MCQ
If $f(x+3)=x^2-7 x+2$, then the remainder when $f(x)$ is divided by $(x+1)$ is
- A8
- B-4
- C20
- D46
✓
Answer
D. 46
We have, $f(x+3)=x^2-7 x+2$. Let $x+3=\alpha$. Then, $x=\alpha-3$.
Replacing $x$ by $\alpha-3$, we obtain
$
f(\alpha)=(\alpha-3)^2-7(\alpha-3)+2 \text { or, } f(\alpha)=\alpha^2-13 \alpha+32
$
Thus, we obtain $f(x)=x^2-13 x+32$.
The remainder when $f(x)$ is divided by $x+1$ is $f(-1)=1+13+32=46$.
ALITER Putting $x+3=-1$ i.e. $x=-4$ in $f(x+3)=x^2-7 x+2$, we obtain: $f(-1)=16+28+2=46$.
We have, $f(x+3)=x^2-7 x+2$. Let $x+3=\alpha$. Then, $x=\alpha-3$.
Replacing $x$ by $\alpha-3$, we obtain
$
f(\alpha)=(\alpha-3)^2-7(\alpha-3)+2 \text { or, } f(\alpha)=\alpha^2-13 \alpha+32
$
Thus, we obtain $f(x)=x^2-13 x+32$.
The remainder when $f(x)$ is divided by $x+1$ is $f(-1)=1+13+32=46$.
ALITER Putting $x+3=-1$ i.e. $x=-4$ in $f(x+3)=x^2-7 x+2$, we obtain: $f(-1)=16+28+2=46$.
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