Answer:
We know that P(x) = 30*x^3 + k*x^2 + 1
a) If (3x - 1) is a factor, let's find the value of k.
Because the degree of the original polynomial is 3, and the degree of the factor is 1, we should multiply it by a polynomial of degree 2.
Then:
30*x^3 + k*x^2 + 1 = (3*x - 1)*(a*x^2 + b*x + c)
30*x^3 + k*x^2 + 1 = 3*a*x^3 + 3*b*x^2 + 3*c*x - a*x^2 - b*x - c
30*x^3 + k*x^2 + 1 = 3*a*x^3 + (3*b - a)*x^2 + (3*c - b)*x - c
The terms with the same exponent should be equal, then:
30*x^3 = 3*a*x^3
30 = 3*a
30/3 = a = 10
k*x^2 = (3*b - a)*x^2
k = (3*b - 10)
0*x = (3*c - b)*x
0 = (3*c - b)
b = 3*c
and
1 = -c
then:
c = -1
Replacing this in the equation: b = 3*c
we get: b = -3
Replacing this in the equation for k, we get:
k = (3*b - 10) = (3*-3 - 10) = -19
k = -19
And we can write:
P(x) = (3*x - 1)*(10*x^2 - 3*x - 1)
b) Now we want to write P(x) as a product of its linear factors.
Then we need to factorize the quadratic part:
We need to solve
10*x^2 - 3*x - 1 = 0
We can use the Bhaskara's equation to get the solutions:
[tex]x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4*10*(-1)} }{2*10} = \frac{3 \pm 7}{10}[/tex]
Then the two zeros of that polynomial are
x = (3 + 7)/10 = 1
x = (3 - 7)/10 = - 4/10 = -2/5
So we can write the polynomial as:
10*(x - 1)*(x - (-2/5))
Replacing this in the equation of P(x) we get:
P(x) = (3*x - 1)*10*(x -1)*(x + 2/5)
= 10* (3*x - 1)*(x -1)*(x + 2/5)
So P(x) is written as a product of lineal factors.
C) The zeros of the polynomial are the values of x such that one of the factors becomes equal to zero.
For the first factor, we have a zero at x = 1/3
for the second, we have a zero at x = 1
for the third one, we have a zero at x = -2/5
