Question 65544: Given that (3x-a)(x-2)(x-7)=3x^3-32x^2+81x-70, determine the value of a...
HELP! Found 2 solutions by venugopalramana, ptaylor:Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! Given that (3x-a)(x-2)(x-7)=3x^3-32x^2+81x-70, determine the value of a...
HELP!
THERE ARE 2 WAYS TO DO THIS.
1.NORMAL WAY IS TO MULTIPLY AND EQUATE THE 2 SIDES TO GET A. BUT EVEN HERE YOU CAN COMPARE ONLY THE CONSTANT TERMS AND GET THE ANSWER IN AN EASY WAY.
2.BUT EASIER WAY IS TO USE THE POWER OF IDENTITY AS AGAINST AN EQN. HOPE YOU KNOW THE DIFFERENCE BETWEEN THE TWO.
THIS BEING AN IDENTITY IN X IT IS TRUE FOR ANY AND EVERY VALUE OF X.SO PUT X=0
WE GET
(-A)*(-2)(-7)= -70
-14A=-70
A=(-70)/(-14)=5
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
