Question

Divide the following polynomials (a4 + 4b4) ÷ (a2 - 2ab + 2b2)

Answer 1

(a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2+2ab+2b^2 =The answer

(a + b)^2 = a^2 + 2ab + b^2 => square of sums
(a - b)^2 = a^2 - 2ab + b^2 => square of deference
and of course one of most important ones:
a^2 - b^2 = (a - b)(a + b) => difference of squares
Best Answer: (a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2 + 2ab + 2b^2
a^4 + 4b^4 => i.e. 4a^2b^2 ,
a^4 + 4a^2b^2 + 4b^4 => a^2 + 2ab + b^2 = (a + b)^2, if : a = a^2 , b = 2b^2:
(a^2 + 2b^2)^2 = a^4 + 4a^2b^2 + 4b^4 => We can't add or subtract the value to the expression.
a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 =>
(a^2 + 2b^2)^2 - 4a^2b^2 =>
(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab) =>
(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)

Greetings!

Answer 2

Answer:

(a^2+2ab+2b^2)

Step-by-step explanation:

(a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)

To factor a^4 - 4b^4 we use square form

$(a^2+2b^2)^2 = (a^2)^2 + 2(a^2)(2b^2) + (2b^2)^2= a^4 + 4a^2b^2 + 4b^4$

$(a^2+2b^2)^2 = a^4 + 4a^2b^2 + 4b^4$

subtract 4a^2 b^2 on both sides

$(a^2+2b^2)^2 - 4a^2b^2= a^4+ 4b^4$

$(a^2+2b^2)^2 - (2ab)^2= a^4+ 4b^4$ ---------> equation we got

Now apply difference of square formula

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

$(a^2+2b^2)^2 - (2ab)^2= (a^2+2b^2+2ab)(a^2+2b^2-2ab)$

Replace the factors in the equation we got

$(a^2+2b^2+2ab)(a^2+2b^2-2ab)= a^4+ 4b^4$

Now replace it in our original equation

$(a^4 + 4b^4) divide (a^2 - 2ab + 2b^2)$

$\frac{(a^2+2ab+2b^2)(a^2-2ab+2b^2)}{(a^2-2ab+2b^2)}$

Cancel out same factors

So answer is $(a^2+2ab+2b^2)$