Functions 12U - Basic Polynomial questions

AI Thread Summary
The discussion centers on solving two polynomial equations. In the first problem, the initial equation is corrected from 1008 to 504, leading to the polynomial 2016 = x^3 + 25x^2 + 150x + 500. Participants emphasize the importance of correctly handling the division and multiplication when rearranging the equation. For the second problem, factoring is suggested using the substitution u = x^2, simplifying the process. The conversation highlights the need for careful attention to detail in polynomial manipulation and solving techniques.
joeseppe
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Problem One
1008 = [(10+x)(10+x)(5+x)] / 2

Solution Attempt
5004=[(10+x)(10+x)(5+x)]

5004= (100+10x+10x+x^2)(5+x)

5004= 500 + 50x + 5x^2 + 100x + 20x^2 + x^3

5004 = x^3 + 25x^2 + 150x + 500

0 = x^3 + 25x^2 + 150x - 4504

Where do I go from here?

Problem Two
Solve.
x^4 - 10x^2 + 16 = 0

Solution Attempt
I have no idea on this one. What kind of factoring can i use when the first and second parts are ^4 and ^2 respectively, with the last part being a regular number?


Thanks in advanced guys, I know it's probably something little I'm overlooking :)
 
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For the second one factor it like it was u^2-10u+16 where u=x^2. For the first one, I think there may be multiple mistakes. How did 1008 turn into 5004?
 
Dick said:
For the second one factor it like it was u^2-10u+16 where u=x^2. For the first one, I think there may be multiple mistakes. How did 1008 turn into 5004?
Alright, thanks for the help :)

Typo on the first one, let me try again ;)

Problem One
1008 = [(10+x)(10+x)(5+x)] / 2

Solution Attempt
504=[(10+x)(10+x)(5+x)]

504= (100+10x+10x+x^2)(5+x)

504= 500 + 50x + 5x^2 + 100x + 20x^2 + x^3

504 = x^3 + 25x^2 + 150x + 500

0 = x^3 + 25x^2 + 150x - 4

Where do I go from here?
 
joeseppe said:
Alright, thanks for the help :)

Typo on the first one, let me try again ;)

Problem One
1008 = [(10+x)(10+x)(5+x)] / 2

Solution Attempt
504=[(10+x)(10+x)(5+x)]

504= (100+10x+10x+x^2)(5+x)

504= 500 + 50x + 5x^2 + 100x + 20x^2 + x^3

504 = x^3 + 25x^2 + 150x + 500

0 = x^3 + 25x^2 + 150x - 4

Where do I go from here?

Again you are making mistakes. Remember its a divide by two on the RHS. So when you shift it to the LHS, it becomes a multiply by two. It thus becomes 1008*2 and NOT 1008/2.

Then, do you know how to factorize third degree polynomials?
 
praharmitra said:
Again you are making mistakes. Remember its a divide by two on the RHS. So when you shift it to the LHS, it becomes a multiply by two. It thus becomes 1008*2 and NOT 1008/2.

Then, do you know how to factorize third degree polynomials?

Righto I missed that one, so the next lines should be:

2016 = x^3 + 25x^2 + 150x + 500

0 = x^3 + 25x^2 + 150x -1516

Then I would just use factor theorem to solve for x, does that work out in this case?
 
solve slowly...dont need to hurry... don't make so many mistakes

(10 + x)*(10+x)(5+x) = (5+x)(x^2 + 20x + 100) = 5x^2 + 100x + 500 + x^3 + 20x^2 + 100x

= x^3 + 25x^2 + 200x + 500...check it!
 

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