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The remainder of dividing $P(x)$ by $(x-1)$ is $5$ and by $(x+2)$ is $2$. Determine the remainder of dividing $P(X)$ by $x^2+x-2$.

I tried writing $P(x)$ as $P(x) = (x-1)(x+2)B(x)+K$ and thought that K should be equal to $P(1)$ or $P(-2)$ which are equal to 5 and 2. But is it possible to find a more exact solution?

Kratch32
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    Looks like you have a typo - I'm guessing you want $x+2$ in the form you tried. You should also make sure you're consistent in using $x$ or $X$. – Chris Lewis Jan 19 '24 at 22:24
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    The form you have is wrong. When you divide a polynomial by a quadratic, the remainder will be linear in $x$; ie the form you should be trying is $P(x)=(x-1)(x+2)B(x)+ax+b$. Now, you know $P(1)$ and $P(-2)$, so what happens when you substitute these values in? – Chris Lewis Jan 19 '24 at 22:28
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    $p(x) = (x-1)q(x)+5$, so $p(-2) = 2 = -3q(-2)+5$ from which we get $q(-2) = 1$. Hence $q(x) = (x+2) w(x) + 1$ for some $w$. Finally, $p(x) = (x-1)(x+2)w(x) + x-1 +5 = (x-1)(x+2)w(x)+x+4$. – copper.hat Jan 19 '24 at 22:39
  • As an alternative $P(x)-(x+4)$ has roots $1,-2$ – Hari Shankar Jan 20 '24 at 07:46

1 Answers1

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When $P(x)$ is divided by $x - 1$, the remainder is $5$. So, using the remainder theorem, $P(1) = 5$. Similarly, $P(-2) = 2$

Now, let $$\begin{align*} &P(x) = (x^2 + x - 2)Q(x) + R(x) \\[0.3cm] \Longrightarrow \,\,\, &P(x) = (x - 1)(x + 2)Q(x) + R(x) \end{align*}$$

As Chris pointed out, the remainder $R(x)$ need not be a constant.

The remainder has to be of a degree smaller then the divisor's. So we assume it's a linear polynomial.

$R(x) = ax + b$,$\,$ for some $a, \, b$.

So,

$$P(x) = (x - 1)(x + 2)Q(x) + ax + b$$

Moving ahead

$$\begin{align*} &P(1) = 0 + a + b \\[0.3cm] \Longrightarrow \,\,\, &5 = a + b \end{align*}$$

Similarly, $-2a + b = 2 \,\,$ [using $P(-2)$]

So, we have two equations in $a$ and $b$. Solving them, we get $a = 1$ and $b = 4$.

Hence,

$$R(x) = x + 4$$

Haris
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