SOLUTION: Find all the rational zeros using the Rational Root Theorem and then graph.
𝑓(𝑥) = (12𝑥^3) + (73𝑥^2) + 5𝑥 − 6
The Rational Root th
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Question 1125475: Find all the rational zeros using the Rational Root Theorem and then graph.
𝑓(𝑥) = (12𝑥^3) + (73𝑥^2) + 5𝑥 − 6
The Rational Root theorem really confused me on this one, so thank you for the help!!
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The first coefficient is 12 and the second is 6
p are factors of the constant 6, which would be +/-1,2,3,6
q are factors of the leading coefficient 12, which are +/-1,2,3,4,6,12
p/q can be a root, so there are multiple possibilities, the integers alone are +/-6,3,2,1.
since factors involving integers, if found, will make the remainder easier, start with them.
one can use synthetic division, but also one may see if the given root makes the polynomial's value 0.
1 clearly does not, for f(1) is not 0. Indeed, any positive integer can't be a root.
with synthetic division, try -6
12---73----5--- -6
12---1---- -1---- 0
(x+6) and (12x^2+x-1) are factors
factor 12x^2+x-1 looking at fractional roots like (1/3 or 1/4). f(1/4) works, so (4x-1) is a factor. That means (3x+1) must also be a factor.
rational zeros are -6,-1/4, and +1/3.
Answer by greenestamps(13215) (Show Source): You can put this solution on YOUR website!
According to the rational roots theorem, the possible roots are
+ or - {1, 2, 3, 6, 1/2, 3/2, 1/3, 2/3, 1/4, 3/4, 1/6, 1/12}
There are many techniques you could use to find the actual roots among all those possibilities. I don't know if it is in general possible to know by looking at a particular polynomial function which method(s) might turn out to be the easiest to use.
So let's think about some things we COULD do....
Nearly always, the first thing to try in finding the roots is to test the possible integer values by evaluating the expression. In this example, trying the positive integer values should quickly show that there are no roots greater than 1.
Testing the integer values also shows that
f(-1) > 0
f(0) < 0
f(1) > 0
Since polynomial functions are continuous, that means there is a root between -1 and 0 and another between 0 and 1.
Testing roots using synthetic division is always an option; but testing fractional roots is often awkward. So let's think about what other tools we might use to find the roots.
One thing that is often helpful is the sum and product of the roots. In this example, the sum of the roots is and the product of the roots is .
At this point, we know...
(1) the sum of the roots is -73/12, which is about -6
(2) the product of the roots is 1/2
(3) there is a root between -1 and 0
(4) there is a root between 0 and 1
The sum of the two roots close to 0 is going to be close to 0; since the sum of all three roots is about -6, the third root has to be -6.
Once we know one root, the usual way to finish the problem would probably be to find the others by using synthetic division, giving us a quadratic equation that we can solve by factoring.
But let's continue with the path we are on to see if we can easily finish the problem from here.
And in fact we can. The sum of all three roots is -73/12, and one of the roots is -6. That means the sum of the other two roots (the ones close to 0) is -1/12. And a look at the list of possible rational roots quickly finds -1/12 = 1/4 - 1/3 -- so the other two roots are 1/4 and -1/3.
And as a check we can verify that the product of all three roots is 1/2:
So the roots of the function are -6, -1/3, and 1/4.
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