A student found that a cubic function has zeros 16 and 1-2i with a leading coefficient of 3. What is the constant term of this polynomial function with real coefficients.

Question

directrix · Answer

A cubic polynomial function has 3 zeros by the fundamental theorem of algebra.

16, 1 - 2i --> you are given

The conjugate of 1 - 2i is also a zero.

So, your zeros are 16, (1-2i), and (1+2i)

That means

(x - 16) (x - 1 + 2i) (x - 1 - 2i) = 0

If you will multiply that out and post the result, I'll check your work and help you move to the answer.

anonymous · Answer

Also, the leading coefficient of three indicates a factor of three times the whole mess on the left hand side.

anonymous · Answer

i got $$x ^{2} + x ^{2} - x + 2x i- x - 2x i -16x + 16 - 32i + 16 + 32i$$

anonymous · Answer

Not sure how you did that.

anonymous · Answer

alllllrighht. i multiplied (x-16)(x-1+2i)(x-1-2i)

anonymous · Answer

Try doing it one bit at a time; multiply the second and third factors.

anonymous · Answer

:( okay

anonymous · Answer

k i got $$(x - 16) (x ^{2} - x - 2x i + 1 + 2i + 2x i - 2i)$$

anonymous · Answer

I think $$(x-1+2i)(x-1-2i)=x ^{2}-2x+5$$

directrix · Answer

I had  x^3 - 18 x^2 + 37 x -80 = 0 for the product of all three factors.

anonymous · Answer

So your function is something like $$f(x)=3(x-16)(x ^{2}-2x+5)$$

anonymous · Answer

Directrix, I think you are correct, except that won't take into account the requirement for a leading coefficient of three.

anonymous · Answer

Thanks btw!

anonymous · Answer

Tiffanee, if you multiply out f(x) in my post above, I think you'll get a constant term of -240.

anonymous · Answer

alrighty