Expert: Sherman D. - 10/9/2004

Question
Hi, I can't seem to solve these two problems.  I was wondering if you could help me find the missing link.

Use the given zero to find all the zeros of the function.

f(x)=2x^3+3x^2+50x+75     zero given: 5i
f(x)=x^3-7x^2-x+87        zero given: 5+2i

I seem to only get half of the problem correct so I know I'm missing something...Can you help and if so, can you explain how to solve.

Thanks.


Answer
f(x) = 2x^3 + 3x^2 + 50x + 75
(2x^3 + 3x^2) + (50x + 75)
x^2(2x + 3) + 25(2x + 3)
since both the x^2 and 25 have the same (2x + 3), they can be factored like this
(x^2 + 25)(2x + 3)
which further factors to
(x + 5i)(x - 5i)(2x + 3)

so the other 2 zeros are

x = -5i and (-3/2)

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f(x) = x^3 - 7x^2 - x + 87

I went to www.quickmath.com to find this one

it gave me x = 5 - 2i and x = -3

it also gave me a factor of (x + 3)(x^2 - 10x + 29)

if you were to take x^2 - 10x + 29, and use the quadratic formula

x = (-b ± sqrt(b^2 - 4ac))/2a

x = (-(-10) ± sqrt((-10)^2 - 4(1)(29)))/2(1)
x = (10 ± sqrt(100 - 116))/2
x = (10 ± sqrt(-16))/2
x = (10 ± 4i)/2
x = 5 ± 2i
x = 5 + 2i or 5 - 2i

sorry i don't know of the other way, since it has been awhile since i have done problems like this.