Factor completely. 150m^2nz+20mn^2c-120m^2nc-25mn^2z

Question

Factor completely.
150m^2nz+20mn^2c-120m^2nc-25mn^2z

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Margaret 2 weeks 2021-09-08T19:47:35+00:00 1 Answer 0

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    2021-09-08T19:48:43+00:00

    Answer:

    The factor form is:

    5mn\left(6m-n\right)\left(5z-4c\right)

    Step-by-step explanation:

    Here we have to find the factor of the expression:

    150m^2nz+20mn^2c-120m^2nc-25mn^2z

    So we need to take out the common terms, as we do for finding the greatest common factors.

    Now the expression that is given can be re-written as:

    150m^2nz+20mn^2c-120m^2nc-25mn^2z\\=150mmnz+20mnnc-120mmnc-25mnnz\\=30\cdot \:5nmmz+4\cdot \:5nmnc-24\cdot \:5nmmc-5\cdot \:5nmnz

    Next, we will find the common terms, as follows;

    30\cdot \:5nmmz+4\cdot \:5nmnc-24\cdot \:5nmmc-5\cdot \:5nmnz\\=5nm\left(30mz+4nc-24mc-5nz\right)\\

    Now  we will factorise the expression:

    \left(30mz+4nc-24mc-5nz\right)\\

    as follows;

    \left(30mz+4nc-24mc-5nz\right)\\=6m\left(5z-4c\right)+n\left(4c-5z\right)\\=\left(-4c+5z\right)\left(6m-n\right)\\

    So the final factor form is:

    150m^2nz+20mn^2c-120m^2nc-25mn^2z=5mn\left(6m-n\right)\left(5z-4c\right)

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45:7+7-4:2-5:5*4+35:2 =? ( )