by Prof Dr. S. Gramlich
4/7/19
in conjuction with "Alg & Trig," 5th ed, by Blitzer, section 3.4
Steps for finding roots or zeros of
polynomial with degree n>=3 when cannot factor:
1) degree n = # roots {use Root Property p.382}.
2) Use Descarte's Rule of Signs (p.384) to find # of + & - real zeros.
3) divide factors of constant term by factors of leading coefficient {Rational Zero Thereom, p.377}.
4) Use synthetic division on EACH step3 quotients as divisors into the original polynomial until find a zero remainder.
{if don't get a zero remainder then roots may be irrational or imaginary.}
5) if get a zero remainder then divisor c in step4 is a factor (x-c) with its respective quotient as another factor of the original polynonial.
6) find roots (x-intercepts) on the quotient from step5 by setting = 0 and solving for x
(i.e. by factoring, quadratic formula, completing the square, or doing steps 1-5 again).
when solving for x in a function f(x) polynomial expression, refer to solution as "zero's."
when solving for x in polynomial equation set = 0, refer to solution as "roots" or "x coordinates of x-intercepts."
EXAMPLE
Similar to 3.4 UtryIt#5 or 3.4 #23 (from "Alg & Trig," 5th ed, by Blitzer)
find roots of x^4 +2x^3 -5x^2 -8x +4 =0 ?
leading coefficient 1 (from x^4) and constant term 4.
STEP1: degree 4 => 4 roots
if decide to graph (not necessary),
p. 350 leading coeff test:
ldg coeff = +1 and degree =4 even so graph rises to left & reight.
STEP2: Descartes Rule
for f(x) =x^4 +2x^3 -5x^2 -8x +4
f(+)= + + - - +, 2 sign changes => 2 +real or 0 +real
f(-)= + - - + +, 2 sign changes => 2 -real or 0 -real
STEP3: possible rational zeros:
(p factors of constant 4) / (q factors of leading coefficient 1)
p/q = (+-1, +-2, +-4) /(+-1)
= +-1, +-2, +-4
STEP4: synthetic division
1] 1 2 -5 -8 4
1 3 -2 -20
-------------------
1 3 -2 -10 R -16
-1] 1 2 -5 -8 4
-1 -1 +6 +2
----------------
1 1 -6 -2 R6
-2] 1 2 -5 -8 4
-2 0 +10 -4
----------------
1 0 -5 +2 R0
STEP5:
since Remainder = 0 for divisor -2, orig polynomial equation becomes
(x -c)*(quotient with respect to x-c) = 0
[x-(-2)](+1x^3 +0x^2 -5x +2) = 0
(x+2)(x^3 -5x +2) = 0
STEP6:
Since Descartes Rule => 2 -real#s there must be 1 more -real root. By further abductive reasoning, since -1 did not yield negative Remainder above then -4 is a possible root (unless -2 is of multiplicity 2). So employ synthetic division with -4 as divisor into remaining factor
1x^3 +0x^2 -5x +2.
-4] 1 0 -5 2
-4 16 -44
----------------
1-4 11 R-42
remainder -42 is not zero so -4 is not a root. So try synthetically dividing -2 into 1x^3 +0x^2 -5x +2.
-2] 1 0 -5 2
-2 4 +2
----------------
1-2 -1 R4
nope. wow, neither -4 or -2 yielded zero remainders so the 2nd -real root could be irrational.
Consequently, this leaves possibility that there are still 2 +real roots.
Since +1 did not yield Remainder 0 above, try +2 or +4.
2] 1 0 -5 2
2 4 -2
----------------
1 2 -1 R0
yes! so orig polynomial further breaks down from (x+2)(x^3 -5x +2) to:
(x+2)(x-2)(x^2 +2x -1) = 0
x^2 +2x -1 is not factorable so use the quadratic equation:
x = {-b +-sqrt[(b^2 -4ac]} /(2*a)
x = {-2 +-sqrt[(-2)^2 -4(1)(-1)]} /(2*1)
=[-2 +-sqrt(4+4)] /2
=[-2 +-sqrt(8)] /2
=[-2 +-sqrt(4*2)] /2
=[-2 +-2sqrt(2)] /2
=-2/2 +-2[sqrt(2)]/2
=-1 +-sqrt(2)
so the 4 roots of (x+2)(x-2)(x^2 +2x -1) = 0
are -2, +2, -1 +sqrt(2), -1 -sqrt(2)