X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), i have to find the value of k,
can you please help me with starting the question, i am completely confuse
(ps of anyone is confuse ^ means power)
X^3 + 2x^3 - 5x - 6 = (x-2)(x+3)(x-k), i have to find the value of k,
can you please help me with starting the question, i am completely confuse
(ps of anyone is confuse ^ means power)
If you are choosing to expand the RHS, then you will have:
$\displaystyle x^3+2x^2-5x-6=(x-2)(x+3)(x-k)$
$\displaystyle x^3+2x^2-5x-6=\left(x^2+x-6\right)(x-k)$
$\displaystyle x^3+2x^2-5x-6=\left(x^3+x^2-6x\right)-\left(kx^2+kx-6k\right)$
$\displaystyle x^3+2x^2-5x-6=x^3+(1-k)x^2-(6+k)x+6k$
Equating coefficients involving k, we find:
$\displaystyle 2=1-k$
$\displaystyle -5=-(6+k)$
$\displaystyle -6=6k$
Take your pick, and solve for k.
Find the value of k in terms of X right?
x^3 + 2x^3 - 5x -6 = (x-2)(x+3)(x-k)
= x^3 + 2x^3 - 5x - 6 = (x^2 + x - 6)(x - k)
= x^3 + 2x^3 - 5x - 6 = (x^3 - (k)x^2 + x^2 - (k)x -6x -(k)6
Subtract x^3 from both sides to get
2x^3 -5x - 6 = (k)x^2 + x^2 - (k)x - 6x - (k)6
Add 6x to both sides to get
2x^3 + x - 6 = (k)x^2 + x^2 - (k)x - (k)6
Subtract x^2 from both sides to get
2x^3 + x^2 + x - 6 = (k)x^2 + (k)x - (k)6
Take k out from the right side to get
2x^3 + x^2 + x - 6 = (k)(x^2 + x - 6) <--------- I may have messed up from this part, check your math too.
Look, we have x^2 + x - 6 on both sides so divide
2x^3 + 1 = (k)