Factorization by grouping (Example 1)

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In this video we will see how to factor by grouping, for this we will separate in pairs and factor each pair by common term.
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[@linulfoilias9270]

Gracias

[@dangrtl]

me puede ayudar con esta factorización? a2+b2+c2+2ab+2ac+2bc

[@pvs_np]

a² + b² + c² + 2ac + 2ab + 2bc
= (a + b + c) (a + b + c)
= (a + b + c)²

[@elisabenites8110]

Así me enseña mi maestra

[@MamAlpha]

mais integrais muito mais por favor

[@MateFacilYT]

+vanuzia souza Ferreira
Claro, más adelante subiré muchas más, me falta subir aun una gran variedad de integrales.

[@luisramirogutierrezhuerta8530]

Creo que el resultado del ejercicio final el (c-d) + (a-d)

[@Luckyrrul]

like.... me gustó :-)

[@valeriavilla9140]

Plis urgenteee3

[@agentesalud1066]

Hola persona que ve este video

[@valeriavilla9140]

Una pregunta me podrian ayudar a factorizar esto: ax-az-by+cx+cy+ay-bx+bz-cz

[@carlostorres-kq6kv]

:0

[@pvs_np]

ax -az - by + cx + cy +ay - bx + bz - cz
Organizando;
ax +ay -az - bx - by + bz + cx +cy - cz
= a (x + y - z) - b (x + y - z) + c (x + y - z)
= (x + y - z) (a - b + c)
El del vídeo; xy + xz + my + mz
= x (y + z) + m (y + z)
= (x + m) (y + z)

[@kev341]

xd me salio al revez (x+m) (y+z) xd

[@amaruv.r9760]

El orden de los factores no altera el producto