SOLUTION: What are all the rational zeros of these polynomials?y=2x^3-10x^2-8x+40 and y=3x^3-24x^2-48x+384

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Question 89876: What are all the rational zeros of these polynomials?y=2x^3-10x^2-8x+40 and y=3x^3-24x^2-48x+384
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
What are all the rational zeros of these polynomials?
y=2x^3-10x^2-8x+40
I graphed it to find a zero at x=-2.
---------------------
Then used synthetic division as follows:
-2....2....-10....-8....40
........2.....-14....20..|..0
================
Remainder: 0
Quotient: 2x^2-14x+20
==========
Find the zeroes of the Quotient:
2(x^2-7x+10)=0
x^2-7x+10=0
(x-5)(x-2)=0
x=5 or x=2
=================
and y=3x^3-24x^2-48x+384
Graphing shows a zero at x=-4
Use synthetic division to find the other two zeroes:
-4)....3....-24....-48....384
.........3.....-36....96...|..0
==========
Remainder: 0
Quotient: 3x^2-36x+96
Find the zeroes of the quotient:
3(x^2-12x+32)=0
x^2-12x+32=0
(x-4)(x-8)=0
x=4 or x=8
================
Cheers,
Stan H.
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
"What are the rational zeros of y=2x^3-10x^2-8x+40?"

Any rational zero can be found through this equation

where p and q are the combinations of the last and first coefficients


So let's list the factors of p (which is 40)



Now let's list the factors of q (which is 2)



Now let's divide them




Now simplify

These are all the possible zeros of the function




To save time, I'm only going to use synthetic division on the possible zeros that are actually zeros of the function.
Otherwise, I would have to use synthetic division on every possible root (there are 32 possible roots, so that means there would be at most 32 synthetic division tables).
However, you might be required to follow this procedure, so this is why I'm using this procedure


If you're not required to follow this procedure, simply use a graphing calculator to find the roots




So we find that the equation has a root at . This means the test zero is -2



Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
-2|2-10-840
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
-2|2-10-840
|
2

Multiply -2 by 2 and place the product (which is -4) right underneath the second coefficient (which is -10)
-2|2-10-840
|-4
2

Add -4 and -10 to get -14. Place the sum right underneath -4.
-2|2-10-840
|-4
2-14

Multiply -2 by -14 and place the product (which is 28) right underneath the third coefficient (which is -8)
-2|2-10-840
|-428
2-14

Add 28 and -8 to get 20. Place the sum right underneath 28.
-2|2-10-840
|-428
2-1420

Multiply -2 by 20 and place the product (which is -40) right underneath the fourth coefficient (which is 40)
-2|2-10-840
|-428-40
2-1420

Add -40 and 40 to get 0. Place the sum right underneath -40.
-2|2-10-840
|-428-40
2-14200

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 3 coefficients (2,-14,20) form the quotient




So


So that means factors to


Factor completely

So the zeros are

, or

==========================================================================

"What are the rational zeros of y=3x^3-24x^2-48x+384?"

Any rational zero can be found through this equation

where p and q are the combinations of the last and first coefficients


So let's list the factors of p (which is 384)



Now let's list the factors of q (which is 3)



Now let's divide them




Now simplify

These are all the possible zeros of the function




To save time, I'm only going to use synthetic division on the possible zeros that are actually zeros of the function.
Otherwise, I would have to use synthetic division on every possible root (there are 64 possible roots, so that means there would be at most 64 synthetic division tables).
However, you might be required to follow this procedure, so this is why I'm using this procedure


If you're not required to follow this procedure, simply use a graphing calculator to find the roots



So we find that the equation has a root at . This means the test zero is -4


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
-4|3-24-48384
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 3)
-4|3-24-48384
|
3

Multiply -4 by 3 and place the product (which is -12) right underneath the second coefficient (which is -24)
-4|3-24-48384
|-12
3

Add -12 and -24 to get -36. Place the sum right underneath -12.
-4|3-24-48384
|-12
3-36

Multiply -4 by -36 and place the product (which is 144) right underneath the third coefficient (which is -48)
-4|3-24-48384
|-12144
3-36

Add 144 and -48 to get 96. Place the sum right underneath 144.
-4|3-24-48384
|-12144
3-3696

Multiply -4 by 96 and place the product (which is -384) right underneath the fourth coefficient (which is 384)
-4|3-24-48384
|-12144-384
3-3696

Add -384 and 384 to get 0. Place the sum right underneath -384.
-4|3-24-48384
|-12144-384
3-36960

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 3 coefficients (3,-36,96) form the quotient




So


So that means factors to


Factor completely

So the zeros are

, or
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