Question

Factor x^3-x^2-5x+2 completely.

Answer 1

is this an algebra class or precalc?

Answer 2

if it's precalc, then we have possible rational roots of +2 or -2.

Testing these roots, we see that -2 works nicely,
(-2)^3 -(-2)^2-5(-2)+2 = -8-4+10+2 = 0.

So, (x - (-2)) = (x + 2) is one factor.

Then, using synthetic division gives:

-2| 1 -1 -5 2
|____-2__6__-2__
1 -3 1 0

Using the bottom line, we have coefficients of 1, -3 and 1 so
x^2 - 3x + 1

Thus, the answer is (x + 2)(x^2 -3x +1).

Answer 3

its college algebra trig

Answer 4

otherwise we have to use trial and error. :(

Answer 5

we havent learned synthetic division yet, it should be simpler i guess

Answer 6

x^3-x^2-5x+2
(x^2.....1)(x......2)
do trial error untill completing its factor

Answer 7

in general though a good trick when you have a cubic equation is to try a factor of (x + factors of the last number) or (x - factors of the last number). In this case the last number was 2, which has factors of 2 and 1, so I tried (x + 2) and (x - 2), but you could have also tried (x + 1), (x-1) they just don't work.

Answer 8

ah alright, i think thats right because we were listing factors in class. thanks.

Answer 9

When you have *any* polynomial with a leading coefficient of 1 and integer coefficients, you should try factoring out \(( x\pm n\)) where n is the various factors of the final term. You can see why this is:

Assume we have a polynomial with roots a, b. \[(x-a)(x-b) = x^2 -ax -bx + ab\]That last term is generated only by the multiplication of the roots. Let's add another root, c. \[(x-c)(x^2 -ax -bx + ab) = x^3 -ax^2 -bx^2 + abx -cx^2 +acx + bcx -abc\]Again, the final term is generated only by the multiplication of the roots.

Answer 10

ah

Answer 11

yeah got it thanks