Jay T. answered • 09/17/18
Tutor
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Retired Engineer/Math Tutor
3x^3 -8x^2 -20x + 16 = 0
For cubic (degree 3) polynomials, use the 'Rational Root Test' to reduce the number of possible roots. This test, which holds for polynomials of any degree, i.e. of the form
anxn + an-1xn-1 + a1x + a0 = 0, an ≠ 0 and a0 ≠ 0
says that each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfies p is an integer factor of the constant term a0, and q is an integer factor of the leading coefficient an.
In the case of 3x^3 -8x^2 -20x + 16 = 0, n=3 and a3 = 3 and a0=16. Since 3 (p) can be factored (i.e. numbers that can be multiplied to equal 3) into 1 and 3, and 16 (q) can be factored into 1, 2, 4, 8, and 16. The list of possible roots are then:
±{1,2,4,8,16} ÷ {1,3}
Thus, the roots are all members of the set of numbers { ±1, ±2, ±4, ±8, ±16, ±1/3, ±2/3, ±4/3, ±8/3, ±16/3 }. From here, there are a few methods of proceeding, but they are either restricted to polynomials of specific degrees, or as complex as just trying each of these roots out one by one.
The fastest way to get the roots, is to use a graph of the function wherein the x-intercepts are the roots. A graphic calculator does this very nicely. Unfortunately, this answer box does not support graphics, but if the plot of the function shows clearly that -2 and 4 are two of the three roots. The third root appears to be 2/3. Plugging that into the equation:
3*(8/27) - 8*(4/9) - 20*(2/3) + 16
= (24 - 96 - 360)/27 + 16
= -432/27 + 16
= -16 + 16
= 0
Thus, 2/3 is a root.
Note: If you are thinking that if a graphic solution is so simple, why bother with something more complicated and time consuming as the Rational Root Test, realize there are two reasons. First, your teacher wanted you to list ALL POSSIBLE roots, which a graph doesn’t do. Second, the graph visually yields approximate answers, while the RRT will give the exact root values. Thus, whether you use the RRT or not depends on how accurate you need the root evaluations to be. The other two equations can be done in the same way. The RRT and graph work just as well for the fourth (or any) degree function you asked about as for the others.
For cubic (degree 3) polynomials, use the 'Rational Root Test' to reduce the number of possible roots. This test, which holds for polynomials of any degree, i.e. of the form
anxn + an-1xn-1 + a1x + a0 = 0, an ≠ 0 and a0 ≠ 0
says that each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfies p is an integer factor of the constant term a0, and q is an integer factor of the leading coefficient an.
In the case of 3x^3 -8x^2 -20x + 16 = 0, n=3 and a3 = 3 and a0=16. Since 3 (p) can be factored (i.e. numbers that can be multiplied to equal 3) into 1 and 3, and 16 (q) can be factored into 1, 2, 4, 8, and 16. The list of possible roots are then:
±{1,2,4,8,16} ÷ {1,3}
Thus, the roots are all members of the set of numbers { ±1, ±2, ±4, ±8, ±16, ±1/3, ±2/3, ±4/3, ±8/3, ±16/3 }. From here, there are a few methods of proceeding, but they are either restricted to polynomials of specific degrees, or as complex as just trying each of these roots out one by one.
The fastest way to get the roots, is to use a graph of the function wherein the x-intercepts are the roots. A graphic calculator does this very nicely. Unfortunately, this answer box does not support graphics, but if the plot of the function shows clearly that -2 and 4 are two of the three roots. The third root appears to be 2/3. Plugging that into the equation:
3*(8/27) - 8*(4/9) - 20*(2/3) + 16
= (24 - 96 - 360)/27 + 16
= -432/27 + 16
= -16 + 16
= 0
Thus, 2/3 is a root.
Note: If you are thinking that if a graphic solution is so simple, why bother with something more complicated and time consuming as the Rational Root Test, realize there are two reasons. First, your teacher wanted you to list ALL POSSIBLE roots, which a graph doesn’t do. Second, the graph visually yields approximate answers, while the RRT will give the exact root values. Thus, whether you use the RRT or not depends on how accurate you need the root evaluations to be. The other two equations can be done in the same way. The RRT and graph work just as well for the fourth (or any) degree function you asked about as for the others.
