Question 1206505: Use the Rational Root Theorem to list all possible rational solutions. Then find the actual rational solutions.
2x^4 - 5x^3 - 17x^2 + 41x - 21 = 0
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20062)   (Show Source):
You can put this solution on YOUR website!
All candidates for rational solutions must have a numerator which
divides evenly into the constant term in absolute value |-21|=21;
and whose denominator divides evenly into the absolute value of
the leading coefficient |2|=2.
candidates for numerators of rational solutions: 1,3,7,21
candidates for denominators of rational solutions: 1,2
candidates for rational solutions:
1/1, 1/2, 3/1, 3/2, 7/1, 7/2, 21/1, 21/2
or
1, 1/2, 3, 3/2, 7, 7/2, 21, 21/2
Try 1, using synthetic division:
1 | 2 -5 -17 41 -21
| 2 -3 -20 21
2 -3 -20 21 0
That factors the left side as
The 0 on the bottom right tells us that 1 is a rational solution
Try 1 again in the quotient because it might have multiplicity
more than 1.
1 | 2 -3 -20 21
| 2 -1 -21
2 -1 -21 0
That factors the left side again as
Again the 0 on the bottom right tells us that 1 is a second solution
of at least multiplicity 2.
or
We already know how to finish factoring, for it is a quadratic:
x-1=0; 2x-7=0; x+3=0
x=1; 2x=7; x=-3
x=7/2
So:
1 is a rational solution with multiplicity 2.
7/2 is a rational solution with multiplicity of multiplicity 1.
-3 is a rational solution with multiplicity of multiplicity 1.
Edwin
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
p = 2 = leading coefficient
q = -21 = constant term at the end of the left hand side
Positive factors of q: 1, 3, 7, 21
Positive factors of p: 1, 2
We divide factors of q over factors of p to form the list of all possible rational roots. For now let's just list the positive version of each. The negative mirrored version will be introduced later.
I find it helps to use a table to organize things.
The columns represent the factors of q.
The rows represent factors of p.
Each inner cell will have q/p inside it.
This is what it looks like to fill out the table.
| 1 | 3 | 7 | 21 | | 1 | 1 | 3 | 7 | 21 | | 2 | 1/2 | 3/2 | 7/2 | 21/2 |
Those of course are just the positive possible rational roots.
List the plus minus of each item mentioned to get the full list of possible rational roots.
The possible rational roots are:
1, -1, 3, -3, 7, -7, 21, -21
1/2, -1/2, 3/2, -3/2, 7/2, -7/2, 21/2, -21/2
Then we plug each of them one at a time into the function
f(x) = 2x^4 - 5x^3 - 17x^2 + 41x - 21
to see which result in zero.
Let's try x = 1.
f(x) = 2x^4 - 5x^3 - 17x^2 + 41x - 21
f(1) = 2(1)^4 - 5(1)^3 - 17(1)^2 + 41(1) - 21
f(1) = 2(1) - 5(1) - 17(1) + 41(1) - 21
f(1) = 2 - 5 - 17 + 41 - 21
f(1) = 0
The result of zero tells us that x = 1 is indeed an actual root.
You can use a TI84 graphing calculator to notice the curve touches the x axis at x = 1.
Desmos and GeoGebra are alternative graphing tools.
What if we tried x = -1.
f(x) = 2x^4 - 5x^3 - 17x^2 + 41x - 21
f(-1) = 2(-1)^4 - 5(-1)^3 - 17(-1)^2 + 41(-1) - 21
f(-1) = 2(1) - 5(-1) - 17(1) + 41(-1) - 21
f(-1) = 2 + 5 - 17 - 41 - 21
f(-1) = -72
The nonzero result tells us that x = -1 is NOT a root.
I'll let you do the scratch work on the other possible roots.
You should find that: f(-3) = 0 and f(7/2) = 0, which proves that x = -3 and x = 7/2 = 3.5 are the other two roots.
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Answers:
Possible rational roots:
1, -1, 3, -3, 7, -7, 21, -21
1/2, -1/2, 3/2, -3/2, 7/2, -7/2, 21/2, -21/2
Actual rational roots: -3, 1, and 7/2
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