Problem of the Week

Updated at Sep 21, 2026 1:16 PM

For this week we've brought you this algebra problem.

How would you find the factors of \(21{n}^{2}-40n+16\)?

Here are the steps:



\[21{n}^{2}-40n+16\]

1
Split the second term in \(21{n}^{2}-40n+16\) into two terms.
\[21{n}^{2}-12n-28n+16\]

2
Factor out common terms in the first two terms, then in the last two terms.
\[3n(7n-4)-4(7n-4)\]

3
Factor out the common term \(7n-4\).
\[(7n-4)(3n-4)\]

Done
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