If x^3+3x^2+kx-24 is divisible by x-3 then it is also divisible by: ?

Answer 1

If it is divisible by x - 3, then (x - 3) is a factor of that polynomial, and if x - 3 = 0, x = 3.

If we substitute x = 3 to the polynomial, then it must yield 0 (since 3 is a root).
x³ + 3x² + kx - 24

Thus,
(3)³ + 3(3)² + k(3) - 24 = 0

Then,
27 + 27 + 3k - 24 = 0

Transpose,
3k = -30

Thus,
k = -10

The polynomial then becomes
x³ + 3x² - 10x - 24 = 0

We then divide it by x - 3, to factor it out. The easiest method goes like this
x³ - 3x² + 6x² -18x + 8x - 24 = 0

Now, we group
(x³ - 3x²) + (6x² - 18x) + (8x - 24) = 0

We factor out from each group
x²(x - 3) + 6x(x - 3) + 8(x - 3) = 0

We then factor out x - 3
(x - 3)(x² + 6x + 8) = 0

We can factor the quadratic trinomial
(x - 3)(x + 2)(x + 4) = 0

Therefore, your polynomial is also divisible by (x + 2) and (x + 4).

^_^

Answer 2

Let f(x) = x^3+3x^2+kx-24
Since x - 3 is a factor f(3) = 0 (by the factor theorem)

ie 27 + 27 + 3k - 24 = 0
So 3k = -30
k = -10
ie f(x) = x^3 + 3x^2 - 10x - 24
= (x - 3)(x^2 + 6x + 8)
= (x - 3)(x + 2) (x + 4)

So it is also divisible by x + 2 and x + 4

Answer 3

Synthetic division:

3 | 1 3 k -24
--- 0 3 18 3(k+18)
---------------
--- 1 6 k+18 0

Formatting is not good, but to make it divisible by 18, -24 + 3(k+18) = 0, so k + 18 = 8, so k = -10.

then if k = -10, the result of dividing by (x-3) would be the quadratic equation:

x² + 6x + 8

Which factors to (x+2)(x+4), so -4 and -2 would also be factors.

Answer 4

It is also divisible by (x-4) and (x-2).

Answer 5

yes because every constant is divisible by 3