Question #83abc

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Question #83abc

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Answer 1 · 2016-10-19T16:34:23

No useful factorization for $f (x, y) = x^{2} + 6 x y + 9 y^{2} - 21 y + 12$ $f(x,y) = x^2 + 6xy + 9y^2 - 21y + 12$

Explanation:

$f (x, y) = x^{2} + 6 x y + 9 y^{2} - 21 y + 12$ $f(x,y) = x^2 + 6xy + 9y^2 - 21y + 12$

If $f (x, y)$ $f(x,y)$ is factorable, that implies in multiple tracing of $f (x, y) = 0$ $f(x,y)=0$

So, if $f (x, y)$ $f(x,y)$ is factorable for instance into $f_{1} (x, y) f_{2} (x, y)$ $f_1(x,y)f_2(x,y)$ this means that

$f (x, y) = f_{1} (x, y) f_{2} (x, y) = 0$ $f(x,y)=f_1(x,y)f_2(x,y)=0$ has a minimum of two null leafs. One for $f_{1} (x, y) = 0$ $f_1(x,y)=0$ and the other for $f_{2} (x, y) = 0$ $f_2(x,y)=0$

In our case $f (x, y)$ $f(x,y)$ represents a slanted parabola having one null leaf, so this $f (x, y)$ $f(x,y)$ is not factorable.

We know that $f (x, y)$ $f(x,y)$ represents a parabola because putting it in the form

$f(x,y)=1/2(x,y)cdot M cdot((x),(y))+(x,y)cdot b + c$

$M=((2, 6),(6, 18))$ has characteristic polynomial

$p_M(s) = s^2-20s=s(s-20)$

which has a null root.

Attached the null trace of $f(x,y)=0$

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Question #83abc

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