Given the function f(x)=x^4+3x^3-7x^2-27x-18 , factor completely

(x - 3), (x + 2), (x + 1) and (x + 3)

Step-by-step explanation:

Using synthetic division, I'd begin by testing various factors of -18 to determine whether any of them will divide into f(x)=x^4+3x^3-7x^2-27x-18 with no remainder.  Note that possible factors of -18 are ±1, ±2, ±3, ±6, ±18.

Let's arbitrarily start with -3.  In synthetic division, will the divisor yield a zero remainder ( which would tell us that -3 is a root of f(x)=x^4+3x^3-7x^2-27x-18 and that (x + 3) is a factor)?

-3   )   1    3    -7    -27    -18

              -3    0      21    +18

         1      0    -7     -6      0  

Since the remainder is zero, -3 is a root and (x + 3) is a factor.  The coefficients of the quotient are shown above:  1  0  -7  -6.  Possible factors of 6 include ±1, ±2, ±3, ±6.  Arbitrarily choose 2.  Is this a root or not?

2    )    1     0     -7    -6

                 2     4     -6

          1       2     -3    -12    

Here the remainder is not zero, so +2 is not a root and (x - 2) is not a factor.

Continuing to use synthetic division, I find that -1 is a root and (x + 1) is a factor, because the remainder of synth. div. is zero.  The coefficients of the quotient are 1, -1 and -6, which represents the quadratic y = x^2 - x - 6, whose factors are (x - 3)(x + 2).

Thus, the four factors of the original polynomial are (x - 3), (x + 2), (x + 1) and (x + 3).

a

step-by-step explanation: