Byron S. answered • 11/02/14
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Math and Science Tutor with an Engineering Background
In general, yes, you set a function equal to zero to find its zeroes. However, in equations like this, they're not easily factorable, so you need to apply other methods to find the first few zeros before you can (hopefully) factor the rest.
Looking at each part of your problem, I'll discuss methods and ideas surrounding each.
f(x)=2x3+5x2-28x-15
Find the maximum number of zeros
The maximum number of zeros a polynomial can have is its degree. This function is a 3rd degree polynomial (x3 is the highest power), so it can have a maximum of 3 zeros. It might have less, possibly only 1, but at most there are 3.
Find the bounds to the zeros
All of the zeros must divide the constant term of the polynomial (15 in this case), so none of them can be larger. The zeros for this function are bounded by -15 ≤ x ≤ 15.
List potential rational zeros
If there are rational zeros, they must be a factor of the constant (15) divided by a factor of the leading coefficient (2). For this problem, the possible rational zeros are:
±(1, 3, 5, 15, 1/2, 3/2, 5/2, 15/2)
Determine the real zeros
Not all zeroes are rational, but rational zeros are real. Zeros that are real but irrational will come from a trinomial that you have to solve using the quadratic equation rather than factoring.
Find the x & y intercepts
The x-intercepts are the zeros you found, listed in point form (3,0), etc.
The y-intercept is found with the constant of the polynomial, as a point (0,15)
I hope this clears up some of your confusion. Please comment if you have more questions.
Byron S.
Since this is a 3rd degree equation, you can't use the quadratic formula yet. First you'll have to find and divide out one of the zeros and then either factor or use the quadratic formula on the quotient, which will be 2nd degree (quadratic.) Synthetic
division is typically the easiest way to do this.
Since you found the answers on your graph, you can skip the guess and check part of finding zeros, and divide out one of the known zeros to find the rest.
I'll divide out the zero 3:
3 | 2 5 -28 -15
6 33 15
2 11 5 | 0
The remainder is then 2x2 + 11x + 5, which you can factor or use the quadratic formula. It factors to (2x+1)(x+5), which gives you the remaining zeros you were looking for. Note that all of these numbers are rational, and are in the list of
possible rational zeros I gave above.
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11/02/14
Krista J.
ohh that makes so much more sense. So I just use synthetic division to check all of the rational zeros and if it ends with a remainder of 0, it is a real zero?
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11/02/14
Byron S.
Yes, that's correct. It won't always be true that all of the zeros are rational, but you will find the rational ones that way.
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11/02/14
Krista J.
11/02/14