To find the roots of the polynomial equation x³ – 2x² + 10x – 136 = 0, we can use a combination of methods, including synthetic division and the Rational Root Theorem.
First, let’s figure out if there are any rational roots possible by applying the Rational Root Theorem. This theorem tells us that any rational root, in the form of p/q (where p divides the constant term and q divides the leading coefficient), could be among the factors of -136 (the constant term) over the factors of 1 (the leading coefficient).
The factors of -136 are: ±1, ±2, ±4, ±8, ±17, ±34, ±68, ±136. Next, we can test these potential rational roots by substituting them into the polynomial.
After testing, we find that when we plug in x = 4, it satisfies the equation:
4³ – 2(4)² + 10(4) – 136 = 64 – 32 + 40 – 136 = -64 ≠ 0
Continuing this process, we can test x = 8:
8³ – 2(8)² + 10(8) – 136 = 512 – 128 + 80 – 136 = 328 ≠ 0
However, when we try x = -4:
(-4)³ – 2(-4)² + 10(-4) – 136 = -64 – 32 – 40 – 136 = -272 = 0
Thus, x = -4 is one root. Now that we have found one root, we can perform synthetic division to divide the original polynomial by (x + 4):
After performing synthetic division, the quotient is:
x² – 6x + 34
Now we need to find the roots of this quadratic. We can use the quadratic formula, which is:
x = (-b ± √(b² – 4ac)) / 2a
For our case, a = 1, b = -6, and c = 34:
x = (6 ± √((-6)² – 4(1)(34))) / (2(1))
This simplifies to:
x = (6 ± √(36 – 136)) / 2
x = (6 ± √(-100)) / 2
Because we are dealing with a negative number under the square root, we will have complex roots:
x = (6 ± 10i) / 2
This gives us:
- x = 3 + 5i
- x = 3 – 5i
In conclusion, the three roots of the polynomial equation x³ – 2x² + 10x – 136 = 0 are:
- x = -4
- x = 3 + 5i
- x = 3 – 5i