9514 1404 393
Answer:
B (correctly marked)
1/3, 1/2, 4
Step-by-step explanation:
The rational root theorem tells you the rational roots will be of the form:
±(divisor of constant)/(divisor of leading coefficient)
Here, that means they will be of the form ±{1, 2, 4}/{1, 2, 3, 6}.
Right away, you can eliminate choices A and C because there cannot be 9 in the numerator, and the number of possibilities goes beyond just 1, 2, and 4.
Possible rational roots are ...
±1, ±2, ±4, ±1/2, ±1/3, ±1/6, ±2/3, ±4/3
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I like a graphing calculator for finding the roots of higher-degree polynomials. A graph of this one shows roots to be 1/3, 1/2, and 4.
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Descartes' rule of signs tells you there are 0 negative real roots, and 1 or 3 positive real roots. The ratio of the first two coefficients suggests that one of the roots may be relatively large, so it might work to try 4.
((6·4 -29)·4 +21)·4 -4 = (-20 +21)·4 -4 = 0
So, x=4 is a root. Synthetic division (2nd attachment) shows the remaining quadratic factor to be 6x^2 -5x +1. Factoring this*, we have ...
6x^3 -29x^2 +21x -4 = (x -4)(6x^2 -5x +1) = (x -4)(3x -1)(2x -1)
The roots associated with the quadratic factor are 1/3 and 1/2.
The real roots are 1/3, 1/2, and 4.
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* To factor ax^2 +bx +c, you look for factors of the product ac that have a sum of b. Here, you're looking for factors of 6·1 = 6 that have a sum of -5.
6 = (-1)(-6) = (-2)(-3) . . . sums are -7 and -5
The numbers we're looking for are -2 and -3. Then the factorization is (6x -2)(6x-3)/6 = (3x -1)(2x -1).
Note that we have identified 'p' and 'q' as the factors of 'ac' that total to 'b'. The factorization is then (ax+p)(ax+q)/a. Here, both factors end up with coefficients of x that are not unity. That is not always the case.
There are other ways to make use of 'p' and 'q'. You can also rewrite the quadratic as ax^2 +px +qx +c, and factor by grouping.
