Polynomials

It is given that 2x^2-x-3 is a factor of f(x) = 2x^3+ax+bx-12.

( a ) Find the values of a and b
( b ) Factorize f (x)
(a)
2x^2-x-3=(x+1)(2x-3)
So, by factor theorem, x=-1 and x=3/2 is a root of f(x)
f(-1)=-2+a-b-12=a-b-14=0...(1)
f(3/2)=27/4+(9/4)a+(3/2)b-12=0
27+9a+6b-48=0
9a+6b-21=0
3a+2b-7=0...(2)
from (1) a=14+b
sub into (2)
3(14+b)+2b-7=0
5b=-35
b=-7
a=14+b=7
(b)
f(x) = 2x^3+ax^2+bx-12
f(x)
= 2x^3+7x^2-7x-12
=(2x^2-x-3)(x+4)
=(x+1)(2x-3)(x+4)

Note that 2x^2 - x - 3 = (2x - 3)(x + 1)
So by factor theorem, (2x - 3) and (x + 1) are factors of f(x)
i.e. f(3/2) = f(-1) = 0
i.e. 2(27/8) + 9a/4 + 3b/2 - 12 = 0 ...(1)
- 2 + a - b - 12 = 0 ...(2)

So, solving (1) and (2) gives a = 7, b = -7

Hence f(x) = (x + 1)(2x - 3)(x + c) for some constant c
By equating constant term, c = 4
So f(x) = (x + 1)(2x - 3)(x + 4)

It should be "f(x) = 2x^3 + ax^2 + bx - 12".

(a)

f(x) = 2x^3 + ax^2 + bx - 12

If it is divisible by 2x^2 - x - 3,
then,

f(x)
= x(2x^2 - x - 3) + 4(2x^2 - x - 3)
= 2x^3 - x^2 - 3x + 8x^2 - 4x - 12
= 2x^3 + 7x^2 - 7x - 12

By comparing coefficients,
a = 7, b = -7.

And it prove that f(x) should not be equal to 2x^3 + ax + bx - 12.


The common method is as follow:

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

Then, by remainder theorem, f(3/2) = 0, f(-1) = 0.

After that, you can form 2 linear equations with 2 variables a and b. And I think you can solve them by yourself.


(b)

f(x) = (x + 4)(2x^2 - x - 3) = (x + 4)(2x - 3)(x + 1)

2007-01-28 17:18:58 補充：
In (a), the common method is done by myisland8132.

I want to ask if f(x) = 2x^3 + ax + bx -12 or f(x) = 2x^3 + ax^2 + bx - 12.