How do you factor $16r^2 - 8r + 1$ ?

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Answer 1 · 2016-05-08T17:11:47

$color(blue)( (4r - 1 ) ( 4r - 1 )$ is the factorised form of the expression.

$16r^2 - 8r +1$

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like $ar^2 + br + c$ , we need to think of 2 numbers such that:

$N_1*N_2 = a*c = 16* 1 = 16$

AND

$N_1 +N_2 = b = -8$

After trying out a few numbers we get $N_1 = -4$ and $N_2 =-4$

$(-4)*(-4) = 16$ , and $( -4 ) +(-4)= - 8$

$16r^2 - 8r +1 = 16r^2 - 4r -4r+1$

$4r ( r - 1 ) - 1 ( 4r -1)$

$(r -1 )$ is a common factor to each of the terms

$= (4r - 1 ) ( 4r - 1 )$

How do you factor 16r^2 - 8r + 116r^2 - 8r + 1?