tell me if this is wrong= 8v²+31v-4 (8v²-1v)(32v-4) 1v(8v-1) 4(8v-1) (v+4)(8v-1)
Its right. :)
\[8v2+31v−4\] \[8v2+32v−1v−4\] \[8v(v+4)−1(v+4)\] \[(8v−1)((v+4)\] ur answer and this answer is same and ur process is also right hence u r right
thank you! what could I do to simplify this: (32v²-4v)(-8v+1)
see here : \[(32v^2-4v)(-8v+1)\] as we know \[( a+b)( c+d)=ac+bc+ad+bd\] hence here \[32v^2(-8v)-4v(-8v)+32v^2(1)-4v(1)\] \[-256v^3+32v^2+32v^2-4v\] \[-256v^3+64v^2-4v\]
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