Please help! If (x+1) and (x-2) are factors f(x)=x^3+ax^2+bx+2, find the values of a and b and find the other factor?

f(x) = (x-2)(x^2-1) = x^3 - 2x^2-x+2
a = -2
b = -1
Missing factor = (x-1)

Method 1 :

Let f(x) = x³ + ax² + bx + 2

Since (x + 1) is a factor of f(x), then f(-1) = 0
(-1)³ + a(-1)² + b(-1) + 2 = 0
-1 + a - b + 2 = 0
a - b = -1 …… [1]

Since (x - 2) is a factor of f(x), then f(2) = 0
(2)³ + a(2)² + b(2) + 2 = 0
8 + 4a + 2b + 2 = 0
4a + 2b = -10
2a + b = -5 …… [2]

[1] + [2] :
3 = -6
a = -2

Put a = -2 into [1] :
(-2) - b = -1
b = -1

Let (x + c) be the other factor.

x³ - 2x² - x + 2
= (x³ - 2x²) + (-x + 2)
= x²(x - 2) - (x - 2)
= (x² - 1)(x - 2)
= (x - 1)(x + 1)(x - 2)

The other factor = x - 1


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Method 2 :

Let (x + c) be the other factor.

x³ + ax² + bx + 2 = (x + 1)(x - 2)(x + c)

Compare the constant terms on the both sides :
2 = 1 * (-2) * c
c = -1

The other factor = (x - 1)

x³ + ax² + bx + 2 = (x + 1)(x - 1)(x - 2)
x³ + ax² + bx + 2 = (x² - 1)(x - 2)
x³ + ax² + bx + 2 = (x² - 1)(x - 2)
x³ + ax² + bx + 2 = x³ - 2x² - x + 2

Compare the coefficients on the both sides :
a = -2
b = -1

F(x) = x3 + ax² + bx + 2

That equation factors are

(x + 1) and (x - 2)

F(-1) = -1 + a - b + 2

F(-1) = a – b + 1…………1

F(2) = 23 + a(2)² + b(2) + 2

F(2) = 8 + 4a + 2b + 2

F(2) = 4a + 2b + 10………….2

Solving 1 and 2

(a - b + 1 = 0)

4a + 2b + 10
---------------
2(a - b + 1 = 0)

4a + 2b + 10 = 0
-----------------
2a - 2b + 2 = 0

4a + 2b + 10 = 0
----------------
6a + 12 = 0

A = -2 sub equation 1

-2 - b + 1 = 0

-1 - b = 0

B = -1

If (x + 1) is factor f(x) = x³ + ax² + bx + 2, it means that: f(- 1) = 0

f(x) = x³ + ax² + bx + 2 → when x = - 1, the result is 0

(- 1)³ + a.(- 1)² + b.(- 1) + 2 = 0

- 1 + a - b + 2 = 0

a - b = - 1

a = b - 1 ← equation (1)



If (x - 2) is factor f(x) = x³ + ax² + bx + 2, it means that: f(2) = 0

f(x) = x³ + ax² + bx + 2 → when x = 2, the result is 0

(2)³ + a.(2)² + b.(2) + 2 = 0

8 + 4a + 2b + 2 = 0

4a + 2b = - 10

2a + b = - 5 → recall (1): a = b - 1

2.(b - 1) + b = - 5

2b - 2 + b = - 5

3b = - 3

→ b = - 1


Recall (1): a = b - 1

a = - 1 - 1

→ a = - 2



f(x) = x³ + ax² + bx + 2 → where: a = - 2 and where: b = - 1

f(x) = x³ - 2x² - x + 2

f(x) = (x³ - 2x²) - (x - 2)

f(x) = x².(x - 2) - (x - 2)

f(x) = (x - 2).(x² - 1)

f(x) = (x - 2).(x + 1).(x - 1) ← you can see that the other factor is (x - 1)

Hello,

if x + 1 is a factor, then x = - 1 is a zero of the polynomial (that is f(- 1) = 0):

f(- 1) = (- 1)³ + a(- 1)² + b(- 1) + 2 = 0

- 1 + a(1) - b + 2 = 0


a - b + 1 = 0 (#)


at the same time, if x - 2 is a factor, then x = 2 is a zero of the polynomial (that is f(2) = 0):

f(2) = (2)³ + a(2)² + b(2) + 2 = 0

8 + a(4) + 2b + 2 = 0

4a + 2b + 10 = 0

(dividing both sides by 2)


2a + b + 5 = 0 (##)


to find out a and b then let's solve the system:

a - b + 1 = 0 (#)
2a + b + 5 = 0 (##)

a = b - 1
2(b - 1) + b + 5 = 0

a = b - 1
2b - 2 + b + 5 = 0

a = b - 1
3b + 3 = 0

a = b - 1
3b = - 3

a = b - 1 = - 1 - 1 = - 2
b = - 1

then the polynomial becomes:

f(x) = x³ + (- 2)x² + (- 1)x + 2 =

x³ - 2x² - x + 2

to find the other factor we need to divide the polynomial by the product
(x + 1)(x - 2):

(x³ - 2x² - x + 2) /[(x + 1)(x - 2)] =

(expanding the denominator)

(x³ - 2x² - x + 2) /(x² - 2x + x - 2) =

(x³ - 2x² - x + 2) /(x² - x - 2) =

(long dividing)


x³ - 2x² - x + 2 | x² - x - 2

(dividing x³ by x²)

x³ - 2x² - x + 2 | x² - x - 2
........ ......... ....| x

(multiplying x by each term of the divisor and writing the results, with opposite signs, beneath the terms of the same degree of the dividend)

..x³ - 2x²... - x + 2 | x² - x - 2
- x³. + x² + 2x...... | x

(adding like terms together on the left)

..x³ - 2x²... - x + 2 | x² - x - 2
- x³. + x² + 2x...... | x
----- ------ ------ -----
. 0... - x².. + x + 2

(dividing - x² by x²)

..x³ - 2x²... - x + 2 | x² - x - 2
- x³. + x² + 2x...... | x - 1
----- ------ ------ -----
. 0... - x².. + x + 2

(multiplying - 1 by each term of the divisor and writing the results, with opposite signs, beneath the terms of the same degree of the trinomial down on the left)

..x³ - 2x²... - x + 2 | x² - x - 2
- x³. + x² + 2x...... | x - 1
----- ------ ------ -----
. 0... - x².. + x + 2
.......+ x²... - x. - 2

(adding like terms together down on the left, obtaining the remainder)

..x³ - 2x²... - x + 2 | x² - x - 2
- x³. + x² + 2x...... | x - 1
----- ------ ------ -----
. 0... - x².. + x + 2
.......+ x²... - x. - 2
------ ------- ------ ---
..........0... + 0 + 0 (remainder)

the other factor is the quotient (x - 1)


in conclusion:

x³ - 2x² - x + 2 = (x + 1)(x - 2)(x - 1)



I hope it's helpful

(x+1)(x-2)(x-c) = x^3+ax^2+bx+2
... + 2c = ... + 2
c = 1

(x+1)(x-2)(x-1) = x^3+ax^2+bx+2
Any multiple of 0 is 0.
When x=-1, (x+1)=0 and therefore x^3+ax^2+bx+2 = 0.
When x=2, (x-2)=0 and therefore x^3+ax^2+bx+2 = 0.
When x=1, (x-1)=0 and therefore x^3+ax^2+bx+2 = 0.

(-1)^3 + a(-1)^2 + b(-1) + 2 = 0
(2)^3 + a(2)^2 + b(2) + 2 = 0
(1)^3 + a(1)^2 + b(1) + 2 = 0

a = -2
b = -1