List all possible rational zeros for the polynomial below. Find all real zeros and factor completely. Please show all your work.
f(x) = 2x^4 + 19x^3 + 37x^2 - 55x - 75
Graph this first to find the zeros, hopefully exactly.
There is a process to this and it's why I am asking for help.
All rational roots will have a numerator which is a factor of 75, and a denominator which is a factor of 2.
±1 ±3 ±5 ±15 ±25 ±75
±1/2 ±3/2 ±5/2 ±15/2 ±25/2 ±75/2
Descartes' Rule of Signs says that there are at most
1 positive real root
2x^4 + 19x^3 + 37x^2 - 55x - 75 has one change of sign
2 negative real roots
2x^4 - 19x^3 + 37x^2 + 55x - 75
So, we either have a double root, or some complex roots.
A little synthetic division reveals that there are no real roots greater than 2 or less than -6.
A little more produces the root -1
(x+1)(2x^3 + 17x^2 + 20x - 75)
(x+1)(2x-3)(x+5)^2
I have done synthetic divion and come up with 3/2 and -1 but still have not gotten it down to a trinomial.
To find the possible rational zeros of a polynomial, we need to use the Rational Zero Theorem. According to the theorem, the possible rational zeros of a polynomial in the form f(x) = ax^n + bx^(n-1) + ... + k are all the numbers that can be expressed as p/q, where p is a factor of k (the constant term), and q is a factor of a (the coefficient of the highest degree term).
In our case, the polynomial f(x) = 2x^4 + 19x^3 + 37x^2 - 55x - 75 has a constant term of -75 and a leading coefficient of 2. Therefore, the possible rational zeros are all the numbers that can be expressed as p/q, where p is a factor of -75 and q is a factor of 2.
To find the factors of -75 and 2, we can use prime factorization.
The prime factors of -75 are:
-75 = -1 x 3 x 5 x 5
The prime factors of 2 are:
2 = 2
Now, let's combine the factors to find the possible rational zeros:
Possible factors for p: ±1, ±3, ±5, ±15, ±25, ±75
Possible factors for q: ±1, ±2
Therefore, the possible rational zeros for the given polynomial f(x) are:
±1/1, ±3/1, ±5/1, ±15/1, ±25/1, ±75/1, ±1/2, ±3/2, ±5/2, ±15/2, ±25/2, ±75/2
Now, to find the real zeros, we can use synthetic division or long division. Alternatively, we can use a graphic calculator or computer software to solve the equation.
If we perform synthetic division using one of the possible rational zeros, say x = 1, and we divide f(x) by (x - 1), the result is:
2 | 2 19 37 -55 -75
2 21 58 3 0
The result shows that 2x^3 + 21x^2 + 58x + 3 is the quotient, and the remainder is 0. This means that x - 1 is a factor of the given polynomial.
Now, we can factor the polynomial using this identified factor and rewrite it as:
f(x) = (x - 1)(2x^3 + 21x^2 + 58x + 3)
To continue factoring the remaining cubic polynomial, we can use other methods like factoring by grouping, synthetic division, or applying the Rational Zero Theorem again.