How do you factor f(x)=x^4+x^3-7x^2-13x-6? and x^5+4x^4+6x^3-8x^2-7x-2?

Question

...by using the rational roots theorem?

Answer 1

use synthetic divsion method
i think this link would help
http://www.purplemath.com/modules/synthdiv.htm

Answer 2

In the first equation, the factors of 6 are 1, 2, 3, and 6 or -1,-2,-3,-6. These factors determine the numerator of possible roots.
The factors of the highest power coefficient (in both cases 1) determine the denominator of possible roots.

Thus possible roots are 1, -1, 2, -2, 3, -3, 6, -6

Check each of these in the equation. If, after plugging one of these into the equation, the result is zero, then you know it is one of the roots of the equation.

For example, let's test 1 in the first equation

1^4 + 1^3 - 7(1)^2 - 13(1) - 6 = 2-7-13-6 = -24 not equal to zero

So 1 is not a root of the equation.

Try -1

(-1)^4 + (-1)^3 -7 (-1)^2 - 13(-1) - 6 = 1 - 1 - 7 + 13 - 6 = 0

So -1 is a root of the equation. Thus one of the factors will be (x+1) = [x-(-1)]

Try -2

(-2)^4 + (-2)^3 - 7 (-2)^2 - 13(-2) - 6 = 16 - 8 - 28 + 26 - 6 = 0

So -2 is a root of the equation. Thus one of the factors will be
(x+2) = [x-(-2)]

Keep doing this to get the remaining factors.

Answer 3

I'll suggest a method here. Apply the same to the second one also. Not sure, if that is by the theorem you mentioned:

x^4 + x^3 -7x^2 - 13x -6
= x^4 + x^3 + [ -7x^2 - 7x -6x -6 ]
= x^3(x+1) + [ -7x(x+1) -6(x+1) ]
= x^3(x+1) - (7x+6)(x+1)
= (x+1)(x^3 -7x -6)

Answer 4

uhm.. let me grab my calculator???????