Free Factorisation 01 Practice Test - 8th Grade 

$\text{Given, the expression is}20l^{2}m+30alm.$

The given expression can be factorised as follows :-

$20l^{2}m+30alm$
$=(2\times 2\times 5\times l\times l\times m)+(2\times 3\times 5\times a\times l\times m)$

Taking out the common factors from both the terms, we get

$=(2\times 5\times l\times m)\left[\right(2\times l)+(3\times a\left)\right]$
$=10lm(2l+3a)$

Given, the expression is
$\frac{4yz(z^2+6z-16)}{2y(z+8)}.$

First, we will factorise $(z^2+6z-16).$

Comparing the above expression with the identity $x^2+(a+b)x+ab$, we note that,$(a+b)=6andab=-16.$

Since, 
$8+(-2)=6and\left(8\right)(-2)=-16$,
the expression, $z^2+6z-16$
$=z^2+8z-2z-16$
$=z(z+8)-2(z+8)$
$=(z-2)(z+8)$ . . . (i)

By substituting (i) in the given expression, we get
$=\frac{4yz(z-2)(z+8)}{2y(z+8)}$
$=\frac{4\times y\times z\times (z-2)\times (z+8)}{2\times y\times (z+8)}$
$=2z(z-2)$

The given expression is

$x(3x^2-27)$

Taking 3 common we get

= $3x(x^2-9)$

[Using $a^2-b^2=(a+b)(a-b)$]

= $3x(x+3)(x-3)$

= $\frac{3x(x+3)(x-3)}{3(x-3)}$ 

= $x(x+3)$

= $x^2+3x$

Hence, the given statement is true

$(a+b{)}^2+(a-b{)}^2=(a^2+b^2+2ab)+(a^2+b^2-2ab)=2(a^2+b^2)$

Dividing by $a^2+b^2$ we get,

= $\frac{2(a^2+b^2)}{(a^2+b^2)}=2$

Thus, $\frac{\left(\right(a+b{)}^2+(a-b{)}^2)}{(a^2+b^2)}=2$

$2x^2+13x+15=2x^2+10x+3x+15=2x(x+5)+3(x+5)=\frac{(2x+3)(x+5)}{(x+5)}=2x+3$

Substituting $x=2$ in resulting expression gives $2\times 2+3=7$.

Given, the expression is
$4x^3+12x^2+5x+15.$

Taking $4x^2$ common from the first two-terms and 5 common from the last two-terms, we get
$4x^2(x+3)+5(x+3)$
$=(x+3)(4x^2+5)$

$\text{Thus,}(x+3)\text{and}(4x^2+5)$
are two factors of the given expression.

$a^2-2ab+b^2-c^2=(a-b{)}^2–c^2$
Using identity $x^2-y^2=(x+y)(x-y)$  we get

$(a-b{)}^2–c^2$ = $\left[\right(a-b)+c]\left[\right(a-b)-c]$​

Hence, factors of $a^2-2ab+b^2-c^2$ are $a-b+c$ and $a-b-c$

$\text{Given:}\frac{a^2+7a+10}{(a+5)}...\left(i\right)$

Comparing $a^2+7a+10$ with the identity $x^2+(a+b)x+ab,$
$\text{we note that,}(a+b)=7andab=10$
$So,5+2=7and\left(5\right)\left(2\right)=10$

Hence,
$a^2+7a+10$
$=a^2+5a+2a+10$
$=a(a+5)+2(a+5)$
$=(a+2)(a+5)$

From (i), we get
$\frac{a^2+7a+10}{(a+5)}=\frac{(a+2)(a+5)}{(a+5)}$
$=(a+2)$

Given, the expression is
$\frac{m^2+7m-60}{(m-5)}.$
Now, comparing the numerator with the identity $x^2+(a+b)x+ab$
We note that,
$(a+b)=7andab=-60.$
So,
$12+(-5)=7and\left(12\right)(-5)=-60$.
Hence,
$m^2+7m-60=m^2+12m–5m–60=m(m+12)–5(m+12)=(m-5)(m+12)$
Therefore,
$\frac{m^2+7m-60}{(m-5)}=\frac{(m-5)(m+12)}{(m-5)}=(m+12)$ 

$\text{Given, the expression is}25+70x+49x^2.$

The given expression can be rewritten as,
$49x^2+70x+25$

$\text{Using the identity:}(a+b{)}^2=a^2+2ab+b^2,\text{we get}$

$49x^2+70x+25=(7x{)}^2+2(5\left)\right(7x)+5^2$
$=(7x+5{)}^2$​
$=(7x+5)(7x+5)$

$\therefore 25+70x+49x^2=(7x+5)(7x+5)$