If x^3 - 3x^2 3x - 7 = (x + 1) (ax^2 + bx + c), then a + b + c =

Question

If $x^3 - 3x^2 3x - 7 = (x + 1) (ax^2 + bx + c),$ then $a + b + c =$

✓

Answer

First multiply
$(x + 1) (ax^2 + bx + c)$
$= ax^3 + bx^2 + cx + ax^2 + bx + c$
$= ax^3 + bx^2 + ax^2 + cx + c$
$= ax^3 + (b + a)x^2 + (c + b)x + c$
Comparing it with
$x^3 - 3x^2 + 3x - 7$
$a = 1$
$b + a = -3 \Rightarrow b + 1 + -3 \Rightarrow b = -4$
$c + b = 3 \Rightarrow c - 4 = 3 \Rightarrow c = 7$
$c = -7$ should be $7$
as if we put $x = -1$ in
$x^3 - 3x^2 + 3x - 7$
$-1 - 3 - 3 - 7 =14$
so $x + 1$ can not be factor so $x + 1$ will be factor if $x^3 - 3x^2 + 3x - 7$ is actually
$x^3 - 3x^2 + 3x + 7$
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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