# How is this equation solved?

1. Dec 31, 2015

### anandzoom

(1+x+x2+x3+x4+x5+x6+x7+x8)(1+x2+x4+x6+x8)(1+x3+x6)(1+x4+x8)(1+x5)(1+x6)(1+x7)(1+x8) = 1+x+2x2+3x3+5x4+7x5+11x6+15x7+22x8+...+x56
Please explain me the step-wise procedure to arrive at the solution.

Last edited: Dec 31, 2015
2. Dec 31, 2015

### SammyS

Staff Emeritus
Are those supposed to be exponents on the variable, x ?

Use that X2 feature in the green bar above the edit/composition box.
This gives something like: (1+x+x2+x3+x4+x5+x6+x7+x8)(1+x2+x4+x6+x8) ...

Furthermore:
You should use the supplied template when starting a thread, as well as showing an attempt at solving/understanding the problem you're presenting.

3. Dec 31, 2015

### FactChecker

The only way I see to solve it is to multiply out the left side. Then move everything to one side and hope that many of the high-exponent terms disappear. Maybe there will only be a low order polynomial left. Look for the zeros of the polynomial.

4. Dec 31, 2015

### HallsofIvy

Staff Emeritus
Assuming those are exponents, that is a polynomial equation of degree 56. There exist "formulas" for polynomial equations of degree 4 or less but there is NO 'step by step" way of solving a general polynomial of degree five or higher and certainly not 56!

5. Dec 31, 2015

### FactChecker

I noticed that the x56 term cancels out. I assume the question has been rigged so that many terms cancel out.

6. Dec 31, 2015

### anandzoom

Everything is in '+' ... terms can cancel out only if there is some '-' if I'm right

7. Dec 31, 2015

### FactChecker

Notice that both sides have +x56. (To see that on the left, scan through all that multiplication and notice that all the highest power terms multiply out to +x56). ) You can subtract x56 from both sides and the rest stays equal. My guess is that a lot of the high-power terms cancel out just like x56.

8. Dec 31, 2015

### SammyS

Staff Emeritus
My guess is that the right hand side is merely the expansion of the left hand side, so they ALL cancel.

OP still has not complied with PF rules.

9. Dec 31, 2015

### Staff: Mentor

In the homework help forums, posts seeking assistance must be accompanied by evidence of the member's attempt at solving the problem. No such evidence has been provided here.