How do you find the number of complex, real and rational roots of #x^3-x^2-2x+7=0#?

1 Answer
Jul 15, 2017

Answer:

Use the discriminant to find that it has #1# irrational real zero and #2# non-real complex zeros.

Explanation:

The cubic equation:

#x^3-x^2-2x+7 = 0#

is in the standard form:

#ax^3+bx^2+cx+d = 0#

Its discriminant #Delta# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=-1#, #c=-2# and #d=7#, so we find:

#Delta = 4+32+28-1323+252 = -1007#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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We were also asked if the roots were rational.

By the rational roots theorem, any rational zeros of this cubic must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #7# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

#+-1, +-7#

None of these work, so our cubic has no rational roots.