Factoring polynomials

[megadeth95]

I need help with the foll. problem:

P(x) = x³ + 2x² - (4K + 5)x - 6 have three distinct zeros. Two of these are identical to the zeros of f(x) = x²+ 5x + K. Compute the value of K.

Attempts:

Using the rational roots theorem:

+-1, +-2, +-3, +-6

I chose -3 because i thought it would work. Then I used synthetic division to find the other zeros.

-3 1 2 (4K + 5) -6
-3 3 6
1 -1 -2 0

(4k+5) must be -5 because -5+3 = -2

next I found the other zeros:

(x² - x - 2) = (x + 1) (x - 2)

So, the zeros of this function are: 2, -1, and -3. As stated before, (4K + 5) = - 5

4k + 5 = -5

k = - 10/4 or -2.5

Now, when I substitute -2.5 to f(x) = x²+ 5x + K, the zeros i get are not the same to the other function...I'm stuck here.

Thanks for helping me

 

[MarkFL]

I would choose to let r be the root the cubic has that the quadratic does not, and set:

\(\displaystyle x^3+2x^2-(4k+5)x-6=(x-r)\left(x^2+5x+k \right)\)

Expanding the right side, we have:

\(\displaystyle x^3+2x^2-(4k+5)x-6=x^3+(5-r)x^2+(k-5r)x-rk\)

Equating coefficients, we obtain the system:

\(\displaystyle 5-r=2\)

\(\displaystyle 4k+5=5r-k\)

\(\displaystyle 6=rk\)

From the first equation, we see \(\displaystyle r=3\) and so from the third equation we find \(\displaystyle k=2\).

 

[soroban]

Hello, megadeth95!

\(\displaystyle P(x) \:=\: x^3+2x^2-(4k+5)x-6\text{ has three distinct zeros.}\)
\(\displaystyle \text{Two of these are identical to the zeros of: }\,f(x) \:=\: x^2+5x+k.\)
\(\displaystyle \text{Find the value of }k.\)

Let \(\displaystyle a,b,c\) be the roots of the cubic.
Let \(\displaystyle a,b\) be roots of the quadratic.

We have: .\(\displaystyle \begin{Bmatrix}a+b+c \:=\:\text{-}2 & [1] \\ ab + bc + ac \:=\: \text{-}(4k+5) & [2] \\ abc \:=\:6 & [3]\end{Bmatrix} \quad \begin{Bmatrix}a+b \:=\:\text{-}5 & [4] \\ ab \:=\:k & [5] \end{Bmatrix}\)


Substitute [4] into [1]: .\(\displaystyle \text{-}5 + c \:=\:\text{-}2 \quad\Rightarrow\quad c \:=\:3\;\;[6]\)


In [2] we have: .\(\displaystyle ab + c(a+b) \:=\:\text{-}4k-5\)

Substitute [5], [6], [4]: .\(\displaystyle k + 3(\text{-}5) \:=\:\text{-}4k - 5 \)

. . . . . \(\displaystyle k - 15 \:=\:\text{-}4k - 5 \quad\Rightarrow\quad 5k \:=\:10\)

Therefore: .\(\displaystyle k \:=\:2\)