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# Find all the real zeros of the polynomial

find all the real zeros of the polynomial
P(x)=x3-5x2-15x+7
its real zeros are
x1=
x2=
x3=
with x1≤x2≤x3

### 2 Answers by Expert Tutors

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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P( X)= X^3 - 5 X^2 - 15X +7      X1 + X2 + X3 = 5

X1 . X2 . X3 =7

Rational roots:

±1 ,± 7

P( 7) = 7^3 - 5 ( 7^2) -15*7 +7

= 343 - 245 - 105  +7 = 0

X3 = 7

X^2 + 2X -1
X - 7 l X^3 -5 X^2 - 15X +7
X^3 ± 7X^2
2X^2 - 15X
2 X^2 ± 14x
-x +7
- X+7
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'     X^2 + 2X -1 =

X = -2/2 ±√(8 ) /2        X1 = -1 - √2  , X2 = -1 +√2

X1< X2 < X3

### -1 - √2 ‹ -1 + √2 < 7

Kenneth G. | Experienced Tutor of Mathematics and StatisticsExperienced Tutor of Mathematics and Sta...
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If the polynomial has degree-1 factors with rational coefficients, then they are (x ± 1) or (x ± 7).

Substituting

P(7) = 0, but the other values are not zeros.

Dividing P(x) by (x-7) gives x2 + 2x - 1 which has roots -1 ± √2

So the zeros are

x1 = -1 - √2

x2  = -1 + √2

x3  =  7