Given one root of the equation, find the others.
a^3-3a^2+a+5=0
one root= 2-1
I have no clue on this one.....
2007-08-27 15:31:06 · 3 answers · asked by aroddick46 1 in Science & Mathematics ➔ Mathematics
If one root is -1, then you need to consider (a + 1) because if a+ 1 = 0, then a is -1.
Divide a^3-3a^2+a+5 by a+1. You'll get a^2-4a+5. Using the quadratic formula, you'll get no solution for a^2-4a+5. So, -1 is actually the only solution.
2007-08-27 15:47:00 · answer #1 · answered by SoulDawg 4 UGA 6 · 1⤊ 0⤋
i don't know what you mean when you said 2-1 but what i found the factors to be are (a+1) and (a^2-4a+5) so that means (a+1)(a^2-4a+5)=0 So, the root is only -1 because the equation (a+1)=0 when solved for a gives a= -1 but the second factor can't be solved for a. So i guess the only root is -1.
2007-08-27 22:49:43 · answer #2 · answered by mazesh 2 · 1⤊ 0⤋
If one root = 2.1
then use long division of a-2.1 into a^3-3a^2+a+5=0
The result will give you a quadratic you may easily solve with known equations or factoring.
2007-08-27 22:35:55 · answer #3 · answered by idiot 3 · 1⤊ 0⤋