Question

if any give incarrect or say I don't know so I report and you lose your points​

Answer 1

Answer:

1; -4;-7

Step-by-step explanation:

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

I hope you like it

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Answer 2

Step-by-step explanation:

x^3 + 3x^2 + 3x - 7

you can also write 7 in the way shown below,

x^3 + 3x^2 + 3x + 1 - 8

(x^3 + 3x^2 + 3x + 1) - 8

as you can see that (x^3 + 3x^2 + 3x + 1) is the formula of (x + 1)^3, so writing that,

(x + 1)^3 - (2)^3

now it has become the formula of (a)^3 - (b)^3 which is (a-b) (a^2 + ab + b^2),

(x + 1 - 2) ( (x+1)^2 + 2 (x+1) + (2)^2)

(x - 1) (x^2 + 1 + 2x + 2x + 2 + 4)

(x-1) (x^2 + 4x + 7)

hope it helps.