Answer:
(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
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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
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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).
