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