^{3}-5x

^{2}-61x-55. The rational roots are formed by the ratios of ±1 ±11 ±5 /±1

^{2}-6x-55, which factors to (x-11)(x+5), whose zeroes are 11 and -5

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???

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f(x)=x^{3}-5x^{2}-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 x^{2}-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

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.

x^{3}-5x^{2}-61x-55=(x+1)(x^{2}-6x-55).

Now let us solve for other roots,

x^{2}-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:

x^{3}-5x^{2}-61x-55=(x+1)(x+5)(x-11)

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

f(x) = X^{3} - 5X^{2} -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.

X^{3} - 5X^{2} - 61 X - 55 , should be divisible by (X +1). You can do the long division and find the

quadratic.

Also : any polynomial of any degree ax^{n} + bX^{n -1} + cX^{n-2} + ......

If the irrational and complex roots are counted, the total number of roots equal to the degree of

polynomial .

Sum of the roots : X_{1} + X_{2 }+ X_{3 + ....}+ X_{n} = -b/a

X_{1} . X_{2} . X_{3} .....X_{n = X}^{0} /a

Like in the case of f(x) = X^{3 }- 5X^{2} - 61X -55

X_{1} + X_{2} + X_{3 = 5}

X_{1 .} X_{2 .
}X _{3 = 55}

X_{1 = -}1 X_{2} + X_{3 = 6}

X_{2} . X_{3 =-55}

Quadratic with roots of X_{2} , X_{2 will be:}

X^{2} - 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.

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