How to factor this?????????

4(16a+4ab+b)

16a^4 + 4a^2b^2 + b^4

= 16a^4 + 8a^2b^2 + b^4 - 4a^2b^2

= 16a^4 + 4a^2b^2 + 4a^2b^2 + b^4 - 4a^2b^2

= (16a^4 + 4a^2b^2) + (4a^2b^2 + b^4) - (2ab)^2

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

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

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

= [(4a^2 + b^2) + 2ab][(4a^2 + b^2) - 2ab]

= [4a^2 + 2ab + b^2][4a^2 - 2ab + b^2]

If that's really the question; I don't think it can be solved.

If you find the d; d = b^2-4ac. If we assume that b is known; then the unknown is 'a'. Thus d = (4b^2)^2 - 4 (16) (b^4) = 16b^4 - 64b^4= -48b^4. And since it's negative, it will be imaginary number; thus it can't be factored.

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If the question is 16a^4 + 8a^2b^2 + b^4; then it can be solved by the following way:
The easiest way is by guessing.

We can see that in 16a^4 + 8a^2b^2 + b^4 there's only powers of even multiply (2 and 4). And there's only 2 unknown (a and b); so, the one that has the largest power is 16a^4 and b^4. Both of them can be easily factored.
By guessing : you will got that 16 = 4*4 ; the 8 = 4+4 ; and the 1 = 1*1.
So, it will be
16a^4 + 8a^2b^2 + b^4 = (4a^2+b^2)*(4a^2+b^2) = (4a^2+b^2)^2
just by guessing. Since the pattern is still easy.


Here's the LONG explanation

If you see the pattern that it has a^4, a^2b^2, and b^4; then you would know that there will be a^2 and b^2 for the factors because there's no odd power (there's only power of 4 and 2). So, it will be
(ma^2+nb^2)(xa^2+yb^2) where m,n,x, and y are real numbers.

From here, you can see that the calculation will form :
(mx)a^4 + (my+nx) a^2b^2 + (ny)b^4.
Thus, this means that
16a^4 + 8a^2b^2 + b^4 = (mx)a^4 + (my+nx) a^2b^2 + (ny)b^4

So, 16 = mx || 4 = my + nx || 1 = ny.

From here, you will find that m=x=4 and n=y=1. It can be guessed easier here.
Thus, the factor for
16a^4 + 8a^2b^2 + b^4 = (4a^2+b^2)*(4a^2+b^2) = (4a^2+b^2)^2

I take it to mean: 16a^4 + (4a^2)(b^2) + b^4.

18a^4 + 4b^4

[4*(a^2) + b^2]^2

16a^4 + 4a^2b^2 + b^4
=(4a^2+b^2)(4a^2+b^2)
http://www.mathway.com/answer.aspx?p=calg?p=16aSMB074SMB15+SMB154aSMB072bSMB072SMB15+SMB15bSMB074?p=213?p=?p=?p=?p=?p=?p=?p=0?p=?p=

(4a² + 2ab + b²)(4a² − 2ab + b²)

There is really no obvious pattern, but if you multiply it out, lots of things cancel and you get the desired result despite the unexpected negative term (actually **because** of the negative term.) Don't confuse this with the difference of two squares since 4a² + b² is not a perfect square.
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Note: (4a² + b²)² = 16a^4 + 8a²b² + b²