# Homework Help: Functions 12U - Basic Polynomial questions

1. Oct 19, 2008

### joeseppe

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 :)

2. Oct 19, 2008

### Dick

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?

3. Oct 19, 2008

### joeseppe

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?

4. Oct 20, 2008

### praharmitra

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?

5. Oct 20, 2008

### joeseppe

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?

6. Oct 20, 2008

### praharmitra

solve slowly....dont need to hurry..... dont 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!!