If $2x^3+4x^2+6x+8=2468$ and $x$ is a positive number, what is the value of $x^3+9x^2+8x+8$

Question

I'm trying to find $x$:

this is what i did so far:

$2x^3+4x^2+6x+8=2468$

Divide by $2$

$x^3+2x^2+3x+4=1234$

Minus $4$, $x^3+2x^2+3x=1230$

factor out an $x$

$x(x^2+2x+3)=1230$

and now, i'm stuck.

Yes · Accepted Answer · 2015-10-09T03:46:45.023

5

If you notice that $2,4,6,8$ correspond to the coefficients of the polynomial $2x^{3} + 4x^{2} + 6x + 8$, then $2x^{3} + 4x^{2} + 6x + 8 = 2468$ is implied by $x =$ ? Using this root and the factor theorem, we can then arrive at $(x-10)(x^{2} + 12x + 123) = 0$, which shows that $x=10$ is the only (real; for complex numbers cannot be ordered like the reals) solution.

edited Oct 09 '15 at 03:46

answered Oct 09 '15 at 03:20

Yes

20,719

Oh, how naive of me. $x=10$!!! – Caddy Heron Oct 09 '15 at 03:25
1

@Gudson, (+1), We need to nullify the rest two roots, right? – lab bhattacharjee Oct 09 '15 at 03:28
@lab if they're negative or 0. – Caddy Heron Oct 09 '15 at 03:30
@CaddyHeron, Yes! Otherwise, you will have multiple value for $x^3+9x^2+8x+8$ – lab bhattacharjee Oct 09 '15 at 03:32
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I believe you can go on from noting that $x = 10$ is a solution; I edited my answer because I think it would be accountable to do so. :) – Yes Oct 09 '15 at 03:54

If $2x^3+4x^2+6x+8=2468$ and $x$ is a positive number, what is the value of $x^3+9x^2+8x+8$

1 Answers1