How do you find the rest of the zeros given one of the zero #c=-1# and the function #f(x)=x^5+2x^4-12x^3-38x^2-37x-12#?

Question

2460 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-01T06:15:04

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

#f(x) = x^5+2x^4-12x^3-38x^2-37x-12#

#f(-1) = -1+2+12-38+37-12 = 0#

So #x = -1# is a zero and #(x+1)# a factor:

#x^5+2x^4-12x^3-38x^2-37x-12=(x+1)(x^4+x^3-13x^2-25x-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:

#x^4+x^3-13x^2-25x-12 = (x+1)(x^3-13x-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:

#x^3-13x-12 = (x+1)(x^2-x-12)#

To factor the remaining quadratic, note that #4*3 = 12# and #4-3=1#, so:

#x^2-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#.