Don't solve, just substitute

Algebra Level 3

$\begin{aligned} \large x^3+8x-2&=& \large 0 \\ \large x^5+10x^3-2x^2+16x+10 &=& \large \ ? \end{aligned}$

3 solutions

Angel Daniel Lopez Flores
May 3, 2015

$x^5+10x^3-2x^2+16x+10=x^2(x^3+8x-2)+2x^3+16x+10$

$x^2(x^3+8x-2)+2(x^3+8x-2)+10+4$

but $x^3+8x-2=0$

so $x^2(0)+2(0)+10+4$

$x^5+10x^3-2x^2+16x+10=\boxed{14}$

Nice problem!Once I saw it I thought to myself that I'm pretty much screwed xD

Oh and by the way you might wanna change the title a little bit ,since it gives the idea of problem's solution away.

Arian Tashakkor - 6 years, 1 month ago

It should be 8x in second line, right?

Cool Math - 6 years, 1 month ago

oh yes , sorry

Angel Daniel Lopez Flores - 6 years, 1 month ago

haha i couldnt solve it xd

Cool Math - 6 years, 1 month ago

Eric Osborn
May 13, 2015

If the first line is A and the second line is B...

$B-x^{2}A-2A=14$

to illustrate step by step:

$[B-x^{2}A]$

$x^{5}+10x^{3}-2x^{2}+16x+10= ?$

$-(x^{5}+8x^{3}-2x^{2})=0$

gives

$0x^{5}+2x^{3}+0x^{2}+16x+10= ?+0$

or just

$2x^{3}+16x+10=?$

then

$[-2A]$

$-(2x^{3}+16x-4)=0$

which gives

$0x^3+0x+14=?+0$

or just

$?=14$

Aravind Vishnu
May 13, 2015

$x^3=2-8x$

$x^5=2x^2-8x^3$

$x^5+10x^3-2x^2+16x+10=2(x^3+8x+5)$

$=2(x^3+8x-2)+14$ $=2 \times 0 +14=14$