Andy flipped a coin once and rolled a six-sided number cube numbered 1 through 6
1 answer:
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Answer:
{1, (-1±√17)/2}
Step-by-step explanation:
There are formulas for the real and/or complex roots of a cubic, but they are so complicated that they are rarely used. Instead, various other strategies are employed. My favorite is the simplest--let a graphing calculator show you the zeros.
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Descartes observed that the sign changes in the coefficients can tell you the number of real roots. This expression has two sign changes (+-+), so has 0 or 2 positive real roots. If the odd-degree terms have their signs changed, there is only one sign change (-++), so one negative real root.
It can also be informative to add the coefficients in both cases--as is, and with the odd-degree term signs changed. Here, the sum is zero in the first case, so we know immediately that x=1 is a zero of the expression. That is sufficient to help us reduce the problem to finding the zeros of the remaining quadratic factor.
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Using synthetic division (or polynomial long division) to factor out x-1 (after removing the common factor of 4), we find the remaining quadratic factor to be x²+x-4.
The zeros of this quadratic factor can be found using the quadratic formula:
a=1, b=1, c=-4
x = (-b±√(b²-4ac))/(2a) = (-1±√1+16)/2
x = (-1 ±√17)2
The zeros are 1 and (-1±√17)/2.
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The graph shows the zeros of the expression. It also shows the quadratic after dividing out the factor (x-1). The vertex of that quadratic can be used to find the remaining solutions exactly: -0.5 ± √4.25.
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The given expression factors as ...
4(x -1)(x² +x -4)
