Factor Theorem Question: Find a and b for P(x) as a factor of T(x)

[Rithikha]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Let (px+q) be a factor of P(x),
p can possibly be 1 and so can q, according to factor theorem,
Hence, factors (x+1) or (x-1)
P(1) = 0, substituting I got -2 as a and P(-1) =0 , I got 2 as a
If either (x+1) or (x-1) is a factor of P(x), it has to be a factor of T(x),
T(1) = 0, I got 14 as b in both cases.
But the correct answer is a=3,-3 and b=-6,6 respectively.

 

[SammyS]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Let (px+q) be a factor of P(x),
p can possibly be 1 and so can q, according to factor theorem,
Hence, factors (x+1) or (x-1)
P(1) = 0, substituting I got -2 as a and P(-1) =0 , I got 2 as a
If either (x+1) or (x-1) is a factor of P(x), it has to be a factor of T(x),
T(1) = 0, I got 14 as b in both cases.
But the correct answer is a=3,-3 and b=-6,6 respectively.

Hello Rithikha. Welcome to PF !

You want $\ p\, x +q \$ to be a factor of $\ T(x)\$ rather than $\ P(x)\$.

 

[Rithikha]

Hello Rithikha. Welcome to PF !

You want $\ p\, x +q \$ to be a factor of $\ T(x)\$ rather than $\ P(x)\$.

But the constant term is b in T(x), we can get q from it. Or can we?

 

[HallsofIvy]

Your error is in assuming that P(x) has a linear factor with rational coefficients (which you do when you use the rational root theorem).

Instead let y= P(x) and rewrite T(x) in terms of y.

 

[Ray Vickson]

But the constant term is b in T(x), we can get q from it. Or can we?

You want $2x^3-16x + b = (d x + c)(x^2 + ax + 1)$. It is easy to see that we must have $d = 2$, so we need $2x^3 - 16 x + b = (2x + c)(x^2 + ax +1)$.