How do you find the rest of the zeros given one of the zero c=−1 and the function f(x)=x5+2x4−12x3−38x2−37x−12?

1 Answer
George C.
Aug 1, 2016

The zeros of f(x) are x=−1 with multiplicity 3, x=4 and x=−3.

Explanation:

f(x)=x5+2x4−12x3−38x2−37x−12

f(−1)=−1+2+12−38+37−12=0

So x=−1 is a zero and (x+1) a factor:

x5+2x4−12x3−38x2−37x−12=(x+1)(x4+x3−13x2−25x−12)

Substituting x=−1 in the remaining quartic factor, we get:

1−1−13+25−12=0

So the quartic also has x=−1 as a zero and (x+1) as a factor:

x4+x3−13x2−25x−12=(x+1)(x3−13x−12)

Substituting x=−1 in the remaining cubic factor, we get:

−1+13−12=0

So the cubic also has x=−1 as a zero and (x+1) as a factor:

x3−13x−12=(x+1)(x2−x−12)

To factor the remaining quadratic, note that 4⋅3=12 and 4−3=1, so:

x2−x−12=(x−4)(x+3)

hence zeros x=4 and x=−3

So the zeros of f(x) are x=−1 with multiplicity 3, x=4 and x=−3.

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