
If it has any rational solutions the numerator must be
a factor of 4 and the denominator must be a factor of 2,
The only possible rational solutions are
,
,
, or
Trying the easiest one x = 1
So we divide by x - 1, either synthetically:
1|2 11 1 -10 -4
| 2 13 14 4
2 13 14 4 0
or by long division
2x³ + 13x² + 14x + 4
x - 1)2x4 + 11x³ + x² - 10x - 4
2x4 - 2x³
13x³ + x²
13x³ - 13x²
14x² - 10x
14x² - 14x
4x - 4
4x - 4
0
You have now factored the original polynomial as
(x - 1)(2x³ + 13x² + 14x + 4)
Next let's do the same with the cubic polynomial in
the second parenthesis:
If it has any rational solutions the numerator must be
a factor of 4 and the denominator must be a factor of 2,
We've already done that. The only possible rational solutions are
,
,
, or
You could go to the trouble of trying all those. But sooner of
later you'd get around to trying to divide by
or
Divide by x + .5, either synthetically:
-.5|2 13 14 4
| -1 -6 -4
2 12 8 0
or by long division
2x² + 12x + 8
x + .5)2x³ + 13x² + 14x + 4
2x³ + x²
12x² + 14x
12x? + 6x
8x + 4
8x + 4
0
And you have now factored further as
(x - 1)(x + .5)(2x² + 12x + 8)
Now we can factor 2 out of the third parentheses:
(x - 1)(x + .5)2(x² + 6x + 4)
2(x - 1)(x + .5)(x² + 6x + 4)
The quadratic will not factor but we can find its roots
by using the quadratic formula:
So the 4 roots are
1,
,
, and
Edwin