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

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146
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+518

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

michaelcai  Sep 11, 2017

#1
+5905
+2

4(x4 + 3x2 + 1)  =  4x4 + 12x2 + 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  =  42 + 122  =  16 + 144  =  160

hectictar  Sep 11, 2017
Sort:

#1
+5905
+2

4(x4 + 3x2 + 1)  =  4x4 + 12x2 + 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  =  42 + 122  =  16 + 144  =  160

hectictar  Sep 11, 2017
#2
+518
+1

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

michaelcai  Sep 11, 2017
#3
+5905
+1

Hmmm...maybe the 4 at the end is considered a coefficient. It's a coefficient of x0 .

If that's the case... then the answer        =  42 + 122 + 42  =  16 + 144 + 16  =  176

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

hectictar  Sep 12, 2017
#4
+1601
+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

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