2. Writing Equations from Roots
(Remember that the conjugate is also a root)
Double root at -2/3, root at 0, and root at 2i
So -2i is also a root
(x-0)(x + 2/3) (x + 2/3) (x – 2i) (x + 2i)
(x)(3x + 2) (3x + 2) (x – 2i) (x + 2i)
(x)(9x2 + 12x + 4)(x – 4i2)(x2 + 4)
(x)(9x4 +12x x2 + 48x + 16)
Y=(9x5 + 12x x x2 + 16x)

3. 3+4i and 2 are roots (x – 2) (x – (3 + 4i)(x – (3 – 4i))
+16
(x – 2) (x2 -3x + 4ix – 3x + 9 – 12i – 4ix + 12i – 16i2 )
(x – 2) (x2 -6x + 25)
(x3 -6x2 + 25x – 2x x- 50)
Y = x3 - 8x2 + 37x - 50

4. Possible Rational Zeros
Possible rational zeros: Factors of Constant
Factors of lead. coeff
Ex: Find possible rational zeros of: 3x4 + ………
1,2,7,14
1, 3
+ { 1, 2, 7, 14, 1/3, 2/3, 7/3, 14/3}

5. Division X6 + ………. 12 = (x + ~)(x5 + ………) (x+~)(x4 + ……)
150 75
3 25
5 5
X ………. 12
= (x + ~)(x5 + ………)
(x+~)(x4 + ……)
(x+~)(x3 + ……)
(x + ~)(x2 + ~~~~~)
Quad. formula
If it’s x6 then find 4 zeros and divide 4 times until x2
If it’s x3 then find 1 zero and divide 1 time until x2

6. Finding possible rational zeros, zeros and factors
1,2,5,10,2550 1
+1, + 2 +,5, + 10, + 50,+ 25
Y = x3 -8x2 + 37x - 50
Graph to find root(s)… Root is at 2 so 1 real, 2 imaginary
Factors
(x – 2)(X2 -6x + 25)
This means that f(2) = 0
X2 -6x + 25
Not factorable so use Q.formula…… i and 3 – 4i
Zeros: 2, 3 + 4i, 3 – 4i

7. G(x) = 48x4 - 52x3 + 13x - 3 Possible Rational : 1, 3
Zeros ,2,3,4,6,8,12,16, 24,48
+{1, 1/2, 1/3, 1/4, 1/6, 1/8, 1/12, 1/6,1/24, 1/48, 3, 3/2, ¾, 3/8, 3/16}
Graph and zeros are at -1/2, ½, ¾, 1/3
-1/
This is 48x3 so keep going ½
(x + ½)(x – ½)4(4x – 3)(3x – 1)
(2x + 1)(2x – 1)(4x-3)(3x – 1)
48X x + 12  4(12x2 -13x + 3)
4(4x – 3)(3x – 1)  1/3 and ¾ or Q.F

8. Using synthetic to find f(~) Just use ~ in synthetic substitution
Given a factor (x + #), find other factors
Just use -~ in synthetic substitution
so it’s 5.
F(x) = x3 - 3x Find f(-2) using synthetic substitution
3x – 2 is a factor of 6x3 – 25x2 + 2x find other factors
so 3x – 2 = x = x = 2/3 2/
6x2 -21x – 12
3(2x2 – 7x – 4)
(2x + 1)(x – 4)