Question

Factorise the following expressions.
(i) $a^2+8a+16$
(ii) $p^2-10p+25$
(iii) $25m^2+30m+9$
(iv) $49y^2+84yz+36z^2$
(v) $4x^2-8x+4$
(vi) $121b^2-88bc+16c^2$
(vii) $(l+m{)}^2-4lm$
(viii) $a^4+2a^2{b}^2+b^4$

Answer 1

(i) Given $a^2+8a+16$
Here, the middle term $8a$ split into two terms $4a$ and $4a$ such that its product $4a\times 4a=16a^2$ is same as product of last term and first term of given equation.
$=a^2+4a++4a+16$
$=a(a+4)+4(a+4)$
$=(a+4)(a+4)$

Similar way,

(ii) $p^2-10p+25$
Here the middle term $-10p=-5p-5p$
$=p^2-5p-5p+25$

$=p(p-5)-5(p-5)$

$=(p-5)(p-5)$

(iii) $25m^2+30m+9$
Here, middle term $30m=15m+15m$
$=25m^2+15m+15m+9$

$=5m(5m+3)+3(5m+3)$

$=(5m+3)(5m+3)$

(iv) $49y^2+84yz+36z^2$
Middle term $84yz=42yz+42yz$

$=49y^2+42yz+42yz+36z^2$

$=7y(7y+6z)+6z(7y+6z)$

$=(7y+6z)(7y+6z)$

(v) $4x^2-8x+4$
Middle term $-8x=-4x-4x$
$=4x^2-4x-4x+4$
$=4x(x-1)-4(x-1)$

$=(4x-4)(x-1)$

$=4(x-1)(x-1)$

(vi) $121b^2-88bc+16c^2$
Here, middle term $-88bc=-44bc-44bc$
$=121b^2-44bc-44bc+16c^2$
$=11b(11b-4c)-4c(11b-4b)$

$=(11b-4c)(11b-4c)$

(vii) $(l+m{)}^2-4lm$

$=l^2+m^2+2lm-4lm$ (Since $(a+b{)}^2=a^2+2ab+b^2$)

$=l^2+m^2-2lm$

It is the form of $a^2+b^2-2ab=(a-b{)}^2$
Replace $a$ by $l$ and $b$ by $m$, we get

$l^2+m^2-2lm=(l-m{)}^2$
$=(l-m)(l-m)$

(viii) $a^4+2a^2{b}^2+b^4$

$=(a^2{)}^2+2a^2{b}^2+(b^2{)}^2$
It is in the form of $x^2+2xy+y^2=(x+y{)}^2$
Replace $x$ by $a^2$ and $y$ by $b^2$, we get

$(a^2{)}^2+2a^2{b}^2+(b^2{)}^2=[a^2+b^2{]}^2$
=$(a^2+b^2)(a^2+b^2)$