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):

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

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

For part (b):

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

 

[ConnorK]

For part (a):

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

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

For part (b):

Since [imath]P(x)[/imath] is a cubic polynomial, [imath]Q(x)[/imath] must be a linear polynomial (of degree 1). Let’s write [imath]Q(x)[/imath] as [imath]Q(x)=x+c[/imath] (with c to be determined).
Thus, we can expand [imath]P(x)[/imath] as: [imath]P(x)=(x−1)(x+2)(x+c)+ax+b[/imath]
We are also told that [imath]x=-1[/imath] is a root of [imath]P(x)[/imath], meaning [imath]P(−1)=0[/imath]. We can use this to solve for [imath]c[/imath].
To show that the equation [imath]P(x)=0[/imath] has no other real solutions besides [imath]x=−1[/imath], 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:
[imath]P(x)[/imath] has a remainder of [imath]3[/imath] when divided by [imath](x-2)[/imath] we know [imath]P(2)=3[/imath]
Using these conditions, we can plug [imath]x=1[/imath] and [imath]x=2[/imath] into the remainder form [imath]P(x)=(x−1)(x-2)Q(x)+ax+b[/imath] and solve for [imath]a[/imath] and [imath]b[/imath].
After you plug in those values you will get the system of two linear equations:
[math]\begin{cases} 2=a+b \\ 3=2a+b \end{cases}[/math]