Question

Find the roots of the polynomial equation.

-x3 + 5x2 - 11x + 55 = 0

Answer 1

yo u go to james madison? which one? my cousin goes there too

Answer 2

A. \[i \sqrt{11}, -i \sqrt{11}, -5\]
B. \[\sqrt{11}, -\sqrt{11}, 5\]
C. \[i \sqrt{11}, -5\]
D. \[i \sqrt{11}, -i \sqrt{11}, \]

Answer 3

start with x = 5

to test it find f(5)

if its equal to zero then (x - 5) is a factor

Answer 4

Yes i go to James Madison! I do online

Answer 5

how do i find f(5)?

Answer 6

then if

(x -5) is a factor then use polynomial division to to find the quadratic factor...

then solve the quadratic

Answer 7

substitute 5 into the equation in place of x....

this is the factor theorem if

f(5) = 0 ... then (x - 5) is a factor

Answer 8

but given the answer choices ... there is only 1 with a root x = 5 which makes it easy

Answer 9

i got B?

Answer 10

makes sense to me...

but its interesting that you don't know about the factor theorem...

and you need to look up the rational roots theorem...

Answer 11

alternative way

-x^3 + 5x^2 - 11x + 55 = 0

(-x^3 + 5x^2) + ( -11x + 55) = 0

-x^2(x - 5) + ( -11x + 55) = 0

-x^2(x - 5) - 11(x - 5) = 0

(-x^2 - 11)(x - 5) = 0

-x^2 - 11 = 0 or x - 5 = 0

x^2 + 11 = 0 or x - 5 = 0

x^2 = -11 or x = 5

x = sqrt(-11), x = -sqrt(-11), or x = 5

x = i*sqrt(11), x = -i*sqrt(11), or x = 5

Answer 12

The possible roots of a polynomial are the factors of C (constant) divided by the factors of A (coefficient of the first term).

So the possible roots are: +1,-1,+5,-5,+55,-55,+11,-11

For any of these to be the roots, the function must equal to zero when we substitute the value for x. We try +5 and it gives us a zero; f(5) = -(5)^3 + 5(5)^2 - 11(5) + 55 = 0
Therefore, x-5 is a factor.

Now we can do long division and divide the original function by x-5.

(-x^3 + 5x^2 - 11x + 55) / x-5 = -x^2 - 11 (Through long division)

Now we can re-write our original function as f(x) = (x - 5)(-x^2 - 11)
Now we factor the quadratic part to obtain additional roots. Note these roots will be imaginary as this quadratic has no real roots.

-x^2 - 11 = 0
-x^2 = 11
x = sqrt(-11) or -sqrt(-11)

We can rewrite each root in terms f "i" like this:

x = i*sqrt(11), -i*sqrt(11), 5