Answer:
{-3, -1, 1, 7}
Step-by-step explanation:
f(x)=x^4−4x^3−22x^2+4x+21 is of the fourth order, and thus we can expect four zeros. There are various ways in which we could identify the zeros. One would be to check each of the given possible answers and determine whether the equation takes on the value 0 in each case; any such value 0 indicates that the possible zero chosen is indeed a zero.
Important: The Rational Root Theorem applies here. We can assume that there are at least some rational roots, which take the form of fractions, whose numerators are plus and minus factors of the constant term, 21, and whose denominators are plus and minus factors of the coefficient of the leading term of the function (which here is just 1). Thus, factors of the given function will be factors, plus or minus, of the constant term, 21, with the coefficient 1 of the highest power term not making any difference. That list of factors is already given: plus or minus 3, plus or minus 1, 0 and 7.
I will arbitrarily choose -1 and determine using synthetic division whether or not this -1 is a zero of the polynomial. Recall that if synthetic div. produces a zero remainder, then the divisor chosen is indeed a zero of the polynomial.
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-1 / 1 -4 -22 4 21
-1 +5 18 -22
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1 -5 -17 22 -1
Because the remainder is -1, not 0, -1 is not a zero of the given function.
We must move on and try others from the given list. Eliminate -1 and 0. Try 3:
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3 / 1 -4 -22 4 21
+3 -3 -75 -213
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1 -1 -25 -71 -192
So we conclude that 3 is not a zero.
Try +1:
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1 / 1 -4 -22 4 21
+1 -3 -25 -21
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1 -3 -25 -21 0
Since the remainder is zero, we know that +1 is a zero of the given function.
Using the same method, we can show that -3, -1, 1 and 7 are zeros.
