9514 1404 393
Answer:
- zeros are -10, and -1 (multiplicity 2)
- zeros are {-3, -2, 1, 3}
Explanation:
For higher-degree polynomials I find it extremely useful to use a graphing calculator to find the zeros.
The usual approaches are ...
- use the rational root theorem
- try values of x=±1
- narrow the search using Descartes' rule of signs
- estimate maximum or minimum root values
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1. Rational roots will be factors of 10, so ±1, ±2, ±5, or ±10. The signs are all positive, so there are no positive real roots. The sum of odd-degree coefficients is the same as the sum of even-degree coefficients, so -1 is one of the roots. Removing that using synthetic division gives you ...
f(x) = (x +1)(x^2 +11 +10)
The quadratic (more or less obviously) factors as (x+1)(x+10), so the complete factorization is ...
f(x) = (x +1)^2(x +10)
and the zeros are {-10, -1, -1}.
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2. Rational roots will be factors of 18, so will be ±1, ±2, ±3, ±6, ±9, ±18. The sum of coefficients is 0, so x=1 is a root. The rule of signs tells you that there are 2 positive real roots, so there is one more positive real root. Synthetic division tells you the factoring so far is ...
f(x) = (x -1)(x^3 +2x^2 -9x -18)
We note that the first two coefficients have the same ratio as the last two, so we can factor by grouping.
x^3 +2x^2 -9x -18 = x^2(x +2) -9(x +2) = (x^2 -9)(x +2)
Factoring the difference of squares gives you the complete factorization, hence all of the roots. (The zeros are the values of x that make the factors zero.)
f(x) = (x -1)(x +2)(x -3)(x +3)
The zeros are {-3, -2, 1, 3}.