How do you factor $27 c^{2} - 6 a t - r^{2}$ $27c ^ { 2} - 6a t - r ^ { 2}$ completely?

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Answer 1 · 2018-06-27T03:31:52

The only factored form I see is $3 (9 c^{2} - 2 a t) - r^{2}$ $3(9c^2-2at)-r^2$

Notice that among the terms, the only common factor within any of the terms is a 3 within the 27 and the 6, so we could factor this way:

$3 (9 c^{2} - 2 a t) - r^{2}$ $3(9c^2-2at)-r^2$

Answer 2 · 2018-06-27T03:51:01

Please see the explanation below.

Just in case it was meant to be:

$27 c^{2} - 6 r c - r^{2}$ $27 c^2 − 6rc − r^2$

Then:

Using AC method for factoring the quadratics in the form of:

$A x^{2} + B x + C$ $Ax^2+Bx+C$

We need to find two numbers that multiply to -27, and add to -6,

those numbers are -9 and 3 thus:

$27 c^{2} - 6 r c - r^{2}$ $27c^2-6rc-r^2$

$= 27 c^{2} - 9 r c + 3 r c - r^{2}$ $=27c^2-9rc+3rc-r^2$

$= 9 c (3 c - r) + r (3 c - r)$ $=9c(3c-r)+r(3c-r)$

$= (9 c + r) (3 c - r)$ $=(9c+r)(3c-r)$

How do you factor 27c2−6at−r227c ^ { 2} - 6a t - r ^ { 2} completely?