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# Use the rational zeros theorem to find all the real zeros of the polynomial function.use the zeros to factor f over the real numbers.

Use the rational zeros theorem to find all the real zeros of the polynomial  function. use the zeros to factor f over the real numbers.  f(x)=x^3-5x^2-61x-55???

### 3 Answers by Expert Tutors

Michael F. | Mathematics TutorMathematics Tutor
4.7 4.7 (6 lesson ratings) (6)
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f(x)=x3-5x2-61x-55.  The rational roots are formed by the ratios of ±1 ±11  ±5 /±1
f(-1)=-1-5+61-55=0  -1 is a zero.  If we divide by x-(-1)=x+1 we get a quadratic
-1|1    -5   -61   -55
-1      6     55
1   -6   -55       0        x2-6x-55, which factors to  (x-11)(x+5), whose zeroes are 11 and -5

Note: -125-125+305-55=0  i.e. f(-5)=0
f(11)=1331-605-671-55=0

Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
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The polynomial of odd degree always has at least one real zero, since complex roots of a polynomial with rational coefficients must always appear in conjugate pairs (like a+bi and a-bi, where i=√-1).

In this case x=-1 is the root. Indeed,
(-1)3-5(-1)2-61*(-1)-55=-1-5+61-55=0

Thus, we can factor (x+1) out using synthetic division.

x3-5x2-61x-55=(x+1)(x2-6x-55).

Now let us solve for other roots,
x2-6x-55=0; It is immediately obvious that, since the product of two roots must be -55 and their sum must be 6, those roots are 11 and -5. Therefore we obtain:

x3-5x2-61x-55=(x+1)(x+5)(x-11)

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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f(x) =  X3 - 5X2 -61X - 55

Possible rational roots : ±1, ±5 ,±11, ±55

try : X=-1

( -1)3 -5 (-1)2 -61( -1) - 55 =  -1-5 + 61 -55 = 0

X= -1 is one of the solution.

X3 - 5X2 - 61 X - 55 , should be divisible by (X +1). You can do the long division and find the

Also : any polynomial  of any degree  axn + bXn -1 + cXn-2 + ......

If the irrational and complex roots are counted, the total number of roots equal to the degree of
polynomial .
Sum of the roots : X1 + X2 + X3 + ....+ Xn = -b/a
X1 . X2 . X3 .....Xn = X0 /a

Like in the case of  f(x) = X3 - 5X2 - 61X -55

X1 + X2 + X3 = 5
X1 . X2  . 3 = 55

X1 = -1        X2 + X3 = 6
X2 . X3 =-55

Quadratic  with roots of X2 , X2 will be:

X2 - 6 X - 55 =0
(X - 11)( X+5 ) =0      X = 11  X = -5
Doing the long division you should come up with the same answer.