Hello guys!
How are you?
I was just wondering if you guys knew how to do this question?
I've done question a) but don't know how to do question b).
Thank you so much.
The polynomial P(x) has a remainder of 2 when divided by x-1 and a remainder of 3 when divided by x-2.The remainder when P(x) is divided by (x-1)(x-2) is ax+b, i.e. P(x) can be written as P(x)=(x-1)(x-2)Q(x)+ax+b.
a) Find the values of a and b
b) i) Given that P(x) is a cubic polynomial with coefficient of x^3 being 1, and -1 is a solution of the equation P(x)=0, find P(x).
ii) Show that the equation P(x)=0 has no other real solutions.
Hey Phoenix11
So from part a, we know:
$$ a = 1, b = 1 $$
$$ \implies P(x) = (x-1)(x-2)Q(x)+x+1 $$
Because P(x) is cubic, Q(x) must be linear. So let Q(x) = cx+d. Furthermore, since the coefficient of x^3 is 1, c = 1, as the cubic term of P(x) comes from the product of the x terms of (x-1), (x-2) and (cx+d), which means cx^3 = x^3.
$$ \implies P(x) = (x-1)(x-2)(x+d)+x+1 $$
If x = -1 is a solution of P(x) = 0:
\begin{align*}
\implies P(-1) &= 0\\
\implies 0 &= (-2)(-3)(d-1)\\
\implies d &= 1\\
\therefore P(x) &= (x-1)(x-2)(x+1)+x+1\\
\end{align*}
P(x) may also be written as:
\begin{align*}
P(x) &= (x+1)((x-1)(x-2)+1)\\
&= (x+1)(x^2-3x+3)
\end{align*}
which shows x = -1 is a solution to P(x)=0. The discriminant of the quadratic term = 9 - 12 = -3. Therefore, x = -1 is the only real solution, as required.
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