f(x) = 3x^3 - 2x^2 + x + 7

g(x) = 2x^2-4

2 QUESTIONS:

Find g of f(x)

AND...

f(x) has two complex zeros. One is in the 4th quadrant. In what quadrant is the other???

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- Dec 7th 2006, 07:07 PMMr_Greenf(x) and g(x) compositions
f(x) = 3x^3 - 2x^2 + x + 7

g(x) = 2x^2-4

2 QUESTIONS:

Find g of f(x)

AND...

f(x) has two complex zeros. One is in the 4th quadrant. In what quadrant is the other??? - Dec 7th 2006, 07:10 PMRuichan
g of f(x)

= 2* (3x^3 - 2x^2 + x + 7)^2 - 4 - Dec 7th 2006, 08:30 PMputnam120
for the second question i suggest that you use the rational root theorem to find a rational root, lets call it $\displaystyle r$. we then have

$\displaystyle 3x^2-2x^2+x+7=(x-r)Q(x)$ where $\displaystyle Q(x)$ is a quadratic.

however if there is not rational root then this becomes quite difficult. - Dec 8th 2006, 04:32 AMtopsquark
Two observations:

The question states there are only two complex zeros.

f(x) is a cubic polynomial so there can be at most two complex zeros.

Either way you look at it, there are ONLY two complex zeros of f(x). Since f(x) is a polynomial with real coefficients the two complex zeros are complex conjugates of each other.

Since one complex zero is in the 4th quadrant, it must have the form a - bi, where a, b > 0. Since the other zero is the complex conjugate of this, the second zero is of the form a + bi, which would be in the first quadrant.

-Dan