How do you find all the zeros of #P(X) = x^3 - 6x^2 + 4x + 16#?

1 Answer
Mar 26, 2016

Answer:

the roots or zeros are: #x_(1,2)=1+-sqrt(5) and x=4# graph{x^3-6x^2+4x+16 [-2.603, 6.86, -1.507, 3.23]}

Explanation:

Using Zero Rational Theorem you see that the potential roots #p/q# are #{1,2,4,8,16}#.
First root that we will try will #x-4#
#(x^3 - 6x^2 + 4x + 16)/(x-4)# apply long division

#4 |bar(1 - 6 + 4 + 16#
#" " |1 + 4 - 8 - 16#
#" " |1 -2 - 4 - 0#

Thus #(x-4)# divides our polynomial without remainder and #:.# it is a factor.
#(x^3 - 6x^2 + 4x + 16)=color(brown)((x^2-2x-4))(x-4)#
Now factor the binomial #color(brown)(x^2-2x-4)#
Use quadratic formula:
#x_(1,2)=-(b+-sqrt(b^2-4ac))/(2a)=(2+- sqrt((-2)^2-4(1)(-4) ))/2#
#x_(1,2)=(2+- sqrt(4+16))/2=(2+- 2sqrt(5))/2=1+-sqrt(5)#

So the roots or zeros are: #x_(1,2)=1+-sqrt(5) and x=4#