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How would you factor this: (r+1)^2 + 8t(r+1) + 16t^2
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Quadratic form. Make a substitution, x=(r+1) then factor as normal for a quadratic trinomial.
Here's a hint: 4t*4t = ? 4t+4t = ?
4t * 4t would give you 8t^2 where as the other would give you 8t? So x^2 +8tx? but then you're stuck with 16t^2
Close.
You want to get it in the form (x+a)(x+b)
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(x+a)(x+b) = x^2 + ax + bx +ab
In your case, ax+bx = 8tx, x=(r+1), and ab = 16t^2
ya but in this case b and a are a constant, here with x replacing r+1 im still stuck with 2 unknown. :/
So i end up doing this : x^2 +8tx + 16t^2 which then gives me (x+4t)(x+4t) do i then just plug in r + 1?
Thank you I did get it! However not to sure if i do understand but thanks nonetheless
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