Anguyennn
  • Anguyennn
How to factor 3x^3-x^2+4
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DangerousJesse
  • DangerousJesse
Do you have a general idea on how to factor equations?
Anguyennn
  • Anguyennn
yes but I'm not sure what to do here like do i factor out x2?
whpalmer4
  • whpalmer4
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

DangerousJesse
  • DangerousJesse
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: \[\frac{(x+1) (3 x^3-x^2+4)}{x+1}\]
whpalmer4
  • whpalmer4
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.
Anguyennn
  • Anguyennn
okay
whpalmer4
  • whpalmer4
@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.
Anguyennn
  • Anguyennn
My teacher showed my difference of squares/cubes and sum of cubes
Anguyennn
  • Anguyennn
I thought i was suppose to just substitute it into one of them
whpalmer4
  • whpalmer4
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.
whpalmer4
  • whpalmer4
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.
Anguyennn
  • Anguyennn
I am just so confused to which one to apply it to
whpalmer4
  • whpalmer4
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: \[a^3-b^3 = (a-b)(a^2+ab+b^2)\]
Anguyennn
  • Anguyennn
so then what would I do?
whpalmer4
  • whpalmer4
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.
Anguyennn
  • Anguyennn
are you able to show me?
Anguyennn
  • Anguyennn
for the equation that I am stuck on please and thanks
whpalmer4
  • whpalmer4
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?
Anguyennn
  • Anguyennn
uhm
Anguyennn
  • Anguyennn
its so different from how I learned it
whpalmer4
  • whpalmer4
do it the way you learned it, if you learned it :-)
whpalmer4
  • whpalmer4
right now, I'm just trying to show you how it is that the RRT allows you to intelligently guess at the potential factors.
whpalmer4
  • whpalmer4
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.