# Thread: Help with functions or whatever it's name is

1. ## Help with functions or whatever it's name is

Hello guys, i've got this report i have to do and would really need your help.. This will propably be a piece of cake for you guys. Here it is:

Which value for the constant a is the polynom Q(x)=2ax^3-x+a^2 divideable with
a) x+1 b)2x+1

2. Originally Posted by FelluRan
Hello guys, i've got this report i have to do and would really need your help.. This will propably be a piece of cake for you guys. Here it is:

Which value for the constant a is the polynom Q(x)=2ax^3-x+a^2 divideable with
a) x+1 b)2x+1
Use the factor theorem: If $(x-p)$ is a factor of $f(x)$then $f(p) = 0$

$Q(-1) = 2a(-1)^3-(-1)+a^2 = a^2-2a+1 = (a-1)^2 = 0$

$\therefore a = 1$

3. Could you please write more specificly what you have done ? (I know i'm stupid )

4. Factor theorem - Wikipedia, the free encyclopedia

$Q(x)=2ax^3-x+a^2$

The factor theorem says that $(x-p)$ is a factor of $f(x)$ only if $f(p) = 0$

As the question says find the value of $a$ for which $x+1$ is a factor of $Q(x)$ this means the factor theorem is applied and so $Q(-1) = 0$

This means we find the value of Q(-1) by substituting $-1$ wherever $x$ is seen and setting $Q(x)$ to $0$

$Q(-1) = 2a(-1)^3 -(-1) + a^2 = 0$

This simplifies to $a^2-2a+1=0$ which in turn simplifies to $(a-1)^2 = 0$ and hence it's clear that a = 1

5. To be precise, $a = 1$ found in this way is the necessary, but not automatically sufficient, condition for $Q(a, x)$ to be divisible by $x + 1$. Therefore, to show that it is sufficient, one can actually verify that $x + 1$ divides $Q(1, x)$ with no remainder.