Question 71792: someone please show or explain the easiest way to understand this
Find the polynomial with leading coefficient 1 and degree 3 that has -1, 1, and 3 as roots.
Possible answers:
x3 - 3x2 -x +3
.x3 -3x2 + x -3
.x3 +3x2 +x +3
.x3 + 3x2 +x +3
Thanks to anyone who can help me
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the polynomial with leading coefficient 1 and degree 3 that has -1, 1, and 3 as roots
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If 1 is a root, x=1, or x-1=0
If -1 is a root, x=-1 or x+1=0
If 3 is a root, x=3, or x-3=0
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Multiply those three results and you get
(x-1)(x+1)(x-3)=0
That is the polynomial you want.
If you multiply it out you get:
=(x^2-1)(x-3)
=x^3-3x^2-x+3
If you are asked for the equation that has those roots and leading coefficient
of 1 you write: y=x^3-3x^2-x+3
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Cheers,
Stan H.
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website! someone please show or explain the easiest way to understand this
Find the polynomial with leading coefficient 1 and degree 3 that
has -1, 1, and 3 as roots.
Possible answers:
x³ - 3x² - x + 3
x³ - 3x² + x - 3
x³ + 3x² + x + 3
x³ + 3x² + x + 3
Thanks to anyone who can help me
Two methods. The first method works only for multiple choice
tests, but not on a test if you have to show your work.
The first way is to plug them all three, -1, 1, and 3, in and
see which one gives you 0 for all three numbers
Plug x = -1 in the first answer:
x³ - 3x² - x + 3
(-1)³ - 3(-1)² - (-1) + 3
-1 - 3 + 1 + 3
0
Plug x = 1 in the first answer:
x³ - 3x² - x + 3
(1)³ - 3(1)² - (1) + 3
1 - 3 - 1 + 3
0
Plug x = 3 in the first answer:
x³ - 3x² - x + 3
(3)³ - 3(3)² - (3) + 3
27 - 27 - 1 + 3
0
Wow, they all three gave 0 for the answer, so that's the
correct choice!
However, on another test it might not have been the first
one.
If you plug -1, 1, and 3 in the 2nd choice you get
-8,-4, and 0. They aren't all 0, so that's not the
answer.
If you plug -1, 1, and 3 in the 3rd choice you get
4, 8, and 60. They aren't all 0, so that's not the
answer.
Notice that the 4th choice is identical to the 3rd choice.
Now for method 2:
This is the real way to work the problem.
Solve the problem in reverse.
We begin with what we would have ended up with if we
had solved it:
x = -1, x = 1, and x = 3
Now ask: What would we have set = 0 to get those, if we had
solved it?
x + 1 = 0, x - 1 = 0, and x - 3 = 0
Next ask: Then what must the factored form of the polynomial
equation have been to have given us that?
(x + 1)(x - 1)(x - 3) = 0
Next ask: What must the original polynomial equation have been?
Let's multiply the left side out and see
[(x + 1)(x - 1)](x - 3) = 0
[x² - x + x - 1)(x - 3) = 0
(x² - 1)(x - 3) = 0
x² - 3x² - x + 3 = 0
That's why it's the first choice.
Edwin
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