Raymond B. answered • 10/09/20
Math, microeconomics or criminal justice
a 3 degree polynomial has potentially 3 real zeroes.
Its local max and minimums can be found by taking the derivative
and setting it =0
3x^2+4x -13 = 0
x = -4 + or - sqr(16+169) all over 6
= -2/3 + or - (1/6)sqr185) = -.67 + or - 13.6/6 = -.67 + or - 2.3 = -3 or 1.63
-3 is a max and 1.63 is a minimum. the graph will cross the x axis somewhere above x=2
and possibly 2 other times when x<0
P(0) = -10
P(1) = -20
P(2) = -20
P(3) = -4
P(4) = 34
one zero is somewhere between x = 3 and x= 4, probably just a little over 3, maybe 3.1
the other two zeroes are on either side of x=-3
P(-4) = +10
P(-3)=-32
another zero is between -3 and -4, closer to -4, so maybe about -3.7
P(-2) = -35
P(-1)=+4 another zero is between -1 and -2, maybe about -1.1
that's a rough estimate, zeroes of 3.1, -1.1 and -3.7
x^3 +2x^2 -3x -10 =0
has one sign change, which means maximum of 1 positive real solution
replace x by -x to get
-x^3 +2x^2 +3x -10 = 0
this has 2 sign changes, which means maximum 2 negative real solutions
since imaginary solutions come in conjugate pairs, there must be
1 real positive solution and either 0 or 2 real negative solutions
about positive 3.1 and negative 1.1 and negative 3.7 are what de cartes' rule of signs expects
either 3 real or 1 positive real and 2 imaginary
or use a graphing calculator and see where the curve crosses the x axis
those x intercepts are the zeros, the roots, or solutions to the equation
