help...help..help...factor theorems...mathematicians..?

Question

The polynomial x^3 + ax^2 + bx +c os exactly divisible by x^2 -x-2. Find the value of a+b.
The polynomial 2x^2 + bx -3 has a factor x -a, where a is not equal to 0.Express b in terms of a.

Answer 1

1. since x^3 +ax^2+bx+C is divisible by x^2-x-2

x^3+ax^2+bx+C=(x^2-x-2)(x+e) e is a constant since cooefficient of x^3 term is 1 there can be only 1.x in the second factor. right hand term upon multiplication be equal to
x^3+(e-1)x^2-(2+e)x-2e = x^3+ax^2+bx+c

we can equate cooeficients now e-1 = a -----(1)
-2 -e=b-------(2)
add (1) +(2) -3=a+b (e s cancel out)

hence a+b is -3

(2)question like the above
2x^2+bx-3=(x-a)(2x+C) c is unknown constant second factor is 2x because the highest term x^2 on left side also has the coefficient 2, one can convince him/herself by putting dx+C as the second factor and equating coefficients

here equating coefficients give 1--- c-2a = b (x coefficients)
2 ---ac = 3 (constant term)

substitute c=3/a from 2 to 1====> 3/a -(2a) = b
a should not be equal to 0 otherwise c will be infinity.

Answer 2

The polynomial x³ + ax² + bx + c is exactly divisible by x² - x - 2. Find the value of a + b

x² - x - 2 = (x - 2)(x + 1)

So if P(x) = x³ + ax² + bx + c is divisible by x² - x - 2 then it is divisible by (x - 2) and (x + 1)

So P(2) = P(-1) = 0

ie 2³ + a.(2²) + b.2 + c = 0
ie 4a + 2b + c = -8 ..... (1)

Also (-1)³ + a.(-1)² + b.(-1) + c = 0
ie a - b + c = 1 ........... (2)

Equation (1) - equation (2)

3a + 3b = -9
So a + b = -3

The polynomial 2x² + bx - 3 has a factor x - a, where a is not equal to 0. Express b in terms of a.

Let Q(x) = 2x² + bx - 3
Since x - a is a factor Q(a) = 0

ie 2a² + ba - 3 = 0
So ba = 3 - 2a²
So b = (3 - 2a²)/a

Answer 3

Let's do these one at a time.
If x^3 + ax^2 + bx + c is divisible by x² - x -2 = (x+1)(x-2)
Then -1 and 2 are roots of x^3 + ax^2 + bx + c = 0.
So let's plug these values into this equation. We get
-1 +a -b + c = 0
a - b + c = 1
8 +4a +2b + c= 0
4a + 2b + c = -8
Subtracting,
3a + 3b = -9
a + b = -3.

2. If x-a is a factor of 2x² + bx -3 then a is a root of
2x² + bx -3 = 0.
So plug a into this equation to get
2a² + ab -3 = 0.
b =(-2a² +3)/a.