Question

write a linear factorization of x^3+4x-5

Answer 1

\(\large \begin{array}{cccllll}
x^3&+4x&-5\\
&\uparrow &\uparrow \\
&5-1&5\cdot -1
\end{array}
\)

Answer 2

I know one real zero is 1

Answer 3

ohhh shoot... is 3rd degree... I see shoot

Answer 4

well... how did you get 1 for a real zero anyway?

Answer 5

i graphed it

Answer 6

then divide the equation by that, you should end up with a quadratic

and then you should be able to do the quadratic either by factoring or using the quadratic formula

Answer 7

So far I have it factored to (x-1)(x+1/2-((sqrt 19)/2)i(x+1/2+((sqrt 19)/2)i)

Answer 8

But they book has the answer 1/4(x-1)(2x+1+(sqrt 19)i(2x+1-(sqrt 19)i) and i don't know where they got that

Answer 9

hmmm well.. do you know how to divide polynomials?

Answer 10

where did they get the two to distribute to the two complex factors? and also the one fourth at the front? and no

Answer 11

well... I think you're expected to divide the polynomial... so you may want to cover that first

firstly you'd need to use the "rational root" test and check for a root using the "remainder theorem"

then divide by a likely root

so you found "1" by graphing.. then you'd need to divide by that root and use the quotient, which will be a quadratic and factor it