Math help needed someone please help me , here is the question that I am having issues with 4xy+16-x^2-4y^?

Question

your help is greatly appreciated can you please explain the steps?

ooooooooooops
4xy+16-x^2-4y^2

Answer 1

I assume you are asking to factorise
4xy+16-x^2-4y^2
=16-[x^2+(2y)^2-2(x)(2y)]
=16-(x-2y)^2 = 4^2-(x-2y)^2
=[4-(x-2y)][4+(x-2y)]
=(4-x+2y)(4+x-2y) (Ans)

Answer 2

Look for a perfect square trinomial: a^2 - 2ab + b^2 = (a - b)^2

4xy + 16 - x^2 - 4y^2
= -x^2 + 4xy - 4y^2 + 16
= -1(x^2 - 4xy + 4y^2) + 16
= -1[ x^2 - 2(x)(2y) + (2y)^2 ] + 16
= -1(x - 2y)^2 + 16

Look for the difference of squares: a^2 - b^2 = (a + b)(a - b)

= 16 - (x - 2y)^2
= 4^2 - (x - 2y)^2
= (4 + (x - 2y))(4 - (x - 2y))
= (4 + x - 2y)(4 - x + 2y)

ANSWER: The factored form of 4xy + 16 - x^2 - 4y^2 is
(4 + x - 2y)(4 - x + 2y).

Answer 3

4xy + 16 - x^2 - 4y^2 =
16 - x^2 + 4xy - 4y^2 =
16 - (x^2 - 4xy + 4y^2) =
4^2 - (x - 2y)^2 =
(4 - (x - 2y)(4 + (x - 2y) =
(4 - x + 2y)(4 + x - 2y)

Answer 4

take 4xy+16-x^2-4y^2=0..

so taking -1 common the eqn becomes
x^2+4y^2-4xy-16=0.

we know that (a-b)^2=(a^2+b^2-2ab)

so your prob becomes
(x-2y)^2 -16=0

again (a+b)(a-b)=a^2-b^2

so (x-2y)^2-(4^2)= (x-2y+4)(x-2y-4).
thats your answer

Answer 5

what power is y to?