Help with cubic functions?

P(x) = x³ + ax² - x + b
Since P(x) is divisible be x – 1, p(1) = 0
Since P(x) is divisible be x + 3, p(–3) = 0
p(1) = (1)³ + a(1)² - (1) + b = 0
1 + a - 1 + b = 0
a + b = 0 Eq. (1)
p(–3) = (–3)³ + a(–3)² – (–3) + b = 0
–27 + 9a + 3 + b = 0
9a + b = 24 Eq. (2)

9a + b = 24 Eq. (2)
a + b = 0 Eq. (1)
----subtract------------------
8a = 24
a = 24/8 = 3

9a + 9b = 0 Eq. (1) x 9
9a + b = 24 Eq. (2)
---subtract------------------
8b = –24
b = –24/8 = –3

(a, b) = (3, –3)

p(x) = x³ + 3x² – x – 3
p(x) = x²(x + 3) – (x + 3)
p(x) = (x + 3)(x² – 1)
p(x) = (x + 3)(x + 1)(x – 1)

Since (x-1) and (x+3) are factors,
So P(x) = x^3 + ax^2 - x + b
can be written as P(x) = (x-1)(x+3)(x+c)
where (x+c) is the third factor.

Noted that each term of the polynomials forming an equation below:
for x^0: b = (-1)(3)(c) = -3c....--------- equation (1)
for x^1: -1 = (-1)(3) + (3)c + (-1)c
...... => -1 = -3 + 2c
...... => c = 1....---------------------------- equation (2)
from equation 1, => b = -3
for x^2: a = (-1) + (3) + c
...........a = 2 + c, ....--------------------- equation (3)
from equation 2, a = 3

So P(x) = x^3 + 3x^2 -x -3

It's pretty simple.Substitute 1 in x, to get a+b=0 or a = -b.Then substitute 3 in x to get, 9a+b+24.Now put a= -b and solve the equation to find b and use that value to get a.Hope you got it.Good luck.

Since we know that it's divisible by x - 1 an d x + 3, the remainders must be zero. Let's divide through them , see what we get for remainders, then set it to zero and see what happens:

. . . . _x²_+_(a + 1)x_+_a___
x - 1 ) x³ + ax² - x + b
. . . . .x³ - x²
. . . . .----------------
. . . . . . . .(a + 1)x² - x + b
. . . . . . . .(a + 1)x² - (a + 1)x
. . . . . . . .----------------------------
. . . . . . . . . . . . . ax + b
. . . . . . . . . . . . . ax - a
. . . . . . . . . . . . . -----------
. . . . . . . . . . . . . . . . .b + a

So we know that:

b + a = 0

Now, do the same with x + 3 :

. . . . ._x²_+_(a - 3)x_+_(8 - 3a)___
x + 3 ) x³ + ax² - x + b
. . . . . x³ + 3x²
. . . . .----------------
. . . . . . . .(a - 3)x² - x + b
. . . . . . . .(a - 3)x² + 3(a - 3)x
. . . . . . . .----------------------------
. . . . . . . . . . . . . (8 - 3a)x + b
. . . . . . . . . . . . . (8 - 3a)x + 3(8 - 3a)
. . . . . . . . . . . . . -----------
. . . . . . . . . . . . . . . . .b + 9a - 24

Now set that to 0, we get:

b + 9a - 24 = 0
b + 9a = 24

Now we have a system of two equations. solve for a and b:

b + a = 0
b = -a

b + 9a = 24
-a + 9a = 24
8a = 24
a = 3

b = -3

So the function is:

P(x) = x³ + 3x² - x - 3