Question

If x³ - 3x² + 3x - 7 = (x + 1)(ax² + bx + c), then find (a + b + c)

Answer 1

If x³ - 3x² + 3x - 7 = (x + 1)(ax² + bx + c), then find (a + b + c)

first multiply

(x + 1)(ax² + bx + c)

= ax³ + bx² + cx + ax² + bx + c

= ax³ + bx² + ax² + cx + bx + c

= ax³ + (b+a)x² + (c + b)x + c

comparing it with

x³ - 3x² + 3x - 7

a = 1

b + a = -3 => b + 1 = -3 => b = -4

c + b = 3 = > c - 4 = 3 => c = 7

c = -7 should be 7


as if we put x = -1 in
x³ - 3x² + 3x - 7
-1 - 3 - 3 - 7 = 14 so x + 1 can not be factor so x +1 will be factor if
x³ - 3x² + 3x - 7 is actually
x³ - 3x² + 3x + 7
then -1 - 3 - 3 + 7 = 0

hence we can say that
a = 1
b = -4
c = 7
so a + b + c = 4

Answer 2

Step-by-step explanation:

ANSWER

Consider the given equation.

x3−3x2+3x−7=(x+1)(ax2+bx+c)                 .............(1)

Since, (x+1) is the factor, then we get

x2−4x+−7=ax2+bx+c

On comparing both sides, we get

a=1,b=−4,c=−7

Therefore,

a+b+c=1−4−7=−10

Hence, this is the answer.