Posted by **Austin** on Thursday, December 27, 2012 at 7:15pm.

Solve the equation in the real number system

X^4+11x^3+24x^2-23x+35=0

Please show work

- calculus -
**Damon**, Thursday, December 27, 2012 at 7:28pm
the 35 tempts me to try -7 and -

divide mess by (x+5)

get

(x+5)(x^3+6x^2-6x+7)

so divide

(x^3+6x^2-6x+7) by (x+7)

get

(x+5)(x+7)(x^2-x+1)

now that last one has no real roots

b^2-4ac = 1-4 = -3

so our only real roots are

(-5,0) and (-7,0)

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