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Title: Factor by grouping examples Post by: Durham_College on Nov 9, 2018 #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: \(\left(m-n+2\right)\left(m+n-2\right)\) |