How do you factor $4 t^{2} + 4 t + 1$ $4t^2+4t+1$ ?

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How do you factor $4 t^{2} + 4 t + 1$ $4t^2+4t+1$ ?

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Answer 1 · 2018-03-09T10:18:52

$(2 t + 1) (2 t + 1)$ $(2t+1)(2t+1)$

Explanation:

we are given:-

$4 t^{2} + 4 t + 1$ $4t^2+4t+1$ resembles $a x^{2} + b x + c$ $ax^2+bx+c$
here $a = 4, b = 4 and c = 1$ $a=4,b=4 and c=1$
and since, sum $= b$ $=b$ ,product $= a c$ $=ac$

factors are the number which when you (multiply) and also when you (add) them give the same value for sum and product.

sum=4
product=4*1=4
factors=2 and 2

so
$4 t^{2} + 2 t + 2 t + 1$ $4t^2+2t+2t+1$
$2 t (2 t + 1) + 1 (2 t + 1)$ $2t(2t+1)+1(2t+1)$

$(2 t + 1) (2 t + 1)$ $(2t+1)(2t+1)$

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How do you factor $4 t^{2} + 4 t + 1$ $4t^2+4t+1$ ?

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