Question 980628: factor the polynomial then solve f(x)= 0. f(x) = x^3-3x^2-18x+40. x to the 3rd power minus 3x to the 2nd power minus 18x plus 40
Answer by josh_jordan(263)   (Show Source):
You can put this solution on YOUR website! To factor this polynomial, we need to first perform the rational zero test, to see all of the possible zeroes of this polynomial. Thankfully, our leading coefficient is one, so we do not have to worry about fractions. The possible zeroes are: 1, -1, 2, -2, 4, -4, 5, -5, 8, -8, 10, -10, 20, -20, 40, -40, which are all factors of the constant in our polynomial, 40.
If we use synthetic division we will find that one of our zeroes is 2. What is left over when we synthetically divide is 1 -1 20, which translates into:
x^2 - x - 20
So, we now have:
(x - 2)(x^2 - x - 20)
Now, we need to factor x^2 - x - 20, which we can do by finding out what factors of 20, when multiplied together, give us -20, and when added together, give us -1:
-1 x 20 = -20; -1 + 20 = 19 DOESN'T WORK
-2 x 10 = -20; -2 + 10 = 8 DOESN'T WORK
4 x -5 = -20; 4 - 5 = -1 WORKS!
So, x^2 - x - 20, factored, is
(x + 4)(x - 5)
Our final factored form of x^3 - 3x^2 - 18x + 40 is
(x - 2)(x + 4)(x - 5)
To find f(x) = 0 for that polynomial, simply replace all of the x's in the polynomial with 0:
(0)^3 - 3(0)^2 - 18(0) + 40 ----->
0 - 0 - 0 + 40 = 40
So, when f(x) = 0, x^3 - 3x^2 - 18x + 40 = 40
Final Answers:
1. x^3 - 3x^2 - 18x + 40, factored, is (x - 2)(x - 5)(x + 4)
2. x^3 - 3x^2 - 18x + 40, when f(x) = 0, is 40.
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