Question

How to factor 3x^3-x^2+4

Answer 1

Do you have a general idea on how to factor equations?

Answer 2

yes but I'm not sure what to do here
like do i factor out x2?

Answer 3

Here you need to do some intelligent guessing. There isn't any single term that you can factor out, like a power of \(x\) or a coefficient. So that means you're going to have something of the form \((x+a)\) where \(a\) is a constant.

Answer 4

3 x^3-x^2+4

The possible rational roots of\[ 3 x ^{3}-x^{2}+4\] are \[x = ±1/3, x = ±2/3, x = ±4/3, x = ±1, x = ±2, x = ±4\]
Of these, \[x = -1\] is a root. This gives x+1 as all linear factors:
\[\frac{(x+1) (3 x^3-x^2+4)}{x+1}\]

Answer 5

We also know that we need to have 3 roots because of that \(x^3\), so we'll either have something like \[(x+a)(x+b)(x+c)\]or\[(x+a)(x^2+bx + c)\]
In either case, the constant term will be the product of all of the roots.

Answer 6

okay

Answer 7

@DangerousJesse is able to use that information to make a list of all possible rational roots.

\[(x+a)(x+b) = x^2 + bx + ax + ab =x^2 + (a+b)x + ab\]
\[(x+c)(x+a)(x+b) = (x+c)(x^2 + (a+b)x + ab) \]\[= x^3 + (a+b)x^2 + abx + cx^2 + c(a+b)x + abc\]
you can see that in all these cases, that constant term is just the product of the constant terms of the factors. If you factor your constant term, you know what your possible factor constant terms are. It's a bit more complicated when you have a coefficient other than \(1\) for the leading term, but that's the general idea.

Answer 8

My teacher showed my difference of squares/cubes and sum of cubes

Answer 9

I thought i was suppose to just substitute it into one of them

Answer 10

When you have a potential factor to test, you can test it by evaluating the polynomial at that value of \(x\). If \(P(a) = 0\), then \(x-a\) is a factor.

Answer 11

Oh, well, if you can remember the form of a difference of cubes or sum of cubes, and your problem is one of those, then you're set! We were showing you a more general approach that works for bigger polynomials.

Once you have verified that you have a good factor, you can use synthetic division to divide by that factor to get a simpler polynomial and repeat the process until you've found all of the factors.

Answer 12

I am just so confused to which one to apply it to

Answer 13

You only have one equation to apply it to: \[3x^3-x^2+4
\]

This does not appear to be either a difference of squares or a sum/difference of cubes.

Difference of squares:
\[a^2-b^2 = (a-b)(a+b)\]

Sum of cubes:
\[a^3+b^3 = (a+b)(a^2-ab+b^2)\]
Difference of cubes:
\[a^3-b^3 = (a-b)(a^2+ab+b^2)\]

Answer 14

so then what would I do?

Answer 15

Well, you would compare the thing you have to factor to the left hand side of those "recipes" and realize that you can't use any of them. At that point, you move on to using the rational root theorem, which is what we were talking about.

If you have a polynomial with integer coefficients, then the possible zeroes are the set of all of the factors of the constant term over all of the factors of the leading coefficient (the coefficient of the term containing the highest power of the variable). There will be more possibilities than correct answers, usually. You have to test them and sort out the good from the bad.

Answer 16

are you able to show me?

Answer 17

for the equation that I am stuck on please and thanks

Answer 18

Maybe a simpler example will help.

\[x^2+x-2\]
Pretend we can't look at that and factor it by inspection.

We know that it must be of the form\[(x+a)(x+b)\]for unknown values of \(a,\ b\). How to find them?

\[(x+a)(x+b) = x*x + x*b + a*x + a*b \]\[= x^2 + (a+b)x + ab\]
Compare that with our original:
\[x^2 + x -2\]

By matching up like terms, we can see that

\[x^2 + x -2 = x^2 + (a+b)x + ab\]
\[x = (a+b)x\]\[-2 = ab\]
so \(1 = a+b\) and \(a*b = -2\)

Can you tell me the values of \(a,\ b\) which make that work?

Answer 19

uhm

Answer 20

its so different from how I learned it

Answer 21

do it the way you learned it, if you learned it :-)

Answer 22

right now, I'm just trying to show you how it is that the RRT allows you to intelligently guess at the potential factors.

Answer 23

but I have to leave, so maybe someone else can carry on with helping you. I'll check back later and make sure you get to the answer.