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Durham_College Durham_College
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11 months ago
#9:   \(x^2+y^2+2xy-4\)

Rearrange

\(x^2+2xy+y^2-4\)

Factor the first three terms as your would a quadratic trinomial by trial and error \(\left(x^2+2xy+y^2\right)-4\). It becomes:

\(\left(x+y\right)\left(x+y\right)-4\)

Write the brackets in exponent form:

\(\left(x+y\right)^2-4\)

This is a difference of squares, so use the pattern: \(a^2-b^2=\left(a+b\right)\left(a-b\right)\)

\(a=x+y\)
\(b=2\) because the square root of \(4\) is \(2\)

Therefore:

\(\left(x+y+2\right)\left(x+y-2\right)\)



#11:   \(m^2-n^2-4+4n\)

Rearrange like this, notice how I grouped them as a trinomial and factored out the negative:

\(m^2-\left(n^2-4n+4\right)\)

Now factor by trial and error:

\(m^2-\left[\left(n-2\right)\left(n-2\right)\right]\)

Write as an exponent:

\(m^2-\left(n-2\right)^2\)

This is a difference of squares:

\(\left(m-\left(n-2\right)\right)\left(m+\left(n-2\right)\right)\)

Clean up more:

\(\left(m-n+2\right)\left(m+n-2\right)\)
Source  Calter, Michael A., Paul Calter, Paul Wraight, Sarah White. Technical Mathematics with Calculus, Canadian Edition, 3rd Edition. John Wiley & Sons (Canada), 2016.
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