4$
$2x = \underline 1 \times 2 \times \underline{\underline x} $
$3{x^2} = \underline 1 \times 3 \times \underline{\underline x} \times x$
$4 = 1 \times 2 \times 2$
Common factors are $1$ and $x$.
$\therefore H.C.F. = 1 \times x = x$

View full question & answer→

Question 562 Marks

Find the common factors of the given term: $14pq, 28p^2q^2$

Answer

$14pq, 28p^2q^2$
14pq $ = \underline 2 \times \underline{\underline 7} \times \mathop p\limits_o \times \mathop q\limits_w $
$28p^2q^2$ $ = \underline 2 \times 2 \times \underline{\underline 7} \times \mathop p\limits_o \times p \times \mathop q\limits_w \times q$
Common prime factors are $2, 7, p$ and $q$.
$\therefore H.C.F. = 2 \times 7 \times p \times q = 14pq$

View full question & answer→

Question 572 Marks

Find the common factors of the given term: $2y, 22xy$

Answer

$2y, 22xy$
$2y = 2 \times y$
$22 = 2 \times 11 \times x \times y$
Common prime factors are $2$ and $y$.
$\therefore H.C.F. = 2 \times y = 2y$

View full question & answer→

Question 582 Marks

Find the common factors of the given term: $12x, 36$

Answer

$12x, 36$
$12x = 2 \times 2 \times 3 \times x$
$36 = 2 \times 2 \times 3 \times 3$
Common prime factors are $2$ (occurs twice) and $3$.
$\therefore H.C.F. = 2 \times 2 \times 3 = 12$

View full question & answer→

Question 592 Marks

Factorise the expression: $x^2+ 5x + 6$

Answer

We have $x^2+ 5x + 6,$
$= x^2+ 3x + 2x + 6 = x(x + 3) + 2(x + 3)$
$= (x + 3)(x + 2)$

View full question & answer→

Question 602 Marks

Factorise: $a^2– 2ab + b^2– c^2$

Answer

$a^2– 2ab + b^2– c^2= (a – b)^2– c^2[$ Using identity, $(x – y)^2= x^2– 2xy + y^2]$
$= {(a – b) – c)}{(a – b) + c}] [$Using identity, $x^2- y^2= (x - y)(x + y)]$
$= (a – b – c) (a – b + c)$
This is the required factorisation.

View full question & answer→

Question 612 Marks

Factorise: $49p^2- 36$

Answer

$49p^2– 36 = (7p)^2- (6)^2$
Now using identity, $a^2- b^2= (a - b)(a + b),$
$= (7p - 6)(7p + 6)$
This is the required factorisation.

View full question & answer→

Question 622 Marks

Factorise: $4y^2– 12y + 9$, using the identity $a^2- 2ab + b^2= (a - b)^2$

Answer

Using identity, $a^2-2 a b+b^2=(a-b)^2$
$4 y^2-12 y+9$
$=(2 y)^2-2 \times 3 \times(2 y)+(3)^2$
$=(2 y-3)^2$
= (2y – 3)(2y – 3)
This is the required factorisation.

View full question & answer→

Question 632 Marks

Factorise: $x^2+8 x+16$

Answer

Using identity, $a^2+2 a b+b^2=(a+b)^2$
$x^2+8 x+16$
$=x^2+2(x)(4)+4^2 $
$ =(x+4)^2$
= (x + 4)(x + 4)
This is the required factorisation.

View full question & answer→

Question 642 Marks

Factorise: $6xy - 4y + 6 - 9x$

Answer

$6xy – 4y + 6 - 9x$
$= 6xy - 4y - 9x + 6$
$= 2y(3x - 2) - 3(3x - 2)$
$= (3x - 2)(2y - 3)$

View full question & answer→

Question 652 Marks

Find the division:$ 7x^2y^2z^2\div 14xyz$

Answer

$7x^2y^2z^2\div 14xyz$ = $\frac {7\times x\times x\times y\times y\times z\times z}{2\times 7\times x\times y\times z}$
= $\frac{x\times y\times z} {2}$= $\frac {1}{2} xyz$

View full question & answer→

Question 662 Marks

Solve the division: $-20x^4$ $\div$ $10x62$

Answer

Now, $-20 \mathrm{x}^4=-2 \times 2 \times 5 \times \mathrm{x} \times \mathrm{x} \times \mathrm{x} \times \mathrm{x}$
and $10 \mathrm{x}^2=2 \times 5 \times \mathrm{x} \times \mathrm{x}$
Therefore, $(–20x^4)$ $\div$ $10x^2=$ $\frac{-2\times2\times5\times x\times x\times x\times x}{2\times5 \times x\times x}$
$ = -2$ $\times$ x $\times$ $x = -2x^2$

View full question & answer→

Question 672 Marks

Obtain the factors of $z^2- 4z – 12.$

Answer

$z^2– 4z –12 = z^2– 6z + 2z –12$
$= z(z – 6) + 2(z – 6 )$
$= (z – 6)(z + 2)$
Thus, the factors of $z^2– 4z – 12$ are $(z – 6)$ and $(z + 2)$

View full question & answer→

Question 682 Marks

Find the factors of $y^2- 7y + 12.$

Answer

$y^2– 7y + 12 = y^2– 3y – 4y + 12$
$= y(y –3) – 4(y – 3) = (y – 3)(y – 4)$
Thus, the factors are $(y – 3)$ and $(y – 4)$.

View full question & answer→

Generate Paper Free All Factorization topics