Y4+5y2+9 factorise please help me​

Solution given;

keeping

.....[I]

we have

a²+b²=(a+b)²-2ab

or

a²+b²=(a-b)²+2ab

same like that

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.

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.