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I know this is a basic math question but still I always get stuck on this.

$(x^3 + 3x^2 + 3x + 1) \times (1 + x + x^2)$

I know this is the solution: $1 + 4x + 7x^2 + 7x^3 + 4x^4 + x^5$

But I want to know how they got there. This has to be solved with distributing the second term over the first one (I guess).

So I get these terms:

$x^3 + 3x^2 + 3x + 1$

$x^4 + 3x^3 + 3x^2 + 1x$

$x^5 + 3x^4 + 3x^3 + 1x^2$

And than I'd have to sum these up. I already highly doubt this is correct, but even if it was correct I wouldn't know how to sum these up, because it's not possible to sum numbers with a different exponent.

Could someone explain step-by-step on how to solve this $(x^3 + 3x^2 + 3x + 1) \times (1 + x + x^2)$

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  • $\begingroup$ $$x^3 + 3x^2 + 3x + 1=(1+x)^3$$ $\endgroup$ Dec 26, 2017 at 12:47
  • $\begingroup$ @labbhattacharjee Yeah, I converted $(1 + x)^3$ to $x^3 + 3x^2 + 3x + 1$ using the Binomium of Newton $\endgroup$
    – O'Niel
    Dec 26, 2017 at 13:21
  • $\begingroup$ @O'Niel using the Binomium of Newton If you know the binomial expansion, you could also write $x^2+x+1=(x+1)^2-x$ then calculate the product as $(x+1)^5 - x\cdot(x+1)^3\,$. $\endgroup$
    – dxiv
    Dec 26, 2017 at 18:31

2 Answers 2

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Yes, you are right, so far. But, why can’t you just sum up terms with same power of $x$, that is, $$(x^3+3x^2+3x+1)(1+x+x^2) $$ $$= (x^\color{red}{5})+(3x^\color{green}{4} + x^\color{green}{4}) + (x^\color{blue}{3}+3x^\color{blue}{3}+3x^\color{blue}{3})+(3x^\color{orange}{2}+3x^\color{orange}{2}+x^\color{orange}{2})+(3x^\color{cyan}{1}+x^\color{cyan}{1})+1 $$ $$= x^\color{red}{5}+4x^\color{green}{4}+7x^\color{blue}{3}+7x^\color{orange}{2}+4x^\color{cyan}{1}+1$$

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  • $\begingroup$ Than the term $4x^4$ in my cursus is wrong? $\endgroup$
    – O'Niel
    Dec 26, 2017 at 13:17
  • $\begingroup$ @O’Niel Sorry, it was a typo. $\endgroup$
    – user371838
    Dec 26, 2017 at 13:23
  • $\begingroup$ Thanks! Am able to solve it now! $\endgroup$
    – O'Niel
    Dec 26, 2017 at 13:54
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You don't sum expressions with different exponents, but you can sum expressions with the same exponent. Look at the three components you have noted of the product.

You have just one constant term $1$

The linear terms are $3x+x=4x$

The quadratic terms are $3x^2+3x^2+x^2=7x^2$

Can you see what is happening there, and how the components of the product are being built up? Can you compute the other terms?

It is probably a lot easier than you are thinking, and when you are used to it, it will become routine.

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  • $\begingroup$ Thanks a lot! Your explanation helped a lot! $\endgroup$
    – O'Niel
    Dec 26, 2017 at 13:54

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