use the given zero to find the remaining zeros of the function
h(x)=x^4-15x^3+60x^2+70x-816
zero: 5-3i
To find the remaining zeros of the function, we will use the fact that complex zeros occur in conjugate pairs. Since the given zero is 5-3i, its conjugate is 5+3i.
So, we have two zeros: 5-3i and 5+3i.
To find the remaining zeros, we can use polynomial division or synthetic division to divide the function by the quadratic factor (x - (5-3i))(x - (5+3i)).
Using synthetic division:
1 | 1 -15 60 70 -816
------------------------------------
1 -14 46 116 | 300
The result of the division is the polynomial 1x^3 - 14x^2 + 46x + 116. We can now solve this polynomial to find the remaining zeros.
However, solving cubic equations can be quite complex and it may not always be possible to find exact solutions analytically. In such cases, numerical methods such as Newton's method or using graphing calculators or computer software can be used to approximate the zeros.
To find the remaining zeros of the function h(x) = x^4 - 15x^3 + 60x^2 + 70x - 816, given the zero 5 - 3i, we can use the conjugate pairs theorem.
The conjugate pairs theorem states that if a polynomial has complex roots, the complex roots occur in conjugate pairs. In other words, if a + bi is a root of a polynomial, then a - bi is also a root of that polynomial.
Since 5 - 3i is given as a root, its conjugate, 5 + 3i, must also be a root of the polynomial.
So, we add the factors for each root (zero) to the equation for h(x):
h(x) = (x - (5 - 3i))(x - (5 + 3i))(other factors)
Expanding the factors:
h(x) = ((x - 5) + 3i)((x - 5) - 3i)(other factors)
Using the difference of squares formula: a^2 - b^2 = (a + b)(a - b)
h(x) = ((x - 5)^2 - (3i)^2)(other factors)
h(x) = ((x - 5)^2 + 9)(other factors)
Now, expanding further:
h(x) = (x^2 - 10x + 25 + 9)(other factors)
Simplifying:
h(x) = (x^2 - 10x + 34)(other factors)
Now, we need to find the remaining roots by factoring the quadratic equation x^2 - 10x + 34.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
For the quadratic equation x^2 - 10x + 34, the coefficients are:
a = 1, b = -10, c = 34
Substituting these values into the quadratic formula:
x = (-(-10) ± √((-10)^2 - 4(1)(34))) / (2(1))
x = (10 ± √(100 - 136)) / 2
x = (10 ± √(-36)) / 2
Since the discriminant (-36) is negative, the roots will be complex numbers.
Calculating:
x = (10 ± √(-36)) / 2
x = (10 ± 6i) / 2
x = 5 ± 3i
Therefore, the remaining zeros of the function h(x) are 5 + 3i and 5 - 3i, which matches the given zero.