Question 65544
(1)  Given that (3x-a)(x-2)(x-7)=3x^3-32x^2+81x-70, determine the value of a...


Lets try the brute force method:
First divide both sides by (x-2)(x-7) and we get:


3x-a=(3x^3-32x^2+81x-70)/((x-2)(x-7)) 
Next subtract 3x from both sides:

-a=(3x^3-32x^2+81x-70)/((x-2)(x-7))-3x  

Now multiply right side by ((x-2)(x-7))/((x-2)(x-7)) to get a common denominator
-a=((3x^3-32x^2+81x-70)-3x(x-2)(x-7))/((x-2)(x-7))
Noting that (x-2)(x-7)=x^2-9x+14 we will now expand the numerator:

-a=(3x^3-32x^2+81x-70-3x^3+27x^2-42x)/((x-2)(x-7)) collecting like terms in the numerator:

-a=(-5x^2+39x-70)/(x^2-9x+14) multiply both sides by (-1)

a=(5x^2-39x+70)/(x^2-9x+14) when we divide the quadratics, we get:

a=5+6x/(x^2-9x+14) factoring the denominator, we get



a=5+6x/((x-2)(x-7))  Where x cannot equal 2  or  7

ck
substitute  for a in (1) and we get:

(3x-5-6x/((x-2)(x-7)))(x-2)(x-7)=3x^3-32x^2+81x-70 and
3x(x^2-9x+14)-5(x^2-9x+14)-6x=3x^3-32x^2+81x-70 expanding left side:
3x^3-27x^2+42x-5x^2+45x-70-6x=3x^3-32x^2+81x-70 collecting like terms on left side:

3x^3-32x^2+81x-70=3x^3-32x^2+81x-70 


Happy holidays---ptaylor