how do you factor equations like x^3+x2+4x+4 and d^3+2d^2+3d+6. also, can some1 tell me how to type with expon

Answer 1

Basically, you just try and clump things together, and if you're lucky it all works out. One thing that is obvious to take out are factors of x or d (or whatever). So if we do that with your sample problems...

x^2 can come out of x^3 and x^2 to produce x^2 (x+1). 4 can come out of the other two to produce 4(x+1). You'll note that you have a repeated term there, so you can group that too, ending up with (x+1)(x^2+4). At this point it doesn't hurt to look and see if you can't simplify it more... if your last term were x^2-4 you could, for example. In this case, I think we're as far as we go.

d^2 can also come out of d^3 and 2d^2 to give you d^2(d+2). 3 can come out of 3d and 6 to make 3(d+2). Here again we have a repeated term. We end up then with (d+2)(d^2+3).

These problems were obviously chosen to allow this kind of simplification... if you were just coming up with equations from observed variables, they would seldom be so accomodating. Hope that helps!

Answer 2

You are typing the expon correctly.

x^3 + x^2 +4x = 4 = x^2(x+1) + 4(x+1) = (x^2 + 4) (x + 1)

d^3 + 2d^2 + 3d + 6 = d^2(d+2) + 3(d+2) = (d^2+3) (d+2)

Key: try to combine the first two terms together, and the last two terms together. Then you can find that there's a common factor.

You might encounter harder questions such as the following:
x^3 + 5x^2 + 8x + 4 (answer: (x+1)(X+2)(x+2) )

I think know how to do those by trial and error -- usually the end result will be (x+a)(x+b)(x+c) while a*b*c = 4 (the costant term in the equation). So possible answers are a=1, b=1, c=4, or a=2, b=2, c= 1. Just multiply them out and you will find the right answer.

Answer 3

For x^3 + x^2 + 4x + 4, this one actually groups; one is lucky if a cubic can be factored by grouping, because there are harder cubics to solve that take longer.

I'm going to factor the first two terms and the last two terms.

x^2(x + 1) + 4(x + 1)

Notice that each term has (x + 1). What we're going to do is, in turn, FACTOR (x + 1) out of the already factored terms, to get

(x + 1) (x^2 + 4)

Answer 4

x² + 4x = ninety six Take the coefficient of the x time period (+4) divide by 2 (+4/2) and sq. the end result (+4/2)² This reduces to 4, so the answer is #3. yet I in no way shrink it thoroughly, (a) because it would want to might want to be undone contained in the subsequent steps, (b) you may want to ignore which signal to apply. as an celebration: x² - 4x = 0 to finish the sq., you upload (-4/2)² to both aspect. that is decreased to 4, yet then you truthfully lose the minus signal that you want later. extra effective to apply (-2)²: x² - 4x + (-2)² = 0 + (-2)² (x-2)² = 4

Answer 5

okay so here you have

x^3+x^2+4x+4

Divide it out:

(x^3+x^2)+(4x+4)

now, factor out from the mini equations
you should get

x^2(x+1)+4(x+1)

okay now here's the tricky part. You see how (x+1) is repeated? Factor THAT (x+1) out so therefore we end up with.

(x+1)( x^2+4)

And there ya go

Answer 6

IF the problem you face is not conviniently groupable, then look into cubic formula (its ugly) if the greatest power in the question is greater than 3 then try something called "newton's method".