Step-by-step explanation:
<em>"Determine the number and type of roots for the equation using one of the given roots. Then find each root. (inclusive of imaginary roots.)"</em>
Given one of the roots, we can use either long division or grouping to factor each cubic equation into a binomial and a quadratic. I'll use grouping.
Then, we can either factor or use the quadratic equation to find the remaining two roots.
1. x³ − 7x + 6 = 0; 1
x³ − x − 6x + 6 = 0
x (x² − 1) − 6 (x − 1) = 0
x (x + 1) (x − 1) − 6 (x − 1) = 0
(x² + x − 6) (x − 1) = 0
(x + 3) (x − 2) (x − 1) = 0
The remaining two roots are both real: -3 and +2.
2. x³ − 3x² + 25x + 29 = 0; -1
x³ − 3x² + 25x + 29 = 0
x³ − 3x² − 4x + 29x + 29 = 0
x (x² − 3x − 4) + 29 (x + 1) = 0
x (x − 4) (x + 1) + 29 (x + 1) = 0
(x² − 4x + 29) (x + 1) = 0
x = [ 4 ± √(16 − 4(1)(29)) ] / 2
x = (4 ± 10i) / 2
x = 2 ± 5i
The remaining two roots are both imaginary: 2 − 5i and 2 + 5i.
3. x³ − 4x² − 3x + 18 = 0; 3
x³ − 4x² − 3x + 18 = 0
x³ − 4x² + 3x − 6x + 18 = 0
x (x² − 4x + 3) − 6 (x − 3) = 0
x (x − 1)(x − 3) − 6 (x − 3) = 0
(x² − x − 6) (x − 3) = 0
(x − 3) (x + 2) (x − 3) = 0
The remaining two roots are both real: -2 and +3.
<em>"Find all the zeros of the function"</em>
For quadratics, we can factor using either AC method or quadratic formula. For cubics, we can use the rational root test to check for possible rational roots.
4. f(x) = x² + 4x − 12
0 = (x + 6) (x − 2)
x = -6 or +2
5. f(x) = x³ − 3x² + x + 5
Possible rational roots: ±1/1, ±5/1
f(-1) = 0
-1 is a root, so use grouping to factor.
f(x) = x³ − 3x² − 4x + 5x + 5
f(x) = x (x² − 3x − 4) + 5 (x + 1)
f(x) = x (x − 4) (x + 1) + 5 (x + 1)
f(x) = (x² − 4x + 5) (x + 1)
x = [ 4 ± √(16 − 4(1)(5)) ] / 2
x = (4 ± 2i) / 2
x = 2 ± i
The three roots are x = -1, x = 2 − i, x = 2 + i.
6. f(x) = x³ − 4x² − 7x + 10
Possible rational roots: ±1/1, ±2/1, ±5/1, ±10/1
f(-2) = 0, f(1) = 0, f(5) = 0
The three roots are x = -2, x = 1, and x = 5.
<em>"Write the simplest polynomial function with integral coefficients that has the given zeros."</em>
A polynomial with roots a, b, c, is f(x) = (x − a) (x − b) (x − c). Remember that imaginary roots come in conjugate pairs.
7. -5, -1, 3, 7
f(x) = (x + 5) (x + 1) (x − 3) (x − 7)
f(x) = (x² + 6x + 5) (x² − 10x + 21)
f(x) = x² (x² − 10x + 21) + 6x (x² − 10x + 21) + 5 (x² − 10x + 21)
f(x) = x⁴ − 10x³ + 21x² + 6x³ − 60x² + 126x + 5x² − 50x + 105
f(x) = x⁴ − 4x³ − 34x² + 76x − 50x + 105
8. 4, 2+3i
If 2 + 3i is a root, then 2 − 3i is also a root.
f(x) = (x − 4) (x − (2+3i)) (x − (2−3i))
f(x) = (x − 4) (x² − (2+3i) x − (2−3i) x + (2+3i)(2−3i))
f(x) = (x − 4) (x² − (2+3i+2−3i) x + (4+9))
f(x) = (x − 4) (x² − 4x + 13)
f(x) = x (x² − 4x + 13) − 4 (x² − 4x + 13)
f(x) = x³ − 4x² + 13x − 4x² + 16x − 52
f(x) = x³ − 8x² + 29x − 52
is the mean, that we already found at point 1), and N=4. Substituting data, we have:
