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How do you factor n^2-10n+9? | HIX Tutor

<mathjax>#n^2-10n+9 = (n-1)(n-9)#</mathjax>
<mathjax>#n*n=n^2#</mathjax> 
<mathjax>#n*-9=-9n#</mathjax> 
<mathjax>#n*-1=-n#</mathjax> 
<mathjax>#-9*-1=9#</mathjax>
When you combine those you get: <mathjax>#n^2-10n+9#</mathjax>

<mathjax>#n^2-10n+9 = (n-1)(n-9)#
<mathjax>#n*n=n^2#</mathjax> 
<mathjax>#n*-9=-9n#</mathjax> 
<mathjax>#n*-1=-n#</mathjax> 
<mathjax>#-9*-1=9#
When you combine those you get: <mathjax>#n^2-10n+9#

Savannah Anderson

To factor \( n^2 - 10n + 9 \), you need to find two numbers that multiply to give you the constant term (9) and add to give you the coefficient of the middle term (-10). In this case, those numbers are -1 and -9. Therefore, you can factor the expression as \( (n - 1)(n - 9) \).

How do you factor #n^2-10n+9#?