P(x) = x^2 + ax + a is divisible by x + b with a and b in real numbers.

(A) b =/= 1 and a = - b / (b - 1)
(B) b =/= 1 and a = - b^2 / (b - 1)
(C) b =/= 1 and a = b^2 / (b - 1)
(D) b =/= 1 and a = b / (b - 1)

It's just for hobby's sake, I just can't figure this one out. It's probably very simple haha.

For polynomials, P being divisible by Q means there exists a polynomial R such that : P = Q . R

As a consequence, degree(P) = degree(Q) + degree(R)

In your example, P = X^2 + aX + a has degree 2, and Q = X + b has degree 1, hence R must have degree 1, which means R = cX + d

Now express P = QR :

X^2 + aX + a = ( X + b )( cX + d ) = cX^2 + (d + bc)X + bd

Identify all coefficients :

1 = c
a = d + bc
a = bd

and solve for c, d :
c = 1
d = a - b = a/b providing b is not zero.

The last equation gives a condition on a and b for this to be true :
a - b = a/b

Solve for a :
a = b^2 / ( b - 1 )
with b =/= 1

In case b = 0, then a = 0 and all the 4 formulas proposed are correct.

Eureka, I comprehend your explanation at last instead of just understanding it. What do you think of this?
Cos (x) = Sin (x) + 1 / Sqrt(3); What is Cos ^3 (x) - Sin ^3 (x)? I know for a fact that the answer has no variable in it, but I keep ending up with one anyway.

Attempts:




/e Mathway can solve cos^(3)(x) - sin^(3)(x) where cos(x) = sin(x) + (1)/(\sqrt(3)) but cannot solve a^(3)-b^(3) where a = b + (1)/(\sqrt(3)) without variables in the answer. I suspect that this means there is a trigonomical identity left for me to apply somewhere.

This post was edited by Forg0tten on Dec 31 2016 04:44pm

c = cos (x)
s = sin (x)

c = s + 1/√3
c-s = 1/√3

c³ - s³ = (c-s)(c²+cs+s²) = (c-s)(1+cs)
c³ - s³ = (1/√3)(1+cs)

On another hand, (c-s)² = c² -2cs + s² = 1 - 2cs = 1/3
hence, cs = 1/3

and

c³ - s³ = (1/√3)(1+1/3) = 4 / (3√3)

math is allways simple =)

Looks like I was close lol, I can't believe I haven't thought of substituting c-s = 1/Sqrt(3)
What have you done at the bold part, just take the square on both factors at c-s = 1/Sqrt(3) --> (c-s)^2 = (1/Sqrt(3))^2 and simplify?
Genius ._. I guess I know all the know-hows, but it's been too long to remember when to use which tool available to me hah.

omg i hate calc ima refresh this