To find all the roots of the polynomial function f(x) = x^3 – 9x^2 + 26x – 24, given that one root is x = 2, we can use the Remainder Theorem.
The Remainder Theorem states that if a polynomial f(x) is divided by (x – r), the remainder of that division is f(r). When r is a root of f(x), then f(r) = 0 and the division will result in another polynomial.
Since we know x = 2 is a root, we can perform synthetic division to divide f(x) by (x – 2).
Using synthetic division:
- Set up the coefficients of f(x): 1 (for x^3), -9 (for x^2), 26 (for x), -24 (constant).
- Perform synthetic division using root 2:
2 | 1 -9 26 -24
| 2 -14 24
----------------------
1 -7 12 0
The result of the synthetic division is x^2 – 7x + 12, and the remainder is 0, confirming that x = 2 is indeed a root.
Next, we can factor the resulting quadratic polynomial: x^2 – 7x + 12 = 0.
To factor it, we look for two numbers that multiply to 12 (the constant term) and add to -7 (the coefficient of x). The numbers -3 and -4 work nicely:
Thus, we can factor:
- x^2 – 7x + 12 = (x – 3)(x – 4)
Now we set the factors equal to zero to find all roots of the original polynomial:
- x – 2 = 0 gives us x = 2
- x – 3 = 0 gives us x = 3
- x – 4 = 0 gives us x = 4
In conclusion, the roots of the function f(x) = x^3 – 9x^2 + 26x – 24 are x = 2, x = 3, and x = 4.