Polynomial Factors: Find a & b | Math Problem

[evanbirch]

This problem came up on a math contest the other day and I drew a blank - now it's bugging me.

"If the polynomial P(x)=x^2+ax+1 is a factor of T(x)=2x^3-16x+b, find the values of a and b."

Anyone have any ideas?

 

[DeadWolfe]

We have that there is a c such that (x^2+ax+1)(2x+c)=2x^3-16x+b. Thus comparing coefficients gives us:

c+2a=0
2+ac=-16
b=c

Solving this yields

a=3 or -3, and thus
b=-6 or 6

 

[tacman]

To solve this problem, we can use the factor theorem which states that if a polynomial P(x) is a factor of another polynomial Q(x), then P(-c) = 0 where c is the root of P(x). In this case, we know that P(x) = x^2+ax+1 is a factor of T(x) = 2x^3-16x+b. This means that when we plug in the root of P(x) into T(x), we should get a result of 0.

So, let's set P(x) = 0 and solve for x:

x^2+ax+1 = 0

Using the quadratic formula, we get:

x = (-a ± √(a^2-4))/2

Since we know that P(x) is a factor of T(x), we can plug in the root we just found into T(x) and set it equal to 0:

2((-a ± √(a^2-4))/2)^3 - 16((-a ± √(a^2-4))/2) + b = 0

Simplifying, we get:

(-a ± √(a^2-4))^3 - 8(-a ± √(a^2-4)) + b = 0

Now, we can expand the cube and simplify further:

-a^3 ± 3a^2√(a^2-4) + 3a(a^2-4) + b = 0

-a^3 ± 3a^3 - 12a + b = 0

Combining like terms, we get:

2a^3 - 12a + b = 0

Now, we have two unknowns (a and b) and one equation. To solve for both, we need another equation. We can use the fact that P(x) = x^2+ax+1 is a factor of T(x) to set up another equation:

T(x)/P(x) = 0

Substituting in the values we found for x and simplifying, we get:

(2x^3-16x+b)/(x^2+ax+1) = 0

2x - 16 + (b-16a)/(x^2+ax+1) = 0

Since P(x) is a factor of T(x