Question

FACTOR 5x^3+17x^2-84x-36

Answer 1

There MUST be a Positive Rational zero for this expression.
1) Do you know why?
2) Can you find it easily if it is Rational?

Answer 2

okay well actually the question was 5x^4+12x^3-101x^2+48+36, and I had to solve it, finding the zeros, I found one of them which was x-1 and I divided that by the question to find the others, but I got 5x^3+17x^2-84x-36, and idk how to factor it

Answer 3

1) No , 2) what does rational mean

Answer 4

1) You just count sign changes on coefficients. This one starts out positive and changes to negative. That's one sign change. There MUST be a REAL and POSITIVE zero.

2) You must know what a RATIONAL number is. Ratios of integers?

3) How did you find that (x-1) is a factor?

Answer 5

1) Ohh okat
2) Oh wait yes I do, I looked through my notes
3) and I subed in one of the factors of 36

Answer 6

I subbedd in 1

Answer 7

then divided the whole thing by x-1

Answer 8

3) Excellent, however, you only got lucky. The possible rational zeros are ALL the factors fo 36 divided by ALL the factors of 5. Anyway, it's always a good idea to try the easy ones first.

Try x = 3.

Answer 9

okay got 0

Answer 10

now what do I do to find the rest?

Answer 11

Reduce it again. Then we can continue out search for Rational Zeros or we can realize that it's just Quadratic and we can use the Quadratic Formula or Completing the Square to beat the other two out of it.

Answer 12

reduce it by dividing it by x-3 ?

Answer 13

Right. Did you use synthetic division to evaluate it? To find that zero on the end?

Answer 14

yes

Answer 15

I got 5x^3+25x^2-26x-28 R -48

Answer 16

Oh, then you already did the division. Do you know this? Those numbers across the bottom are the coefficients you need.

You can't get a remainder. You just told me the remainder is zero. How did you get a different result?

Answer 17

oh yeah, hm maybe I made a mistake, let me re-do it

Answer 18

Don't redo it. Where is the table where you manged zero? That is the one you need.

Answer 19

okay I did make a mistake. now I got 5x^3+27x^2-20x-12

Answer 20

wait im lost, which table? We never did a table in class for this type of questions :o

Answer 21

?? Can't be. It should be only quadratic.

Start with \(5x^{4}+12x^{3}-101x^{2}+48+36\)

Divide the first one. \(\dfrac{5x^{4}+12x^{3}-101x^{2}+48+36}{x-1} = 5x^{3}+17x^{2}-84x-36\)

Divide the second one we found. \(\dfrac{5x^{3}+17x^{2}-84x-36}{x-3} = 5x^{2}+32x+12\)

Don't go back to the beginning and start over. Work with the new, reduced polynomial.

Answer 22

You should have a table if you used "Synthetic Division".

Answer 23

Ohh like that, yeah I did that ! For the zeros right?

Answer 24

dont we call that quotient form ?

Answer 25

Sure, you can call it that. You can call it "Steve" if you like. Just don't go back to the start.

Answer 26

How shall we find the last two zeros?

Answer 27

um, not sure