I need to show that $x-a$ is a factor of $x^4-4a^3x+3a^4$. I've tried long polynomial division but stopped in the middle of it as it didn't seem to work. How else could one approach it?
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5To show that $x-a$ is a factor you only have to show that $x^{4}-4a^{3}x+3a^{4}=0$ when $x=a$. – Kavi Rama Murthy Dec 16 '20 at 09:53
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If you use long division, don't forget those missing terms.
Think about the place-holder $0$ as in $1\color{red}{0}3$.
\begin{array}{ccccccc} & & & x^3 & +ax^2 & +a^2 x & -3a^3 \\ & & -- & --- & --- & --- & --- \\ x-a & ) & x^4 & \color{red}{+0x^3} & \color{red}{+0x^2} & -4a^3x & +3a^4 \\ & & x^4 & -ax^3 \\ & & -- & --- & --- \\ & & & ax^3 \\ & & & ax^3 & -a^2x^2 \\ & & & --- & --- & --- \\ & & & & a^2x^2 & -4a^3x \\ & & & & a^2x^2 & -a^2x & \\ & & & & --- & --- & --- \\ & & & & & -3a^3x & +3a^4 \\ & & & & & -3a^3x & +3a^4 \\ & & & & & --- & --- \end{array}
Ng Chung Tak
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That's helped, thank you. There's a typo in the quotient, should be $x^3+ax^2+a^2x-3a^3$ – super.t Dec 16 '20 at 14:35
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Hint: If polynomial division is asked for, following is how it can be factorised. Otherwise proving that $x-a$ is a factor is a trivial application of the factor theorem. $$x^4-4a^3x+3a^4=(x^4-ax^3)+(ax^3-a^2x^2)+(a^2x^2-a^3x)+(-3a^3x+3a^4)$$
Paras Khosla
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Dividing the expression by $a^4$ you obtain
$(x/a)^4-4x/a+3=(x/a)^4-4x/a+4-1=(x/a-2)^2-1$
Now, using the fact that $t^2-1=(t+1)(t-1)$ yelds to
$(x/a-2)^2-1=(x/a-1)(x/a-3)=1/a^2(x-a)(x-3a)$
obtaining what you are asking for.
Sabino Di Trani
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