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So I'm a high school student and I'm stuck on a question. Please help.

$f(x)=(x-1)^2Q(x)+3x+1$

$f(x)=(x+2)Q(x)+4$

$f(x)=(x-1)^2(x+2)Q(x)+R(x)$

My first approach was Making $R(x)=ax^2+bx+c$, I soon found out that there are only two equations not three to find $a, b, c$

I don't understand how to solve this.

Lucas Kim
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    I guess it's not the same $Q(x)$ on the three lines? You should find a less confusing way to write it. Then remember where the extra equation would come from when the divisor has a double root. – dxiv May 24 '21 at 02:26
  • Notice that $R(x) $ divided by $(x-1)^2$ leaves remainder $3x+1$ and dividing by $x+2$ leaves remainder $4$. That should help you to figure out $a, b, c$. – Paramanand Singh May 24 '21 at 02:29
  • To expand on the comment from @dxiv: differentiate. – Gerry Myerson May 24 '21 at 03:00

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Here's another approach. $$f(x)=(x-1)^2Q_1(x)+3x+1 \tag 1 $$ $$f(x)=(x+2)Q_2(x)+4 \tag 2 $$ $$f(x)=(x-1)^2(x+2)Q_3(x)+R(x) \tag 3 $$ From (1) and (3), $(x-1)^2$ divides $R(x)-3x-1.$ Therefore, $$R(x)=k(x-1)^2+3x+1 \tag 4$$ By (2), $f(-2)=4.$ By (4) and (3), $$R(-2)=9k-6+1=f(-2)=4.$$ Thus, $k=1$ and $$R(x)=x^2+x+2.$$

David
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