Question

can someone help me factor this out please
2x^3 + 8x^2 + 7x - 8

Answer 1

lets pull out common factors

Answer 2

ok

Answer 3

how do i find them?

Answer 4

how many variable have an x?

Answer 5

3?

Answer 6

yup, so pull it out and rewrite

Answer 7

what do you mean pull out?

Answer 8

What am i pulling out? 3? x?

Answer 9

are you sure you've got the problem copied correctly?

Answer 10

yes

Answer 11

dont think its factorable

Answer 12

:/

Answer 13

ok

Answer 14

and maybe that's the answer you're expected to find...but it's always worth asking if there might be an error!

Answer 15

however I don't see any obvious changes to that polynomial that make it factorable

Answer 16

but if you want one that looks similar that is factorable, try \[2x^3-11x^2+10x+8\]

Answer 17

how would this be factored out?

Answer 18

you can set x = to 1... if the equation comes out as 0, then u can use (x-1) as a factoring tool

Answer 19

for example this one is : (x-4)(x-2)(2x+1)

Answer 20

Im so bad at this i do not understand at all

Answer 21

(x-1)?

Answer 22

if the equation = 0

Answer 23

x-1 = 0

Answer 24

ohhhh

Answer 25

but for the one he put up... i ended up at x = 4 and plugged it in. The equation became 0

Answer 26

okay, there are a couple of things going on here. one is called the rational root theorem, and it basically takes advantage of the fact that the leading term and the constant term of the polynomial come from nothing by multiplying together the leading and constant terms of all of the factors.

\[(x+a)(x+b)= x^2+ax+bx+ab = x^2 + (a+b)x + ab\]the leading term comes from multiplying the x's together. the ab term comes from multiplying the constants together. doesn't matter how many factors there are, this is always true.

as a result of that, you can make intelligent guesses as to the rational factors — they'll be the possible factors (or a subset of the possible factors) of the constant term.

Answer 27

it's a little more complicated when the high order term doesn't have 1 as a coefficient...but the idea is the same. have a look at onmath.com/pdfFiles/30.RationalRootTheorem.pdf for a nice writeup

Answer 28

once we have a root, we can divide it out and get a simpler polynomial to work with for the other roots. this is because the polynomial can be written \((x-r_1)(x-r_2)...(x-r_n)\) if the roots are \(r_1,r_2,...r_n\). Dividing out one of the roots in this fashion doesn't change the others, but we get a smaller polynomial to work with.

Answer 29

root?

Answer 30

finally, the polynomial evaluates to 0 at each root, so you can use the suggestions from the rational root theorem, testing each one by plugging it into the formula to see if it gives you a result of 0, in which case it is a root. Then you divide it out and repeat the process.

root <-> zero

Answer 31

in this context, at least. just a value of x such that f(x) = 0

Answer 32

ok i need to re do the question