What is the sum of the squares of the coefficients of $4(x^4 + 3x^2 + 1)$?

michaelcai Sep 11, 2017

#1**+2 **

Best Answer

4(x^{4} + 3x^{2} + 1) = **4**x^{4} + **12**x^{2} + 4

As best I can tell, the "4" at the end is not considered a coefficient.

So....the coefficients are just 4 and 12 .

The sum of the squares of 4 and 12 = 4^{2} + 12^{2} = 16 + 144 = 160

hectictar Sep 11, 2017

#2**+1 **

yeah, I got 160 as well, but my answer checking thing says its incorrect...

michaelcai
Sep 11, 2017

#3**+1 **

Hmmm...maybe the 4 at the end* is *considered a coefficient. It's a coefficient of x^{0} .

If that's the case... then the answer = 4^{2} + 12^{2} + 4^{2} = 16 + 144 + 16 = 176

Does it say that's the right answer? I'm curious now....

hectictar
Sep 12, 2017

#4**+1 **

No, 4 is **NOT **a coefficient. 4, in this biquadratic expression (fancy name for a quartic expression that excludes terms with an odd degree and is written in the form \(ax^4+cx^2+e\)) is a constant.

Remember that coefficients are numbers that multiply a variable. There is no variable that alongside the number 4, so it is a plain and boring constant.

TheXSquaredFactor
Sep 13, 2017