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Polynomial p is defined by $$p(x) = x^3+5x^22x24 $$ has a zero at x = 2. Factor p completely and find its zeros.
Solution
p(x) has a zero at x = 2 and therefore x  2 is a factor of p(x). Divide p(x) by x  2
p(x) / (x  2) = (x^{3} + 5 x^{2}  2 x  24) / (x  2) = x^{2} + 7 x + 12
Using the division above, p(x) may now be written in factored form as follows:
p(x) = (x  2)(x^{2} + 7 x + 12)
Factor the quadratic expression x^{2} + 7 x + 12.
p(x) = (x  2)(x + 3)(x + 4)
The zeros are found by solving the equation.
p(x) = (x  2)(x + 3)(x + 4) = 0
For p(x) to be equal to zero, we need to have
x  2 = 0 , or x + 3 = 0 , or x + 4 = 0
Solve each of the above equations to obtain the zeros of p(x).
x = 2 , x =  3 and x =  4

The polynomial $$p(x)=3x^4+5x^317x^225x+10$$ has irrational zeros at x = ~+mn~ √5. Find the other zeros.
Solution
Zeros at x = ~+mn~ √5, correspond to the factors.
(x  √5) and (x + √5)
Hence polynomial p(x) may be written as
p(x) = (x  √5)(x + √5) Q(x) = (x^{2}  5)Q(x)
Find Q(x) using long division of polynomials
Q(x) = p(x) / (x^{2}  5) =
(3 x^{4} + 5 x^{3}  17 x^{2}  25 x + 10) / (x^{2}  5)
= 3 x^{2} + 5 x  2
Factor Q(x) = 3 x^{2} + 5 x  2
Q(x) = 3 x^{2} + 5 x  2 = (3x  1)(x + 2)
Factor p(x) completely
p(x) = (x  √5)(x + √5)(3x  1)(x + 2)
Set each of the factors of p(x) to zero to find the zeros.
x = ~+mn~√ 5 , x = 1 / 3 , x =  2

Polynomial p is given by $$ p(x) = x^4  2x^3  2x^2 + 6x  3
$$
a) Show that x = 1 is a zero of multiplicity 2.
b) Find all zeros of p.
c) Sketch a possible graph for p.
Solution
a) If x = 1 is a zero of multiplicity 2, then (x  1)^{2} is a factor of p(x) and a division of p(x) by (x  1)^{2} must give a remainder equal to 0. A long division gives
p(x) / (x  1)^{2} = (x^{4}  2x^{3}  2x^{2} + 6x  3) / (x  1)^{2} = x^{2}  3
The remainder in the division of p(x) by (x  1)^{2} is equal to 0 and therefore x = 1 is a zero of multiplicity 2.
b) Using the division above, p(x) may now be written in factored form as follows
p(x) = (x  1)^{2}(x^{2}  3)
Factor the quadratic expression x^{2}  3.
p(x) = (x  1)^{2} (x  √3) (x + √3)
The zeros are found by solving the equation.
p(x) = (x  1)^{2} (x  √3) (x + √3) = 0
For p(x) to be equal to zero, we need to have
(x  1)^{2} = 0 , or (x  √3) = 0 , or (x + √3) = 0
Solve each of the above equations to obtain the zeros of p(x).
x = 1 (multiplicity 2) , x = √3 and x =  √3
c) With the help of the factored form of p(x) and its zeros found above, we now make a table of signs.
.
We use the zeros of p(x) which graphically are shown as x intercepts, the table of signs and the y intercept (0 , 3) to complete the graph as shown below.
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Use the Rational Zeros Theorem to determine all rational zeros of the polynomial $$p(x) = 6x^313x^2+x+2 $$.
Solution
Rational Zeros Theorem: If p(x) is a polynomial with integer coefficients and if m / n (in lower terms) is a zero of p(x), then m is a factor of the constant term 2 of p(x) and n is a factor of the leading 6 coefficient of p(x). Find factors of 2 and 6.
factors of 2: ~+mn~ 1 , ~+mn~ 2
factors of 6: ~+mn~ 1 , ~+mn~ 2 , ~+mn~ 3 , ~+mn~ 6
possible zeros: divide factors of 2 by factors of 6: ~+mn~ 1 , ~+mn~ 1 / 2 , ~+mn~ 1 / 3 , ~+mn~ 1 / 6 , ~+mn~ 2 , ~+mn~ 2 / 3
Because of the large list of possible zeros, we graph the polynomial and guess the zeros from the location of the x intercepts. Below is the graph of the the given polynomial p(x) and we can easily see that the zeros are close to 1/3, 1/2 and 2.
.
We now calculate p(1/3), p(1/2) and p(2) to finally check if these are the exact zeros of p(x).
p(1/3) = 6(1/3)^3  13(1/3)^2 + (1/3) + 2 = 0
p(1/2) = 6(1/2)^3  13(1/2)^2 + (1/2) + 2 = 0
p(2) = 6(2)^3  13(2)^2 + (2) + 2 = 0
We have used the rational zeros theorem and the graph of the given polynomial to determine the 3 zeros of the given polynomial which are 1/3, 1/2 and 2.
