If a polynomial with real coefficients is of degree 3. Can we say there must be at least 1 real root? a. Yes, because complex roots must appear in conjugate pairs, leaving one root real. b. No. we can’t say anything about this case
c. Yes, because if there is one real root, there must be two additional real roots
d. No, because such a curve never crosses the x-axis
e. Sometimes yes and sometimes no

A polynomial of degree 3 has 3 linear factors… (1): (x - i ) (x + i ) (x + a) . If any 2 factors are conjugate as here first 2 factors are conjugate with complex roots. So at least one root should be real.

Step-by-step explanation:

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

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step-by-step explanation:

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