How do you factor completely $4 r^{2} - 21 + 5$ $4r^2 -21+5$ ?

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Answer 1 · 2016-06-20T15:55:09

$(2 r + 4) (2 r - 4)$ $(2r+4)(2r-4)$

To factor $4 r^{2} - 21 + 5$ $4r^2-21+5$ completely we would begin by combining like terms.

$4 r^{2} - 16$ $4r^2-16$

We next would recognize that the binomial is a difference (subtraction) of two squares.

$4 r^{2} = (2 r) (2 r)$ $4r^2 = (2r)(2r)$
and
$16 = (4) (4)$ $16 = (4)(4)$

The difference of two squares always factors in this pattern

$a^{2} - b^{2}$ $a^2-b^2$

$(a + b) ((a - b)$ $(a+b)((a-b)$

So our binomial doctors as

$4 r^{2} - 16$ $4r^2-16$

$(2 r + 4) (2 r - 4)$ $(2r+4)(2r-4)$

How do you factor completely 4r2−21+54r^2 -21+5?