Find the value of a and b such that x - 2 is a factor of both

x^3 + (a+3)x^2-(a+6)x+4b and x^4-ax^3+(b-2)x^2-(a-1)x-(b-1)

By the Factor Theorem, if x - c is a factor of f(x), then f(c) = 0.

If f(x) = x^3 + (a + 3)x^2 - (a + 6)x - 4b and g(x) = x^4 - ax^3 + (b - 2)x^2 - (a - 1)x - (b - 1), then:
f(2) = g(2) = 0.

By evaluating f(x) and g(x) at x = 2:
(i) f(2) = 0
==> 2^3 + (a + 3)(2)^2 - (a + 6)(2) - 4b = 0
==> 4(a + 3) - 2(a + 6)-+ 4b + 8 = 0
==> 4a + 12 - 2a - 12 - 4b + 8 = 0
==> 2a - 4b + 8 = 0
==> a - 2b = -4.
(ii) g(2) = 0
==> 2^4 - a(2)^3 + (b - 2)(2)^2 - (a - 1)(2) - (b - 1) = 0
==> 16 - 8a + 4(b - 2) - 2(a - 1) - (b - 1) = 0
==> 16 - 8a + 4b - 8 - 2a + 2 - b + 1 = 0
==> -10a + 3b + 11 = 0
==> 10a - 3b = 11.

Solving (i) and (ii) simultaneously gives a = 2 and b = 3.

I hope this helps!

Then 2 is root for both:

(2)^3 + (a+3)(2)^2-(a+6)(2)-4b = 0

(2)^4-a(2)^3+(b-2)(2)^2-(a-1)(2)-(b-1) = 0

8 + (a+3)(4) -2(a+6) - 4b = 0

16-8a + 4(b-2) -2(a-1)-(b-1) = 0


8 + 4a +12 -2a - 12 - 4b = 0

16-8a + 4b - 8 -2a + 2 -b + 1 = 0

8 + 2a - 4b = 0

11 - 10a + 3b = 0

2a - 4b = -8

- 10a + 3b = -11

2a = 4b - 8

- 10a + 3b = -11

a = 2b - 4

- 10a + 3b = -11

a = 2b - 4

- 10(2b-4) + 3b = -11

- 20b +40 + 3b = -11

- 17b = -51

b = 51/17

b = 3

a = 2(3) - 4

a = 2

Check:

x^3 + 5x^2- 8x - 12

(2)^3 + 5(2)^2- 8(2) - 12 = 0?

8 + 20 - 16 - 12 = 0?

0 = 0

x^4 - 2x^3 + x^2 -x -2

(2)^4 - 2(2)^3 + (2)^2 -(2) -2 = 0?

16 - 16 + 4 -2 -2 = 0?

0 = 0

This checks.