0

# Find all the real zeros of the polynomial

Find all the real zeros of the polynomial
P(x)=x4-4x3-6x2+4x+5
its real zeros are
x1=
x2=
x3=
x4=
with x1≤x2≤x3≤x4

### 2 Answers by Expert Tutors

Kenneth G. | Experienced Tutor of Mathematics and StatisticsExperienced Tutor of Mathematics and Sta...
0
Possible degree-1 factors with rational coefficients are of the form (x+b) where b = ±1 or ±5

P(1) = 0, p(-1) = 0, p(5) = 0,  so 1, -1 and 5 are roots.

dividing P(x) by (x-1)(x+1)(x-5) give (x+1),  so the root x = -1 occurs twice.

x1 = -1, x2= -1, x3 = 1, x4 = 5.

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
0
P( X ) = X^4 - 4X^3 - 6X^2 + 4X +5       X1 + X2 + X3 + X4 = 4
X1 + X2 + X3 +X4 = 5      ( 1)

Given :
aX^4 + bX^3 + c X^2 + dX + e

Roots: X1,X2, X3, X4

X1 +X2 +x3 + X4 =-b/a

X1 . X2 . X3 . X4 = c/a

Roots   ±1 , ±5

Test:

P( -1) = ( -1)^4 - 4( -1) ^3 - 6 ( -1)^2 + 4 ( -1) +5

= 1 + 4 - 6 -4 + 5 =0

X1 = -1

P( 1) = ( 1) ^4 - 4 ( 1)^3 - 6( 1)^2 + 4 ( 1) +5 =

=  1 - 4 - 6 +4 + 5 =0

From Equation1 :  X3 + X4 = 4 ,   X3 . X4 = -5

then:
( X3 +5) ( X4 - 1 ) = 0

X 3 = -5    X4 = 1

X3 =-5 < X1= -1< X2, X4 = 1

X2 = +1