Asked by ladybug
find all zeros of the following polynomial. write the polynomial in factored form.
f(x)=x^3-3x^2+16x-48
f(x)=x^3-3x^2+16x-48
Answers
Answered by
drwls
Try + and - the prime factors of 48:
1, 2, 3, 4, 6, 8, 12 and 16.
When you find one that works, call it 'a' and divide the f(x) polymonial by (x-a) to get the quadratic factor. The other two roots can easily be found with that.
It looks to me like one root is +3. Therefore x-3 is a factor.
Now evaluate [x^3-3x^2+16x-48]/(x-3)
It should give you a quadratic equation with no remainder. Factor that.
1, 2, 3, 4, 6, 8, 12 and 16.
When you find one that works, call it 'a' and divide the f(x) polymonial by (x-a) to get the quadratic factor. The other two roots can easily be found with that.
It looks to me like one root is +3. Therefore x-3 is a factor.
Now evaluate [x^3-3x^2+16x-48]/(x-3)
It should give you a quadratic equation with no remainder. Factor that.
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