mathmath333
  • mathmath333
question
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{if}\ x^2+ax+b\ \text{leaves same remainder } 5\hspace{.33em}\\~\\ & \normalsize \text{when divided by}\ (x-1)\ \text{or}\ (x+1)\hspace{.33em}\\~\\ & \normalsize \text{then the values of}\ a\ \text{and}\ b\ \text{are}\hspace{.33em}\\~\\ \end{align}}\)
freckles
  • freckles
\[f(x)=x^2+ax+b \\ f(1)=1^2+a(1)+b=5 \\ f(-1)=(-1)^2+a(-1)+b=5\]
freckles
  • freckles
\[a+b=4 \\ -a+b=4\]

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mathmath333
  • mathmath333
is \(f(1)\) for \((x-1)\)
freckles
  • freckles
yep
freckles
  • freckles
f(-1) is for dividing by x-(-1)
mathmath333
  • mathmath333
what is this method called
freckles
  • freckles
remainder theorem
freckles
  • freckles
\[\frac{P(x)}{x-1}=Q(x)+\frac{R}{x-1} \\ P(x)=Q(x)(x-1)+R \\ \text{ plugin 1} \\ P(1)=Q(1)(1-1)+R \\ P(1)=Q(1)(0)+R \\ P(1)=0+R \\ P(1)=R\]
freckles
  • freckles
where our R=5 here
freckles
  • freckles
and we can then divide P(x) by (x+1) and plug in -1 and you see we will get R there too
mathmath333
  • mathmath333
thnx
mathmath333
  • mathmath333
the previus one is easy

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