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y4+5y2+9 factorise please help me​

Y45y29 Factorise Please Help Me class=

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[tex]\displaystyle\ y^4 +5y^2+9=(y^2+3)^2-y^2=(y^2-y+3)(y^2+y+3)[/tex]

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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.

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