Answer the following questions about the equation below
12x^3 +77x^2 -48x+7=0
a. List all rational roots that are possible according to the Rational Zero Theorem:
To get them we look at the last number and the leading coefficient, 7 and 12
We take their absolute values, |7| = 7 and |12| = 12
We list the factors of 7 {1,7}
We list all the factors of 12 {1,2,3,4,6,12}
We make all the fractions that have a factor of 7 as a numerator and
a factor of 12 as a denominator:
We reduce any of them that will reduce, and if there were any
duplicates we would eliminate them, and place the "positive or
negative" sign, ±, before each:
These are all the potential rational zeros.
b. Use synthetic division to test several possible rational roots in order
to identify one actual root:
We try x=1
1 | 12 77 -48 7
| 12 89 41
12 89 41 48
No that leaves remainder 48 not 0.
We try x=-1
-1 | 12 77 -48 7
| -12 -65 113
12 65 -113 120
No that leaves remainder 120 not 0.
We try x=1/2
1/2 | 12 77 -48 7
| 6 83/2 -13/4
12 83 -13/2 15/4
No that leaves remainder 3/4 not 0.
We try x=-1/2
-1/2 | 12 77 -48 7
| -6 71/2 -25/4
12 71 -25/2 3/4
No that leaves remainder 3/4 not 0.
We try x=1/3
1/3 | 12 77 -48 7
| 4 27 -7
12 81 -21 0
Hoo-pee! finally! that leaves remainder 0.
One rational root of the given equation is:
1/3
c. Use the root from part(b) to solve the equation
Since we know that x=1/3 is a zero and that is
equivalent to x-1/3=0, we know that x-1/3 is a factor,
and the other numbers on the bottom of that last
synthetic division are the coefficients of the divisor,
which is of one less degree than the given equation,
so we now have factored the left side of
12x^3+77x^2-48x+7 = 0
as (x-1/3) times a polynomial of degree 3-1 or 2,
a quadratic:
(x-1/3)(12x^2+81x-21) = 0
We can factor 3 out of that quadratic and get
(x-1/3)(3)(4x^2+27x-7) = 0
We may as well multiply that 3 into (x-1/3) and get
(3x-1)(4x^2+27x-7) = 0
We can factor that quadratic and the final factorization
is:
(3x-1)(4x-1)(x+7) = 0
Use the zero-factor principle:
3x-1=0; 4x-1=0; x+7=0
3x=1; 4x=1; x=-7
x=1/3; x=1/4
So the three zeros are 1/3, 1/4, and -7.
The solution set is { }
{1/3, 1/4, -7}
Edwin
