Solve the polynomial equation, to obtain the first root.
x^3+7x^2+19x+13
and
2x^4-19x^3+74x^2-127x=0
I don't know how to start these, or finish them
The first thing you wrote is not an equation. x = 0 is an obvious root of the second equation.
To solve polynomial equations, we use a combination of factoring, the rational root theorem, and synthetic division. Let's go through each equation step by step and find the first root for each.
1. x^3 + 7x^2 + 19x + 13:
To find the first root, we'll start by looking for potential rational roots using the rational root theorem. The rational root theorem states that if a polynomial equation has a rational root, it must be in the form of p/q, where p is a factor of the constant term (in this case, 13) and q is a factor of the leading coefficient (in this case, 1).
The potential rational roots are:
±1, ±13 (since factors of 13 are 1 and 13, and factors of 1 are 1 and -1).
Now, we'll use synthetic division to test these potential roots. Synthetic division involves finding the remainder when the polynomial is divided by a specific root.
Let's start with x = -1:
-1 | 1 7 19 13
| −1 -6 −13
| 1 6 13 0
Since the remainder is 0, x = -1 is a root of the polynomial.
Now we have a quadratic equation left (the result after synthetic division): x^2 + 6x + 13. To find the remaining roots, we can use the quadratic formula.
The quadratic formula is: x = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic equation x^2 + 6x + 13, a = 1, b = 6, and c = 13.
x = (-6 ± √(6^2 - 4(1)(13))) / (2 * 1)
x = (-6 ± √(36 - 52)) / 2
x = (-6 ± √(-16)) / 2
Since we have a negative value under the square root, this quadratic equation has imaginary roots. Therefore, the first (and only) real root of the polynomial equation x^3 + 7x^2 + 19x + 13 is x = -1.
2. 2x^4 - 19x^3 + 74x^2 - 127x = 0:
We'll follow the same process as above.
First, we need to factor out common terms. In this case, we can factor out an x:
x(2x^3 - 19x^2 + 74x - 127) = 0
Now, we have x = 0 as a root.
Next, we can work with the remaining cubic equation: 2x^3 - 19x^2 + 74x - 127.
We'll once again use the potential rational roots from the rational root theorem:
±(1, 127) and ±(127, 2).
Let's test x = -1 using synthetic division:
-1 | 2 -19 74 -127
| -2 21 -95 221
| 2 -21 95 -348
The remainder is not 0, so x = -1 is not a root.
If we test the other potential roots, we find that none of them are roots either. Therefore, we can conclude that there is no rational root for this equation.
To find the roots, we can use numerical methods such as graphing or software tools like calculators or computer programs.
In summary, for the equation 2x^4 - 19x^3 + 74x^2 - 127x = 0, we found x = 0 as a root through factoring. However, we couldn't find the first non-zero root (real or rational) using the rational root theorem. To find the remaining roots, numerical methods or more advanced techniques need to be employed.