Functions 12U - Basic Polynomial questions

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Homework Help Overview

The discussion revolves around solving polynomial equations, specifically focusing on a cubic equation derived from a problem involving a product of binomials and a quartic equation that requires factoring techniques.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the transformation of an equation involving a product of binomials and question the arithmetic involved in shifting terms. There is also discussion about factoring a quartic polynomial by substituting variables.

Discussion Status

Some participants have provided guidance on correcting arithmetic mistakes and suggested factoring methods for the quartic equation. The conversation reflects a collaborative effort to clarify misunderstandings and explore different approaches to the problems.

Contextual Notes

There are indications of typographical errors in the original problem setup, and participants are addressing these while also discussing the implications of dividing and multiplying terms in the equations.

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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