Free Factorisation 02 Practice Test - 8th Grade 

The given expression can be written as,
$49p^2-36=(7p{)}^2-(6{)}^2$
$\text{[Using the identity:}{a}^2-b^2=(a+b)(a-b)]$
$=(7p+6)(7p-6)$

$\text{Hence,}(7p+6)and(7p-6)$
are the factors of $49p^2-36.$ 

Comparing the coefficients of the different elements and dividing, we get:

$\frac{24}{6}=4$

$\frac{x}{1}=x$

$\frac{y^2}{y}=y$

$\frac{z^2}{z^2}=1$

Hence, we have $24x{y}^2{z}^2÷6y{z}^2=4xy$.

$14pq$ itself is a factor of $28p^{2}q$. Thus, any factor of $14pq$ will also be a factor of $28p^{2}q$.

Divide all the terms of the expression $(x^3+2x^2+3x)$ separately by $2x$. 
$\frac{x^3}{2x}=\frac{x^2}{2}$;
 $\frac{2x^2}{2x}=x$;
$\frac{3x}{2x}=\frac{3}{2}$.
Therefore, $\frac{(x^3+2x^2+3x)}{2x}=\frac{x^2}{2}+x+\frac{3}{2}=\frac{(x^2+2x+3)}{2}$

Given,

$\frac{x(x+1)(x+2)(x+3)}{x(x+1)}$
On cancelling the common terms $xand(x+1)$ , we get 
$\Rightarrow (x+2)(x+3)$

We need to regroup ax  + bx - ay - by

We get x(a+b) - y(a+b)

Taking (a+b) common from both the terms we get (a+b)(x-y)

Thus, (a+b) and (x-y) is the factors of  ax  + bx - ay - by

$2t,3t^2\text{and}4\text{can be factorized as,}$
$2t=1\times 2\times t$
$3t^2=1\times 3\times t\times t$
$4=1\times 2\times 2$ 
Hence, 1 is the only common factor among all three.

The given expression can be rewritten as,
$(2y{)}^2-2(2y\left)\right(3)+3^2.$
Comparing with the identity
$(a-b{)}^2=a^2-2ab+b^2$
we get,
$(2y{)}^2-2(2y\left)\right(3)+3^2$
$=(2y-3{)}^2$
$=(2y-3)(2y-3)$

$4y(y+3)$ can be written as

$4\times y\times (y+3)$ 

$=4y\times (y+3)=4(y+3)\times y$
$=4\times y(y+3)$ 
$=4\times (y^2+3y)$
$=4y^2+12y$ 

So, 4$y^2$ is a term in the expansion of $4y(y+3)$. It is not a factor.

$4x$($x+3$) = $4x^2$ + 3
So, $4x^2$ is a term of a given expression. $4x^2$ + 3 is not divisible by $4x^2$.
So, $4x^2$ is not a factor of $4x^2+3$.
The factors of  $4x$($x+3$) are 
$4,xand(x+3)$