If f(x)=x^3−5x^2−2x+24 and (x+2) is a factor, what are the remaining factors? Factor 1:
Factor 2:
Factor 3:

(x+2), (x-4), (x-3)

Step-by-step explanation:

Use the rational root theorem to get started, then factor the remaining quadratic to find:

x^3 − 5x^2 − 2x + 24 = (x + 2)(x − 4)(x − 3)

Explanation:

Let  f(x) =x^3 − 5x^2 − 2x + 24

By the rational root theorem, any rational zeros of  f(x)  must be expressible in the for p/q for integers p, q with p a divisor of the constant term 24 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are the factors of 24, namely:

± 1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 24

Try each in turn:

f(1) = 1 − 5 − 2 + 24 = 18

f(−1) = −1 − 5 + 2 + 24 = 20

f(2) = 8 − 20 − 4 + 24 = 8

f(−2) = −8 − 20 + 4 + 24 = 0

So  x = −2  is a zero and (x + 2) is a factor.

x^3 − 5x^2 − 2x + 24 = (x + 2)(x^2 − 7x + 12)

We can factor  

x^2 − 7x + 12 by noting that 4 × 3 = 12 and 4 + 3 = 7, so:

x^2 − 7x + 12 = (x − 4)(x − 3)

Putting it all together:

x^3 − 5x^2 − 2x + 24  =  (x + 2)(x − 4)(x − 3)

center= (-2,0)

radius= sqrt2/2

step-by-step explanation:

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step-by-step explanation: