Free Factorisation 03 Practice Test - 8th Grade 

Given, the expression is
$3m^2+9m+6.$
Taking 3 common from the above expression, we get
$3(m^2+3m+2)...\left(i\right)$

Now comparing $m^2+3m+2$ with the identity $x^2+(a+b)x+ab.$
$\text{We note that,}$
$(a+b)=3andab=2.$
$So,$
$2+1=3and\left(2\right)\left(1\right)=2$

Hence,
$m^2+3m+2$
$=m^2+2m+m+2$
$=m(m+2)+1(m+2)$
$=(m+2)(m+1)$

Now from (i), we get
$\Rightarrow 3m^2+9m+6=3(m^2+3m+2)=3(m+2)(m+1)$
Therefore,
3, (m+2) and (m+1) are the 3 irreducible factors of the given expression.

The given expression is

$x^2-2x-15$

We have to split the middle term in such a way the sum is -2 and product is -15 , which can be done in following way

$=x^2-5x+3x-15=x(x-5)+3(x-5)$
$=(x+3)(x-5)$

So, $x+3$ is one of the factors of the given expression.

Since 14 is HCF of 4p and 21q, then both of them should have 7 and 2 as prime factors.
4p = 2 x 2 x p 
21q = 3 x 7 x q
Since 14 is the HCF, p should be 7 and q should be 2.

Given, the expression is
$5pq(p^2-q^2).$ 

$\text{[Using the identity:}{a}^2-b^2=(a+b)(a-b)]$

$\Rightarrow 5pq(p^2-q^2)=5pq(p+q)(p-q)$

Now dividing the above expression by
$2p(p+q),\text{we get}$

$\frac{5pq(p+q)(p-q)}{2p(p+q)}$

$=\frac{5q(p-q)}{2}$

The given expression is
$(2x{)}^2+5x$

Substituting $x$ = 3 in the above expression we get

$(2\times 3{)}^2+5\times 3$
$6^2+5\times 3$
36 +15  = 51

Hence, the given statemnet is true

$\frac{3x^2+1}{3x^2}$

$=\frac{3x^2}{3x^2}+\frac{1}{3x^2}$

$=1+\frac{1}{3x^2}$

Hence, the given statement is false.

$5(x+2y)=5x+5\times 2y=5x+10y$

$\frac{72}{8}=9\frac{y^4}{y^2}=y^2\frac{z^6}{z^3}=z^3$

Therefore, $\frac{72x^2{y}^4{z}^6}{8x^2{y}^2{z}^3}=9y^2{z}^3$

Substituting $x=-3$ in $x^2-5x$ gives 
$(-3{)}^2-5(-3)=9+15=24$
 

The given algebraic equation is $x^2-6x-16$.
We can factorize this using the following identity using,
$(x+a)(x+b)=x^2+(a+b)x+ab$
So, we have to first find the factors of 16.
$8\times 2=4\times 4=16$
We can see that, $8-2=6$.
So, we split the expression as shown
$x^2+2x-8x-16$
$=x(x+2)-8(x+2)$
$=(x+2)(x-8)$
Hence $x-8$ is a factor of the given expression.