9514 1404 393
Answer:
- -1, -1/2, 3
- -3, -√5, √5
- -√6, -1/2, 2, √6
- 2-√3, 1, 2+√3
Explanation:
Personally, the "work" I like to do when finding higher-degree polynomial zeros is writing the equation into a graphing calculator. The rational zeros are usually fairly obvious (for school problems), and the irrational ones often can be arrived at without a lot of trouble.
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1. The graph in the first attachment shows the zeros are -1, -1/2, 3. (You expect a zero with 2 in the denominator because of the leading coefficient.)
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2. The graph in the first attachment shows the zeros are -3, -2.236, and +2.236. Factoring the function by grouping shows you the factors to be ...
f(x) = (x^3 +3x^2) -(5x +15) = x^2(x +3)-5(x +3) = (x +3)(x^2 -5)
The quadratic factor can be factored using the form for "difference of squares":
f(x) = (x +3)(x -√5)(x +√5)
The value of √5 is about 2.236, matching the numbers shown on the graph.
The zeros are -3, -√5, +√5.
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3. The graph in the second attachment shows the zeros are -2.449, -1/2, 2, +2.449. (As in problem 1, you expect a zero with 2 in the denominator.) Factoring out the rational zeros* gives ...
f(x) = (2x +1)(x -2)(x^2 -6) = (2x +1)(x -2)(x -√6)(x +√6)
The zeros are -1/2, 2, -√6, +√6.
* The "work" of the synthetic division is shown in the third attachment. The coefficients of the quadratic are reduced by a factor of 2 from the result of division by (x+1/2) so they effectively show the quotient from division by (2x+1).
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4. The graph in the fourth attachment shows the zeros are 0.268, 1, 3.732. Factoring out the rational zero gives ...
f(x) = (x -1)(x^2 -4x +1)
The zeros of the quadratic factor can be found any of several ways. One of my favorite is to use the graph to discover the vertex. For the vertex form ...
y = (x -h)^2 +k
the roots are h±√(-k). The vertex form of the quadratic term is
y = (x -2)^2 -3
so the roots are 2±√3.
The zeros are 1, 2-√3, 2+√3.