Unknown Remainder from Polynomial Division

[ConnorK]

Hello

I have been stuck on this question for a while. I have no working out to show as i don't know how to even approach this question. All help appreciated.

 

[Pedja]

For part (a):

• $P(x)$ has a remainder of $2$ when divided by $x-1$ we know $P(1)=2$
• $P(x)$ has a remainder of $6$ when divided by $x+2$ we know $P(-2)=6$

Using these conditions, we can plug $x=1$ and $x=-2$ into the remainder form $P(x)=(x−1)(x+2)Q(x)+ax+b$ and solve for $a$ and $b$.

For part (b):

Since $P(x)$ is a cubic polynomial, $Q(x)$ must be a linear polynomial (of degree 1). Let’s write $Q(x)$ as $Q(x)=x+c$ (with c to be determined).
Thus, we can expand $P(x)$ as: $P(x)=(x−1)(x+2)(x+c)+ax+b$
We are also told that $x=-1$ is a root of $P(x)$, meaning $P(−1)=0$. We can use this to solve for $c$.
To show that the equation $P(x)=0$ has no other real solutions besides $x=−1$, we would analyze the discriminant or check for other real roots via technique like factoring.

 

[ConnorK]

For part (a):

• $P(x)$ has a remainder of $2$ when divided by $x-1$ we know $P(1)=2$
• $P(x)$ has a remainder of $6$ when divided by $x+2$ we know $P(-2)=6$

Using these conditions, we can plug $x=1$ and $x=-2$ into the remainder form $P(x)=(x−1)(x+2)Q(x)+ax+b$ and solve for $a$ and $b$.

For part (b):

Since $P(x)$ is a cubic polynomial, $Q(x)$ must be a linear polynomial (of degree 1). Let’s write $Q(x)$ as $Q(x)=x+c$ (with c to be determined).
Thus, we can expand $P(x)$ as: $P(x)=(x−1)(x+2)(x+c)+ax+b$
We are also told that $x=-1$ is a root of $P(x)$, meaning $P(−1)=0$. We can use this to solve for $c$.
To show that the equation $P(x)=0$ has no other real solutions besides $x=−1$, we would analyze the discriminant or check for other real roots via technique like factoring.

• P(x) has a remainder of 6 when divided by x+2 we know P(−2)=6

I don't understand where you got the remainder of 6 from or the x+2 from or how i would plug in those values of x in the third equation and solve for a and b.

 

[Pedja]

Sorry , I misread a question. It should read:
$P(x)$ has a remainder of $3$ when divided by $(x-2)$ we know $P(2)=3$
Using these conditions, we can plug $x=1$ and $x=2$ into the remainder form $P(x)=(x−1)(x-2)Q(x)+ax+b$ and solve for $a$ and $b$.
After you plug in those values you will get the system of two linear equations:
$$\begin{cases} 2=a+b \\ 3=2a+b \end{cases}$$