Factoring n(n+1)2x2−x−2 for n∈Z

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sa3b4or9i9 · Accepted Answer

Step 1To get factors, writen(n+1)2x2−x−2=n(n+1)2(x2−2n(n+1)x−2n×2n+1)=n(n+1)2(x2−(2n−2n+1)x−2n⋅2n+1)=n(n+1)2((x−2n)(x+2n+1))Can you proceed

zevillageobau · Accepted Answer

Step 1Alternative factoring method:When B2−4AC≥0, with A&amp;gt;0, then Ax2+Bx+C factors intoA(x2+BAx+CA)=A[(x+B2A)2−B24A2+CA]=A[(x+B2A)2−B2−4AC4A2]=A[(x+B2A)2−(B2−4AC2A)2]Under the assumptions, this factors into the difference of two squares as1) =A[(x+B2A+B2−4AC2A)×(x+B2A−B2−4AC2A)]With the posted problem, you have thatA=n(n+1)2B=−1C=−2From this you can immediately conclude (sinceA&amp;gt;0,C&amp;lt;0 that B2−4AC&amp;gt;0. Therefore, the formula in (1) above applies.B2A=−1n(n+1).B2−4AC2A=1+4n(n+1)n(n+1)=(2n+1)n(n+1).Therefore, Ax2+Bx+C factors inton(n+1)2[(x+−1n(n+1)+(2n+1)n(n+1))×(x+−1n(n+1)−(2n+1)n(n+1))].n(n+1)2[(x+−1n(n+1)+(2n+1)n(n+1))×(x+−1n(n+1)−(2n+1)n(n+1))].This simplifies into12n(n+1){[n(n+1)x+2n]×[n(n+1)x−2(n+1)]}This further simplifies to2) 12×[(n+1)x+2]×[nx−2].In (2) above, either (n+1) or n will be even, thus allowing the factor of 12 to be cleared.Therefore, you can forgo any consideration of the floor or ceiling functions, and simply divide the formula mentioned in (2) above into two cases: either n is odd or n is even.Finally, while I showed the derivation of the formula in (2) above, my answer could have been significantly shorter, if I had instead provided a (sanity-checking) verification of the formula in (2) above.That is, if you manually multiply the factors in (2) above, the product will be Ax2+Bx+C, where A, B, C are as specified in your original question.

Factoring \(\displaystyle{\frac{{{n}{\left({n}+{1}\right)}}}{{{2}}}}{x}^{{{2}}}-{x}-{2}\) for \(\displaystyle{n}\in{\mathbb{{{Z}}}}\)

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