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Question Number 12255 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17

Answered by mrW1 last updated on 16/Apr/17

$$(x+1)(x+2)(x+3)(x+4)+1$$$$=(x+\frac{5}{2}-\frac{3}{2})(x+\frac{5}{2}-\frac{1}{2})(x+\frac{5}{2}+\frac{1}{2})(x+\frac{5}{2}+\frac{3}{2})+1$$$$=[{(x+\frac{5}{2})}^2-{\left(\frac{3}{2}\right)}^2][{(x+\frac{5}{2})}^2-{\left(\frac{1}{2}\right)}^2]+1$$$$={(x+\frac{5}{2})}^4-[{\left(\frac{1}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2]{(x+\frac{5}{2})}^2+{\left(\frac{1}{2}\right)}^2{\left(\frac{3}{2}\right)}^2+1$$$$={(x+\frac{5}{2})}^4-2\times \frac{5}{4}\times {(x+\frac{5}{2})}^2+{\left(\frac{5}{4}\right)}^2$$$$={[{(x+\frac{5}{2})}^2-\frac{5}{4}]}^2$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17

$$={(x^2+5x+5)}^2$$$$nice.thankyousomuch.$$

Answered by ajfour last updated on 17/Apr/17

$$f\left(x\right)=(x+1)(x+4)(x+2)(x+3)+1$$$$=(x^2+5x+4)(x^2+5x+6)+1$$$$lett=x^2+5x$$$$f\left(x\right)=(t+4)(t+6)+1$$$$=t^2+10t+25$$$$={(t+5)}^2={(x^2+5x+5)}^2.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Apr/17

$$sobeautiful.thanksalot.$$

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