how to factorize: x^9-3x^8+x^6-9x^4-4x^2-16x+84 = 0

Question

campbell_st · Answer

lol... good luck... I used my calculator... and its a very difficult problem

anonymous · Answer

no just group using paretheses

anonymous · Answer

@Spidermonkeyy  please show the working

anonymous · Answer

its a lot, but im trying to work it out

anonymous · Answer

i got it but it's a lot of work so i will just tell you how to do it okay?

anonymous · Answer

Basically you take two terms in the equation that would factor together well like x^9 and x^6 because you could make it x^6(x^3+1) and that would be one factor, then do -3x^8 and -9x^4 and factor out a -3x^4 to get -3x^4(x^4+3) that's a factor then do the rest.

anonymous · Answer

You should get $$x ^{6}(x ^{3}+1) -3x ^{4}(x ^{4}+3)-4(x ^{2}+4x-21)$$

anonymous · Answer

now make x^6 and -3x^4 a factor like so
$$(x ^{6}-3x ^{4})(x ^{3}+1)(x ^{4}+3)-4(x ^{2}+4x-21)$$

anonymous · Answer

finally factor again to get$$x ^{4}(x ^{2}-3)(x ^{3}+1)(x ^{4}+3)-4(x ^{2}+4x-21)$$

anonymous · Answer

make x^4 and -4 a factor again$$(x^4-4)(x^2-3)(x^3+1)(x^4+3)(x^2+4x+21)$$
was supposed to be a +21 the whole time my bad, have fun getting those lol

campbell_st · Answer

well I think the constant term is incorrect 

-4 * -3 * 1 * 3 * 21 isn't 84      

36 * 21 = 756

so a slight error there...