How do you find all the zeros of f(x)=x^3 + 4x^2 - 5f(x)=x3+4x2−5?

1 Answer
Leland Adriano Alejandro
Mar 20, 2016

x_1=1x1=1 and x_2=(-5+sqrt5)/2x2=−5+√52 and x_3=(-5-sqrt5)/2x3=−5−√52

Explanation:

From the given:
y=x^3+4x^2-5y=x3+4x2−5

Perform synthetic division

x^3" " " "x^2" " " "x^1" " " " "x^0x3 x2 x1 x0

1" " " " " 4" " " " "0" " " -51 4 0 −5 trial divisor =1=1
underline(" " " " " " "1" " " " " "5" " " " " "5)
1" " " " " 5" " " " " "5" " " " " " 0larrremainder

x_1=1 is a zero

depressed equation:

x^2+5x+5=0

by quadratic formula

x=(-b+-sqrt(b^2-4ac))/(2a)

Let a=1 and b=5 and c=5

x=(-b+-sqrt(b^2-4ac))/(2a)

x=(-5+-sqrt(5^2-4(1)(5)))/(2(1))

x=(-5+-sqrt5)/2

the other zeros are:
x_2=(-5+sqrt5)/2

x_3=(-5-sqrt5)/2

God bless....I hope the explanation is useful.

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