i expanded and simplified the RHS to $x^4+Bx^3+4x^2+Ax^3+ABx^2+2Ax+2Bx+4$
i don’t know where to go from here.
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1You should combine like terms. For instance, instead of having two $x^3$ terms, you should have just $(A+B)x^3$ – The Chaz 2.0 May 02 '20 at 11:34
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Then compare/equate coefficients. For example, there are no $x^2$ on the left. So $ 0 = AB + 4$, which is the coefficient of $x^2$ on the right. – The Chaz 2.0 May 02 '20 at 11:36
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Lastly, you can look at the zeroes of the LHS (there are no real roots) and conclude that the factors on the RHS can't have real roots. By looking at determinants, you should conclude that $A^2 < 8$, and likewise for $B^2$. – The Chaz 2.0 May 02 '20 at 11:40
2 Answers
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After your expansion
$$x^4+Bx^3+2x^2+Ax^3+ABx^2+2Ax+2x^2+2Bx+4$$
Just like how you grouped the $2x^2+2x^2$, we can group like terms (same $x^i$)
$$x^4+(A+B)x^3+(AB+4)x^2+(2A+2B)x+4\equiv x^4+4$$
Now since this is true for all $x$, all the matching coefficients must be equal! (Might not be obvious)
$$x^4+(A+B)x^3+(AB+4)x^2+(2A+2B)x+4\equiv x^4+0x^3+0x^2+0x+4$$
$$A+B=0,AB+4=0,2A+2B=0$$
$$\implies A=-B,AB+4=0$$
$$\implies -B^2+4=0$$
$$A,B=2,-2$$
$$\therefore x^4+4=(x^2+2x+2)(x^2-2x+2)$$
Gareth Ma
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i understand everything up until -B^2+4=0. how do you get this? and where does the A go? – Bruxner Games May 04 '20 at 02:02
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After $-B^2+4=0$ you get $B=\pm2$, and since $A=\pm B$ they're the two solutions $2,-2$. – Gareth Ma May 04 '20 at 02:17
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sorry, i meant, $-B^2+4=0$ is where i don’t understand. i have no idea how you get $-B^2$ – Bruxner Games May 04 '20 at 02:21
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oh, just realised you solve simultaneously. sorry for my stupidity. thanks a lot – Bruxner Games May 04 '20 at 02:31
