How can I factor x^3+7x^2+14x+8=0? I know the answer is (x+4)(x+2)(x+1), but how do I get that?

Answer 1

You do it in two stages. First factor out x+1 that would give you:

(x^2+6x+8) (x+1)=0

Then factor x^2+6x+8 further:

(x+2)(x+4)(x+1)=0

Answer 2

If this polynomial has a rational factor of the form (x+r) r will be a factor of the constant. Also note the coefficient of the highest degree term, since that is one. r will be an integer, otherwise the denominator of r would be a factor of the coefficient of the leading term.
try the following values for r:
1, -1, 2, -2, 4, -4, 8 and -8
Start by using the factor theorem if f(r) = 0 then (x-r) is a factor.
f(1) = 1 +7 +14+8>0
Clearly we should skip the positives and try a negative:
f(-1)= -1 +7 -14 +8 = 0
So (x+1) is a factor:
(x^3+7x^2+14x+8)/(x+1) = x^2 +6x +8
Factor the above and you have all three factors.

Answer 3

You have to start with a guess at a factor. The factors will be the factors of the last coefficient (8) divided by the factors of the first coefficient (1). So possible choices are:

{+/-1, +/-2, +/-4, +/-8}

Here you have to try some numbers until you find one that works. Trying -1, you get an answer of 0. So we know that (x + 1) is a factor.

Now do synthetic division:
....................... x² + 6x + 8
x + 1 ) x^3 +7x² + 14x + 8
.......... x^3 + x²
.................... 6x² + 14x
.................... 6x² + 6x
................... .......... 8x + 8
................... .......... 8x + 8

Now you have (x+1)(x² + 6x + 8) and you can easily factor the last part:
( x+1 )( x+2 )( x+4 ) = 0

Answer 4

Hello,

Put in -1 and we have (-1)^3 +7*(-1)^2 +14*(-1) +8 and we have -1+7-14+8 = 0 so (x+1) is a factor now divide this into the original polynomial and we have. x^2 + 6x + 8 and tis factors into (x+4)(x+2) so we have (x+1)(x+4)(x+2).

Hope This Helps!