9514 1404 393
Answer:
- {±1/2, ±1, ±3/2, ±2, ±3, ±9/2, ±6, ±9, ±18}
- {±1, ±2, ±3, ±4, ±6, ±8, ±12 ±24}
- -2, -1, 2
- -4, 1, 2
Step-by-step explanation:
Possible rational zeros are of the form ...
±{divisor of the constant term}/{divisor of the leading coefficient}
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1. Rational zeros will be one of ±{1, 2, 3, 6, 9, 18}/{1, 2}, so will be ...
{±1/2, ±1, ±3/2, ±2, ±3, ±9/2, ±6, ±9, ±18}
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2. Rational zeros will be one of {±1, ±2, ±3, ±4, ±6, ±8, ±12 ±24}.
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3. The pairs of coefficients suggest factoring by pairs:
f(x) = (2x^3 +2x^2) -(8x +8) = 2x^2(x +1) -8(x +1) = 2(x^2 -4)(x +1)
f(x) = 2(x -2)(x +2)(x +1)
Zeros of f(x) are 2, -2, -1, values of x that make the factors zero.
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4. The sum of coefficients is zero, so x=1 is a zero. Factoring that out by any of several means, you find ...
f(x) = (x -1)(x^2 +2x -8)
The quadratic is factored by looking for factors of -8 that have a sum of 2. (4 and -2 are those)
f(x) = (x -1)(x +4)(x -2)
Zeros of f(x) are 1, -4, 2.
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The attached graph confirms our factoring of the polynomials of questions 3 and 4. For cubic and higher degree, I find a graphing calculator to be quite helpful finding zeros.
