f(x) = 4x³ - 7x² - 11x + 5 = 0
By the rational root theorem, all the rational roots of this polynomial are of the form a/b where the numerator a is a factor of the constant 5 and the denominator b is a factor of the leading term 4.
So choices for the numerator are -5, -1, 1, or 5 and for the denominator 1,2,4 (and the negatives, but we don't have to count those separately if we've already included the negative ones for the numerator).
That's 12 possible rational roots; we might as well list them,
-5, -1, 1, 5, -5/2, -1/2, 1/2, 5/2, -5/4, -1/4, 1/4, 5/4
We just need to try these until we find a root or exhaust the list.
f(-5) = 4(-5)³ - 7(-5)² - 11(-5) + 5 = -615
f(-1) = -4 - 7 + 11 + 5 = 5
f(1) = 4 - 7 - 11 + 5 = -9
f(5) = 4(5)³ - 7(5²) - 11(5) + 5 = 275
f(-5/2) = 4(-5/2)³ - 7(-5/2)² - 11(-5/2) + 5 = -295/4
f(-1/2) = 4(-1/2)³ - 7(-1/2)² - 11(-1/2) + 5 = 33/4
f(1/2) = 4(1/2)³ - 7(1/2)² - 11(1/2) + 5 = -7/4
f(5/2) = 4(5/2)³ - 7(5/2)² - 11(5/2) + 5 = -15/4
f(-5/4) = 4(-5/4)³ - 7(-5/4)² - 11(-5/4) + 5 = 4(-125/64) - 7(25/16) + 55/4 + 5
= -125/16 - 175/16 + 220/16 + 80/16 = 0
Found one! x=-5/4 is a root. Phew, that was a lot of work.
Once we have a rational root we can factor. We know (x - (-5/4)) is a factor of our polynomial. We can make things easier by clearing the fraction and factoring out
4x + 5
We'll do long division,
x² - 3x + 1
4x + 5 | 4x³ - 7x² - 11x + 5
4x³ + 5x²
-12x² - 11x + 5
-12x² -15x
4x + 5
4x + 5
0
So we found
4x³ - 7x² - 11x + 5 = (4x + 5)(x² - 3x + 1)
That quadratic doesn't seem to factor; we'll get the remaining roots by the quadratic formula:
x = -5/4 or x= (1/2)(3 ± √9-4(1)) = (1/2)(3 ± √5)
x=-5/4 or x=(3 + √5)/2 or x=(3 - √5)/2
