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# Factorization!

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Factor \(-16x^4+x^2+2x+1\) into two quadratic polynomials with integer coefficients. Submit your answer in the form \((ax^2+bx+c)(dx^2+ex+f)\), with  a>d.

Dec 29, 2018
edited by mathtoo  Dec 29, 2018

#1
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Factor the following:
-16 x^4 + x^2 + 2 x + 1

Factor -1 out of -16 x^4 + x^2 + 2 x + 1:
-(16 x^4 - x^2 - 2 x - 1)

The factors of 1 - 16 x^4 that sum to -2 are x (-4 x^2 - 1) and x (4 x^2 - 1).
So, (16 x^4 - 1) - x^2 - 2 x = -x^2 + x (-4 x^2 - 1) + x (4 x^2 - 1) + (4 x^2 - 1) (4 x^2 + 1):
-(-x^2 + x (-4 x^2 - 1) + x (4 x^2 - 1) + (4 x^2 - 1) (4 x^2 + 1))

-x^2 + x (-4 x^2 - 1) + x (4 x^2 - 1) + (4 x^2 - 1) (4 x^2 + 1) = x ((4 x^2 - 1) - x) + (4 x^2 + 1) ((4 x^2 - 1) - x):
-x (4 x^2 - x - 1) + (4 x^2 + 1) (4 x^2 - x - 1)

Factor 4 x^2 - x - 1 out of x (4 x^2 - x - 1) + (4 x^2 + 1) (4 x^2 - x - 1), resulting in (4 x^2 - x - 1) ((4 x^2 + 1) + x):

-(4 x^2 - x - 1) (4 x^2 + x + 1)

Dec 29, 2018
#2
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That is correct! Now, try to solve it when a>d.

mathtoo  Dec 29, 2018
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If you apply the negative sign to the parentheses, -4x^2 is less than 4x^2, right?

So a

CoolStuffYT  Dec 29, 2018
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How about saying thank you Mathtoo ... Melody  Dec 29, 2018
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Thank you!

mathtoo  Dec 29, 2018
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-16x^4 + x^2 + 2x + 1       =

x^2 + 2x + 1 - 16x^4    =

(x + 1)^2 - 16x^4   =

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

[  4x^2 + x + 1 ]    [ -4x^2 + x + 1 ]

And   a > d   Dec 29, 2018
edited by CPhill  Dec 29, 2018
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Thank you, everyone! Dec 29, 2018