Dayanara Terry

2022-07-08

Find all roots of
$\left(x+1\right)\left(x+3\right)\left(x+4\right)\left(x+6\right)=720$

Sariah Glover

Expert

Actually you can use your method: $\left({x}^{2}+7x+6\right)\left({x}^{2}+7x+12\right)=720$ and now let ${x}^{2}+7x+9=u$
Then $\left(u-3\right)\left(u+3\right)=720$ which gives two values for u. This is easier than the other method I suggested.

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