Answer:
Solution given;
[tex] y^{4}+5y²+9[/tex]
keeping [tex]y^{4} and 9 together [/tex]
[tex]y^{4}+9+5y²[/tex]
[tex](y²)²+3²+5y²[/tex].....[I]
we have
a²+b²=(a+b)²-2ab
or
a²+b²=(a-b)²+2ab
same like that
[tex](y²)²+3²=(y²+3)²-6y² or (y²-3)²+6y²[/tex]
remember that while adding or subtracting the left term 5y² either adding 6y²or subtracting 6y²
should make the term perfect square
while subtracting it makes perfect square
so
we take
(y²)²+3²=(y²+3)²-6y²
again
substituting value of
(y²)²+3² in equation 1 and it becomes
(y²+3)²-6y²+5y²
solve like terms
(y²+3)²-y²
again
we have
a²-b²=(a+b)(a-b)
by using this.
(y²+3+y)(y²+3-y)
rearrange it
(y²+y+3)(y²-y+3) is a required factorisation form.
