College Algebra & Trigonometry, 2018a

134 CHAPTER 2. POLYNOMIAL AND RATIONAL FUNCTIONS
We can see roots at x = −5, 3, which means that (x+5) and (x−3) are both factors
of this polynomial. We’ll need to divide by both of these factors to break down
the polynomial. First, we divide by (x − 3):
And then by (x +5):
3 3 5 −45 19 −30
↓ 9 42 −9 30
3 14 −3 10 0
3 3 5 −45 19 −30
↓ 9 42 −9 30
−5 3 14 −3 10 0
↓ −15 5 −10
3 −1 2 0
Now we know that 3x 4 +5x 3 − 45x 2 +19x − 30 = (x +5)(x − 3)(3x 2 − x +2)and
so, to finish the problem:
3x 4 +5x 3 − 45x 2 +19x − 30 = 0
(x +5)(x − 3)(3x 2 − x +2)=0
x +5=0 x − 3=0 3x 2 − x +2=0
x = −5 x =3 x ≈ 1 6 ± 0.799i
Next, let’s look at an example where there is a root that is not a whole number:
Example
Find all real and complex roots for the given equation. Express the given polynomial
as the product of prime factors with integer coefficients.
3x 3 + x 2 +17x +28=0