Find the rational zeros and then the other zeros solve F(x)=0 then factor F(x) into linear factors?

Question

f(x)=5x^4 -4x^3=19x^2-16x-4
I found the rational zero to be -1 f(-1)=0, so f(x)=(x+1)
My book gives me then: (x+1) (x-3-2(2i^2))(x-3+2(2i^2))
2(2i^2) being 2 times the square root of 2i in case it looked confusing, it answers that the other zero is 3+-2(2i^2) I DON"T GET how they factord x+1 out of the original equation to get that mumbo jumbo with square roots...........

f(x)=x^3-5x^2+11+17 I posted the wrong math problem to the question
all possible zeros being +- 1 and +-17
-1 is the rational zero, so what is the linear factor to this question?

Answer 1

They got that factor out the hard way! Anyway, I believe that 1, not -1, is a rational zero. I suggest doing synthetic division to break down the polynomial.
Use -1 with the coefficients 5 -4 19 -16 -4
You end up with: 5x^3+x^2+20x+4
Then factor by grouping:
(5x^3+x^2)+(20x+4)
x^2(5x+1)+4(5x+1)
(5x+1)(x^2+4)
(5x+1)(x+2i)(x-2i) If you rearrange these they'll end up being the same as in your book. I just think that this is more clear cut.

Answer 2

There is confusion because you have put an = between the
-4x^3 and the 19x^2. If it is meant to be - 19 then neither f(1) nor f(-1) = 0 so I suspect that it is supposed to be + 19 in which case f(1) = 0 as given by a previous answer.
Also your last paragraph seems to be suggesting that there are three complex zeros which can't be the case with real coefficients. I think that you should check the whole thing again carefully.

Answer 3

For the 1st one you don't need to multiply it out... Just take each part of the expression and set it equal to zero so... X=0 And... 6x - 1=0 6x= 1 X= 1/6 {0, 1/6} As for the second one... 7x^2+19x-6=0 (7x-2)(x+3)=0 Then you'd do the same thing as #1... 7x-2=0 7x=2 X=2/7 Second multiplier-thingy X+3=0 X= -3 {-3, 2/7} Message me again if you still have questions! Hope this helped!

Answer 4

ur question seems to be wrong pls correct it ,x= -1 is not a factor. all coefficients of the equation being real the solutions could be all four real roots ,two real and two imaginary roots or all four imaginary roots equation being order of 4. however ur euation with x= -1 as factor is wrong.

Answer 5

Guess how many readers will understand your question!

1) You have stated rational zeros

2) Other zeros

Could you please state all possible 'zero-states' and difference in between each one of them?

It will help readers to give logical answer to your question!

Regards,