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A polynomial P(x) is divided by x2+x-2. P(-2)=5, P(1)=20. Find the remainder of the division said above.
 

 Mar 21, 2019
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A polynomial P(x) is divided by \(x^2+x-2\).

\(P(-2)=5,\ P(1)=20\).

Find the remainder of the division said above.

 

\(\text{Let $x^2+x-2 = (x-1)(x+2)$} \)

\(P(x)\) is divided by \((x-1)(x+2)\) it's is a polynomial of degree two.
Reminder should be a polynomial of degree less then 2. Say... \(r(x)=ax+b\)

 

\(\text{Let $r(x)=ax+b$ is the remainder of the division } \)

 

Now... \(P(x) = (x-1)(x-2) Q(x) + r(x)\)

 

\(\begin{array}{|lrcll|} \hline & \mathbf{P(x)} & \mathbf{=} & \mathbf{(x-1)(x+2) Q(x) + r(x)} \\ \hline x=-2: & P(-2)=5 & =& (-2-1)(-2+2) Q(-2) + r(-2) \\ & 5 & =& 0 + r(-2) \\ & \mathbf{r(-2)} &\mathbf{=}& \mathbf{5} \\ \hline x=1: & P(1)=20 & =& (1-1)(1+2) Q(1) + r(1) \\ & 20 & =& 0 + r(1) \\ & \mathbf{r(1)} &\mathbf{=}& \mathbf{20} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{r(x)} &\mathbf{=}& \mathbf{ax+b} \\ \hline r(-2)=5 &=& a\cdot (-2)+b \\ 5 &=& -2a+b \\ \hline r(1)=20 &=& a\cdot 1 +b \\ 20 &=& a+b \\ \hline \end{array}\)

 

Solve two equations and get the value of \(a\) and \(b\)

\(\begin{array}{|lrcll|} \hline & 5 &=& -2a+b \\ (1) & b&=& 5+2a \\\\ (2) & 20&=& a+b \quad | \quad b= 5+2a \\ & 20&=& a+5+2a \\ & 15&=& 3a \\ & \mathbf{a} &\mathbf{=}& \mathbf{5} \\\\ & b&=& 5+2a \\ & b &=& 5+2\cdot 5 \\ & \mathbf{b} &\mathbf{=}& \mathbf{15} \\ \\ & r(x) &=& ax+b \\ & \mathbf{ r(x) } &\mathbf{=}& \mathbf{5x+ 15} \\ \hline \end{array}\)

 

The remainder of the division is \(\mathbf{5x+ 15}\)

 

laugh

 Mar 21, 2019
edited by heureka  Mar 21, 2019

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