Factoring Problem?

Question

Can someone help me factor this i've been having trouble with it all week x^3 - x^2 - 5x - 3. Thanks a lot

Answer 1

I think you are meant to look at the magnitudes of the coefficients, that's (1, 1, 5, 3), and see that with the appropriate signs they could add to zero. So is there a value of x that makes the terms appear as (-1, -1, 5, -3) or as (1, 1, -5, 3)? x = 1 won't do it, but x = -1 will - check it and see.

So if the polynomial is zero when x is -1, it has (x + 1) as a factor. This is a particular instance of something you were supposed to know the general case of. Other answerers have told you what it's called.

x^3 - x^2 - 5x - 3 = (x + 1)(x^2 + Qx + ...). Look at what we have for the x^2 term, because the rest of the second factor won't alter it. We have +x^2 already, so we need Q = -2 on the right to give us -x^2 on the far left.

x^3 - x^2 - 5x - 3 = (x + 1)(x^2 - 2x + Q). For the x term we have -2x already, so we need Q = -3 to make it come out as -5x. For the constant term to come out as -3, we need Q = -3, which is just as well. It is BOUND to come out right if we have started with a proper factor. If we haven't, we will end up needing two different values for the last Q. This is called synthetic division.

I presume you can factor (x^2 - 2x - 3) for yourself.

Answer 2

x^3 - x^2 -5x -3, using the remainder theorem, if x-1 is a factor, then f(x) = 0
substituting
1>3 - 1>2 - 5(1) - 3 = -5 -3 = -8, so x-1 is not a factor
lets try x+1, we have to substitute for x = -1
(-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0
this implies that x+1 is a factor

we now have to divide the original polynomial with x+1 to get a quadratic equation we can solve to get the remaining factors, i can't do the division here now cos limitations with keyboard, but this is what u'll get x^2 - 2x - 3 which we can easily factorise to
x^2 + x - 3x - 3 = x(x+1) - 3(x+1) = (x-3)(x+1) we now have two factors ie x- 3 and x + 1, since we are expecting 3 factors, it means that our polynomial has a repeated factor in x + 1, our polynomial can be represented (x-3)(x+1)(x+1)

Answer 3

You may write f(x) = x^3 - x^2 - 5x - 3.

Observe that f(-1) = 0 => ( x + 1) is a factor of f(x). [Using Remainder Theorem].

Similarly note that f(3) = 0 => ( x - 3) is another factor of f(x). [Again by Remainder Theorem].


The last (and third) factor can be easily found to be ( x + 1).

Hence f(x) = x^3 - x^2 - 5x - 3 = ( x - 3)( x + 1)^2. And we are done!

Answer 4

You just have to test numbers until you find one that works, then reduce it to a quadratic.

I saw that x=3 is a solution, which tells me (x-3) is one factor. The problem reduces to (x-3)(x^2+2x+1). Now I factor the quadratic and I have the answer: (x-3)(x+1)(x+1)

Answer 5

First find the roots. Possible roots are: +/-1, +/-3. By trial you will find that -1 (double root) and +3 are the three roots, so:

x^3 - x^2 - 5x - 3 = (x +1)(x + 1)(x - 3)

Answer 6

(x - 3)(x + 1)^2 = x^3 - x^2 - 5x - 3, or
(x - 3)(x + 1)(x + 1)

Answer 7

Cubics can always be solved. Here is the formula (kind of complicated though):

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Answer 8

I don't see a way that this polynomial is factorable. Sorry.

Answer 9

(x-3)(x+1)^2

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