F4 Maths Polynomials

Consider f (x) = 2x^3+x^2-2x-1 and g (x) = 6x^3 - 13x^2+4

a) Factorize f(x) ans: (2x+1)(x+1)(x-1)
f(x) = 2x^3+x^2-2x-1
= (x-1)(3x^2-x-1)
* Try x=1, then put into the function, if =0,
it is divisible by 1, therefore x-1 is a factor,
then do long division to find another factor
= (x-1)(2x+1)(x+1)
Bi) Find g(-1/2) ans: 0
g (x) = 6x^3 - 13x^2+4
= 6(-1/2)^3 - 13(-1/2)^2+4
= -3/4-13/4+4
= 0
ii) factorize g(x) ans: (2x+1)(3x-2)(x-2)
g (x) = 6x^3 - 13x^2+4
= (2x+1)(3x^2-8x+4)
* From bi), we've found that when x= -1/2,
g(x)=0, then it is divisible by -1/2,
therefore 2x+1 is a factor,
then do long division to find another factor
=(2x+1)(3x-2)(x-2)

c) Hence or otherwise, factorize f(x) + g(x) ans: (2x+1)(2x-1)(2x-3)
f(x) + g(x) = (x-1)(2x+1)(x+1)+(2x+1)(3x-2)(x-2)
= (2x+1)[(x-1)(x+1)+(3x-2)(x-2)]
= (2x+1)[x^2-1+3x^2-8x+4]
= (2x+1)(4x^2-8x+3)
= (2x+1)(2x-1)(2x-3)

(a) f(1)=2+1-2-1=0 (x-1) is a factor
f(-1)=-2+1+2-1=0 (x+1) is a factor

So 2x^3+x^2-2x-1=(ax+b)(x^2-1)=ax^3+bx^2-ax-b
Comparing the coefficients, a=2,b=1
2x^3+x^2-2x-1=(2x+1)(x-1)(x+1)

(b) g(-1/2)=6(-1/2)^3-13(-1/2)^2+4=-6/8-13/4+4=0

(ii) (x+1/2) or 2x+1 is a factor of g(x) by b(i)

(c) g(x)=(2x+1)(ax^2+bx+c)=>c=4,a=3,b=-8
Since 3x^2-8x+4=(3x-2)(x-2)

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

f(x)+g(x)=(2x+1)(x+1)(x-1)+(2x+1)(3x-2)(x-2)
=(2x+1)[(x+1)(x-1)+(3x-2)(x-2)]
=(2x+1)(x^2-1+3x^2-8x+4)
=(2x+1)(4x^2-8x+3)
=(2x+1)(2x-1)(2x-3)

不選擇我請出聲..
thanks~