1. Writing Polynomial Functions
I. Writing factors from zeros (and conjugate pairs).
A) Zeros are written in the form (x – h).
1) Where “h” is the zero. (Change the sign)
B) Remember: conjugates come in pairs. So if you have one imaginary zero, you have another of the opposite sign. Same with radicals.
Example: zeros are 6, –8, 2i write as a factored polynomial.
y = (x – 6) (x + 8) (x – 2i) (x + 2i) (the i’s are conjugate pairs)
C) Writing complex zeroes as factors. Same rule.
Example: zeros are 2 ± 3i, write as linear factors.
y = (x – [2 + 3i]) (x – [2 – 3i])  y = (x – 2 – 3i)(x – 2 + 3i)

2. Writing Polynomial Functions
II. Writing factors given the degree and the leading coefficient.
A) Remember: zeros are written in the form (x – h).
1) Where “h” is the zero. (Change the sign)
B) But we want the leading term to be the value “a”.
1) So write it as … a(x – h1)(x – h2)(etc.)
2) Distribute the “a” term to any ONE of the (x – h).
3) If you need it in standard form, FOIL all the (x – h)s.
Example: Leading coeff = 5; Degree 3; zeroes of 2, 3/5 & –½
5(x – 2) (x – 3/5) (x + ½) distribute the 5 to (x – 3/5) to get …
(x – 2) (5x – 3) (x + ½) and then FOIL (x – 2)(5x – 3) to get …
(5x2 – 13x + 6)(x + ½) and distribute for standard form of …
f(x) = 5x3 – 21/2x2 – ½x + 3

3. Writing Polynomial Functions
III. Writing factors from a graph.
A) Find the zeroes of the graph (the x-intercepts).
1) Change the signs to get them into (x – h) form.
B) Look at the x-intercepts. Did it bounce or pass through?
1) Bounces mean (x – h)2 [or other even exponent].
2) Pass through means simply (x – h) if the graph went
straight through the x-intercept.
a) If the graph “flattened out” at the x-intercept before
passing through then you have (x – h)odd.
C) To get the leading coefficient “a”, write the function in factored form with the “a” in front of the factors, then find any point on the graph and plug those values in for x and y. Solve for “a” to find your leading coefficient.
D) If you don’t know any x-intercepts, try to write the equation in shifted format. y = A f (B(x – C)) + D.

4. Writing Polynomial Functions
Example 1: A bounce at -3, a pass thru at 5 and the y-intercept is (0,-5). Write the equation in factored form.
The bounce means an even exponent (2).
y = a(x + 3)2(x – 5) plug in (0,-5)  -5 = a(3)2(-5)  -5 = -45a
solve for a and you get 1/9 so f(x) = 1/9(x + 3)2(x – 5)
Example 2: A bounce at 4, a “flattened” pass thru at -2 and the y-intecept is (0,–7). Write the equation in factored form.
The bounce is (x-h)even, the “flattened” pass thru is (x-h)odd
y = a(x–4)2(x+2)3 plug in (0,-7)  -7 = a(-4)2(2)3  -7 = 128a
solve for a and you get = –1/16 so f(x) = –1/16(x–4)2(x+2)3
Example 3: The graph has no known x-intercepts, but you know the graph is cubic (opposite end directions) and there is a y-int at (0,19) and “flattened” value at (-2,-5).
Shift rules y = a(x+2)3 – 5  plug in (0 , 19)  19 = a(2)3 – 5, solve for and you get so f(x) = 3(x + 2)3 – 5