Question
If $x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c),$ then $a + b + c =$
✓
Answer
The given equation is
$x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$
This can be written as
x$^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$
$= x^3 - 3x^2 + 3x - 7 = ax^3 + bx^2 + cx + ax^2 + bx + c$
$= x^3 - 3x^2 + 3x - 7 = ax^3 + (a + b)x^2 + (b + c)x + c$
Comparing the cofficients on both sides of the equation.
We get,
$a = 1 ...(1)$
$a + b = 3 ...(2)$
$b + c = 3 ...(3)$
$c = -7 ...(4)$
Putting the value of a form $(1)$ in $(2)$
We get,
$1 + b = 3,$
$b = -3 - 1$
$b= -4$
So the value of $a, b$ and $c$ is $1, -4$ and $-7$ respectively.
Therefore,
$a + b + c = 1 - 4 - 7 = -10$
$x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$
This can be written as
x$^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$
$= x^3 - 3x^2 + 3x - 7 = ax^3 + bx^2 + cx + ax^2 + bx + c$
$= x^3 - 3x^2 + 3x - 7 = ax^3 + (a + b)x^2 + (b + c)x + c$
Comparing the cofficients on both sides of the equation.
We get,
$a = 1 ...(1)$
$a + b = 3 ...(2)$
$b + c = 3 ...(3)$
$c = -7 ...(4)$
Putting the value of a form $(1)$ in $(2)$
We get,
$1 + b = 3,$
$b = -3 - 1$
$b= -4$
So the value of $a, b$ and $c$ is $1, -4$ and $-7$ respectively.
Therefore,
$a + b + c = 1 - 4 - 7 = -10$
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