Question

Y4+5y2+9 factorise please help me​

Answer 1

`\displaystyle\ y^4 +5y^2+9=(y^2+3)^2-y^2=(y^2-y+3)(y^2+y+3)`

Answer 2

Solution given;

`y^(4)+5y²+9`

keeping
`y^(4) and 9 together`

`y^(4)+9+5y²`

`(y²)²+3²+5y²`.....[I]

we have

a²+b²=(a+b)²-2ab

or

a²+b²=(a-b)²+2ab

same like that

`(y²)²+3²=(y²+3)²-6y² or (y²-3)²+6y²`

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.