How do you solve $3x^5 = 6x^4 - 72x^3 =$ ?

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Answer 1 · 2016-07-22T16:37:08

$(1) :$ The Soln., in $RR, x=0;$ & in $CC, x=0, x=1+-isqrt23$ .

$(2) :$ The Reqd. Value, (i.e., "=?" of the problem) in $RR, =0;$ and, in $CC, 0, or, -288(-11+-isqrt23)$ .

Given that, $3x^5=6x^4-72x^3$

$rArr3x^5-6x^4+72x^3=0$

$rArr 3x^3(x^2-2x+24)=0$

$rArr x^3=0, or, x^2-2x+24=0$

$x^3=0 rArr x=0..........(1)$

$x^2-2x+24=0................(2)$

$rArrx^2-2x=-24$

$rArrx^2-2x+1=-24+1$

$rArr (x-1)^2=-23$ , which is not possible in $RR$

However, in $CC, (x-1)^2=-23 rArr x-1=+-isqrt23$

$rArr x=1+-isqrt23$

Hence, in $RR$ , the reqd. value (i.e., "=?" of the problem) $=0$

While, in $CC$ , the reqd. value $=3x^5=0, if x=0$ , or,

if, $x=1+-isqrt23$ , the reqd. value (i.e., "=?" of the problem)

$6x^4-72x^3=6x^2(x^2-12x)=6(2x-24)(-24).........[by (2)]$

$=-24*6*2(x-12)=-288(1+-isqrt23-12)=-288(-11+-isqrt23)$

How do you solve 3x^5 = 6x^4 - 72x^3 =3x^5 = 6x^4 - 72x^3 =?