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Polynomoal p is given by $$p(x) = (x  1)^2(x  √3) (x + √3) $$
Make a sign table of p and sketch a possible graph for p.
Solution
We first find the zeros of p.
p(x) = (x  1)^{2} (x  √3) (x + √3) = 0
For p(x) = 0, 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 using:
(x  1)^{2} is positive for all x except at x = 1
x  √3 > 0 for x > √3
x + √3 > 0 for x >  √3
We put each factor in the table and use the rules of multiplication of signs to complete the sign for p as shown below.
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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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f(x) is a polynomial of degree six with a negative leading coefficient. f has a zero of multiplicity 1 at x = 1, a zero of multiplicity 3 at x = 1, and a zero of multiplicity 2 at x = 3. Make a sign table for the polynomial f.
Solution
We first write the factors of polynomial f with their multiplicity.
zero of multiplicity 1 at x = 1 : factor: x + 1
zero of multiplicity 3 at x = 1 : factor: (x  1)^{3}
zero of multiplicity 2 at x = 3 : factor: (x  3)^{2}
Let k (negative) be the leading coefficient of f. Using all the above factors, we write f(x) as
f(x) = k (x + 1)(x  1)^{3}(x  3)^{2}
We first study the sign of the different factors of f.
x + 1 > 0 for x >  1
(x  1)^{3} > 0 for x > 1
(x  3)^{2} > 0 for all x except x = 3
Below is shown the table of signs of each factor and of the polynomial f(x) in the bottom row.
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