Multiplying Polynomials

[Unit_02]

Given the exercise:


Why can't i solve it like this: WRONG FINAL ANSWER



Instead of this: RIGHT FINAL ANSWER



Shouldn't i be able to get the same result by multiplying all the therms straight away instead of first multiplying the first two and then multiply it for the third one? What am i missing here?

 

[Dr.Peterson]

Shouldn't i be able to get the same result by multiplying all the terms straight away instead of first multiplying the first two and then multiply it for the third one? What am i missing here?

You can't multiply all the terms, because it doesn't work! For one thing, your first step treats (2y+1)(4y^2+1) as if it were (2y+1+4y^2+1), which is it not.

Taking a general case, if we multiply (a+b)(m+n)(u+v), the result will be (am+an+bm+bn)(u+v) = amu+amv+anu+anv+bmu+bmv+bnu+bnv.

Your wrong method would give am+an+au+av+bm+bn+bu+bv + mu+mv+nu+nv.

Surely you see that these are entirely different; you've only multiplied two terms at a time, while the correct answer is a sum of products of three terms.

You could get each term in the result directly, by finding all possible products of three terms: one term from (2y-1), one term from(2y+1), and one term from (4y^2+1). This would give 8 terms to add, but it's hard to keep track of things. It's just much easier to work with one factor at a time, as they show.

You can do it in any order, however; for example, you could multiply the last two factors first, then multiply that by the first.

 

[topsquark]

You can do it all at once, but keeping track of things can get a little messy. For example, in your approach you completely missed the term $$(2y)(2y)(4y^2) = 16y^4$$.

$$(2y - 1)(2y + 1)(4y^2 + 1)$$
$$= (2y)(2y + 1)(4y^2 + 1) + (-1)(2y+ 1)(4y^2 + 1)$$
$$= (2y) \left ( (2y)(4y^2 + 1) + (1)(4y^2 + 1) \right ) + (-1) \left ( (2y)(4y^2 + 1) + (1)(4y^2 + 1) \right )$$
$$= (2y) \left ( 8y^3 + 2y + 4y^2 + 1 \right ) + (-1) \left ( 8y^3 + 2y + 4y^2 + 1 \right )$$
$$= (16y^4 + 4y^2 + 8y^3 + 2y) + (-8y^3 - 2y - 4y^2 - 1)$$
$$= 16y^4 - 1$$
Not only is it tedious but it's easy to make a mistake. And if you look at it you see you have to calculate $$(2y + 1)(4y^2 + 1)$$ in any event. It's better to multiply these out 2 factors at a time.

-Dan