When you factorise something like 243z^5 + 32 = 0, how would you determine the real factors after finding the roots of the equation? What makes it a real factor?
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Real factor of a function of order $n$ means $a_0+a_1x+a_2x^2+\cdots+a_rx^r$ where $a_i$s are real and $1\le r<n$
We have $(3z)^5-(-2)^5=0$
Now $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
What if $x-1=0$
Else use Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
lab bhattacharjee
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