MCQ
If $x^3-3 x^2 3 x-7=(x+1)\left(a x^2+b x+c\right)$, then $a+b+c=$
- ✓$4$
- B$-10$
- C$12$
- D$3$
✓
Answer
Correct option: A.
$4$
First multiply
$(x+1)\left(a x^2+b x+c\right)$
$=a x^3+b x^2+c x+a x^2+b x+c$
$=a x^3+b x^2+a x^2+c x+c$
$=a x^3+(b+a) x^2+(c+b) x+c$
Comparing it with
$x^3-3 x^2+3 x-7$
$a=1$
$b+a=-3 \Rightarrow b+1+-3 \Rightarrow b=-4$
$c+b=3 \Rightarrow c-4=3 \Rightarrow c=7$
$\mathrm{c}=-7$ should be 7
as if we put $x=-1$ in
$x^3-3 x^2+3 x-7$
$-1-3-3-7=14$ so $x+1$ can not be factor so $x+1$ will be factor if $x^3-3 x^2+3 x-7$ is actually
$x^3-3 x^2+3 x+7$
then $-1-3-3+7=0$
Hence, we can say that
$a=1$
$b=-1$
$c=7$
so, $a+b+c=4$
$(x+1)\left(a x^2+b x+c\right)$
$=a x^3+b x^2+c x+a x^2+b x+c$
$=a x^3+b x^2+a x^2+c x+c$
$=a x^3+(b+a) x^2+(c+b) x+c$
Comparing it with
$x^3-3 x^2+3 x-7$
$a=1$
$b+a=-3 \Rightarrow b+1+-3 \Rightarrow b=-4$
$c+b=3 \Rightarrow c-4=3 \Rightarrow c=7$
$\mathrm{c}=-7$ should be 7
as if we put $x=-1$ in
$x^3-3 x^2+3 x-7$
$-1-3-3-7=14$ so $x+1$ can not be factor so $x+1$ will be factor if $x^3-3 x^2+3 x-7$ is actually
$x^3-3 x^2+3 x+7$
then $-1-3-3+7=0$
Hence, we can say that
$a=1$
$b=-1$
$c=7$
so, $a+b+c=4$
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